COMPUTABLE ASPECTS OF CLOSED SETS
By
PAUL BRODHEAD
A DISSERTATION PRESENTED TO THE GRADUATE SCHOOL
OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT
OF THE REQUIREMENTS FOR THE DEGREE OF
DOCTOR OF PHILOSOPHY
UNIVERSITY OF FLORIDA
2008
2008 Paul Brodhead
To my Family, the Artisans of Life
ACKNOWLEDGMENTS
I give thanks, first and foremost, to my Ph.D. advisor Douglas Cenzer, who must be
noted in this generation as among the most patient and irrevocably kind of men. His sense
of humor also trumps the cloudiest of di , be they even the clouds of hurricanes which,
in fact, occurred as my studies progressed in Florida. These characteristics, combined with
my determination, helped me gain an understanding, broad and indepth, of computability
that allowed me to successfully undertake and develop the research presented here. The
road to this point began long ago, but reached its pinnacle at Florida under his direction.
The road began at the University of Wisconsin, in my first semester; I was first in
spired to pursue mathematics while taking Algebra and Trigonom( I1: Professor Arnold
Johnson made the course enjo1,bl' and (! i11. ii,;. making it unlike any mathematics
course I ever took. Him I thank, for as have I ahvi had high goals, his influence was
critical as it was at this point in life that I decided that I was to pursue a Ph.D. in math
ematics. Later, in my last semester at Wisconsin, he helped me develop a mathematical
maturity beyond expectations, as he led me in an undergraduate research project in his
area, Algebra. Yet, between these experiences, many others must be thanked.
During my undergraduate years, many inspiring teachers pushed the mathematical
envelope, delivering the endeavor of the ages, the enjoi,bl'! mental fray of the breaking
of mathematical barriers. I thank: Steffen Lempp, Reed Solomon, H. Jerome Keisler,
Arnold Miller, Maury Bramson, Steve Bauman, Simon Hellerstein, Daniel Rider, Daniel
Shea, and Jennifer Ziebarth. In particular, I thank Steffen Lempp for introducing me to
computability theory while I was in his linear algebra class, as it was at this moment that
I determined my Ph.D. specialization. My mindset was ever afterwards directed this way.
For example, I thank Arnold Miller for his patience in listening to my numerous attempts
in using computability theory in his yearlong abstract algebra class. I thank H. Jerome
Keisler for listening to me in his logic course. I also thank Reed Solomon for helping
me even further in his graduate logic course and his subsequent guidance in my first
undergraduate research project, the study of Hilbert's Tenth Problem. I thank for Daniel
Shea (also the undergraduate advisor) for accepting my destined course in mathematics as
I sat in his abstract analysis course, vigorously driven under a department mathematics
scholarship, when years prior I entered his office whilst still in Algebra and Trigonometry,
demanding help in setting up my undergraduate curriculum for an eventual Ph.D. in
mathematics.
Upon the closing of my Wisconsin experience, excellence in mathematics, a higher
level still, evolved at the University of Puerto RicoHumacao during the immediate
summer. I participated in the NSFfunded undergraduate research experience, the
Summer Institute of Mathematics for Undergraduates. I thank: Reinhard Laubenbacher,
Abdul Jarrah, Rebecca Garcia, Malarie Cummings, Cora Seidler, Ivelisse Rubio, and
Herbert A. Medina. Herbert, Ivelisse, and Reinhard, in particular, helped me to see the
urgency in the tackling of the problems at hand. I am greatly appreciative of this help to
this dv.
My research rigor and core mathematical talent bloomed in graduate school at
Florida. I thank: Douglas Cenzer, Bill Mitchell, Jindrich Zapletal, Jean Larson, Alexandre
Turull, Paul Robinson, James Keesling, Jonathan King, Scott McCullough, Stephen
Summers, and Shari Moscow. I especially thank Alexandre Turull. I had the privilege
of having six semesters with him; he helped me to see a problem, its value, and its
complexities. I thank my dissertation committee: Douglas Cenzer, Rick Smith, Bill
Mitchell, Murali Rao, and Beverly Sanders. I also thank Bernard Mair, Rick Smith,
Murali Rao, and Juan Liu; my experiences with them convinced me to move in valuable
directions that took me around the globe and beyond.
My collaborators, near and abroad, you I thank: Douglas Cenzer, Angsheng Li,
Weilin Li, George Barmpalias, Seyyed Dashti, Rebecca Weber, Jeffrey Remmel, Rod
Downey, Noam Greenberg, Keng Meng Ng, and Bji, rn KjosHanssen. Many many more
do I also thank I have collaborated with many people, if even for a short time, and their
contributions are appreciated. I thank: Peter Hinman, Denis Hirschfeldt, Carl Jockusch,
Russell Miller, Johanna Franklin, Bakhadyr Khoussainov, Jan Riemann, Ted Slaman,
Andr6 Nies, Robert Soare, Steve Simpson, Antonio Montalbdn, Barbara Csima, Joe Miller,
Ver6nica Becher, and Wolfgang Merkle.
I thank the people and institutions who contributed their support through funding:
the Southeast Alliance for Graduate Education and the Professoriate, the National Science
Foundation, the University of Florida, the Chinese A( i'I i,: of Sciences, the Association
for Symbolic Logic, and others.
Moii people deserve credit far beyond the brief, or even omitted, acknowledgement
here; I am extremely grateful to you. Of special consideration is my advisor. The mag
nitude of his support is above any measurable scale. His support in all my endeavors,
professionally and beyond, will surely continue to help shape me. This dissertation and the
work therein, and the experiences associated with the attainment thereof are a witness to
this.
To the many people who helped me along this journey, which is actually just a
beginning, you I thank.
TABLE OF CONTENTS
page
ACKNOWLEDGMENTS .......................
A BSTR A CT . . . . . . . .
1 INTRODUCTION ........................
1.1 General Overview .. ..................
1.2 Classical Computability .................
1.3 Closed Sets in Computability ..............
2 EFFECTIVELY CLOSED SETS AND ENUMERATIONS ..
2.1 Introduction . . . . . . . . .
2.2 The Family of 1 Classes ............................
2.2.1 N 11i ili. ii,; in the Literature . . . . . .
2.2.2 Equivalence of the Numberings .....................
2.2.3 Equivalence of the Numberings (Alternate Proof) .. ........
2.2.4 Injective Computable N, 11i11. i i, .. . . .........
2.2.4.1 Original c.e. sets argument .. ...............
2.2.4.2 Ordered tuples of disjoint c.e. sets .. ............
2.2.4.3 Results for effectively closed sets .. ...........
2.3 String Verifiable Families of 1 Classes .. ................
2.3.1 Definition and Examples .. ....................
2.3.2 Computable and Effective Nu11.11 li., . ..........
2.3.3 Families Containing the Clopen Classes .. .............
2.4 N i,,, ,1 Fam ilies of 1 Classes . . . . . . .
2.4.1 Homogeneous Classes .. .....................
2.4.2 Decidable Classes . . . . . . .
2.4.2.1 An injective computable numbering (Alternate proof) .
2.4.2.2 Trees without dead ends: A numbering result .. .....
2.4.2.3 Trees with dead ends: A necessity .. ............
2.4.3 Thin and Perfect Thin Classes .. .................
2.4.3.1 The MartinPour El Construction .. ............
2.4.3.2 Nonexistence of computable numbering .. ........
2.4.4 Small, Very Small, and Nondecidable Classes .. ...........
2.4.4.1 Numberings and high/noncomputable sets .. .......
2.4.4.2 Nonexistence of effective numbering .. ..........
3 RANDOM CLOSED SETS .. . .......................
3.1 O verview . . . . . . . . .
3.2 Effective Randomness of Reals .. .....................
3.2.1 Introduction . . . . . . . .
3.2.2 Constructive Martingale Randomness .. ..............
3.2.3 Prefixfree Randomness .. .....................
3.2.4 MartinLof (n)randomness .. ...................
3.3 MartinLof Randomness of Closed Sets
3.3.1 The HitorMiss Topology on C
3.3.2 Toward a Measure ........
3.3.3 Canonical Coding and Measure
3.3.4 Ghost Coding .. ........
3.3.5 Coding Equivalance .......
3.3.6 Coding and Joins of Closed Sets
3.4 Members of Random Closed Sets .
32 4 1 Pnsitive RPsuilts
3.4.2 Negative Results ....................
3.5 Measure and Dimension .. ...............
3.5.1 M measure . . . . . .
3.5.2 D im ension . . . . . .
3.6 PrefixFree Complexity of Closed Sets .. ..........
3.6.1 Lower Complexity Bounds .. ...........
3.6.2 Upper Complexity Bounds .. ...........
3.7 Other Notions of Randomness for Closed Sets .......
3.7.1 Randomness with Regular Probability Measures .
3.7.2 Randomness with the Inclusion of Trees with Deads
3.8 Random Closed Sets and Effective Capacity .. ......
3.8.1 Computable Capacities .. .............
3.8.2 Regular Measures and Capacities of Closed Sets .
Ends
4 RANDOM CONTINUOUS FUNCTIONS .. ...........
4.1 Overview .. .. .. .. ... .. .. .. .. . . .
4.2 Definining Randomness for Continuous Functions . ..
4.2.1 Representing Functions . ............
4.2.2 Representing Sequences . ............
4.2.3 A Sound Definition . ...............
4.3 Random Continous Functions and Images . ......
4.3.1 Perfect Images, in every instance . .......
4.3.2 Noninjective Images, in every instance . ....
4.3.3 Nonsurjective Images, in instances . ......
4.3.4 Images of computable elements ....... . .
4.4 Random Closed Sets arising from random continuous functions
4.4.1 A Positive Result: Inverse Images of 0 . ....
4.4.2 A Negative Result: Images, in general . ....
4.5 PseudoDistance Functions . ..............
4.6 nRandom ness . ... .. .. .. .. ... .. .. .. .
4.7 Future W ork . . . . . . .
5 CONTINUITY OF CAPPING IN CBT .. . .........
5.1 Introduction . . . . . . .
5.2 Continuity Results . ..................
5.2.1 Continuity Results in C . ...........
5.2.2 Continuity Results in CbT and Main Result . ..
94
95
95
96
97
99
99
100
103
104
106
106
109
110
111
112
113
114
114
115
5.3 Requirements and Strategies ............... .... .. 118
5.3.1 The requirements ............... ....... ..118
5.3.2 A Pstrategy .................. ............ .. 119
5.3.3 An Rstrategy .................. ........... 120
5.3.4 An Sstrategy .................. ........... .. 121
5.4 The Priority Tree .................. ........... .. 125
5.5 The Construction .................. ........... .. 128
5.6 The Verification .................. .............. 132
REFERENCES .................. ................ .. .. 144
BIOGRAPHICAL SKETCH ........... ........ . ... 150
Abstract of Dissertation Presented to the Graduate School
of the University of Florida in Partial Fulfillment of the
Requirements for the Degree of Doctor of Philosophy
COMPUTABLE ASPECTS OF CLOSED SETS
By
Paul Brodhead
May 2008
C(!i r: Douglas Cenzer
Major: Mathematics
A closed set in {0, 1}" may be viewed the set of infinite paths through a tree; a set A
is computable if there is a computer program which halts and gives the correct answer on
every query to the membership predicate for A. A numbering, or enumeration, is a map
from N onto a countable collection of objects; if there is a computable numbering onto a
set A C N then we iv that A is computably enumerable (or c.e.). The set of infinite paths
through a computable, or equivalently a coc.e., tree is called an effectively closed set.
In this work, we investigate: (1) numberings for different families of effectively
closed sets, (2) notions of randomness for nonempty closed subsets of 2", (3) notions of
randomness for continuous functions from 2N to 2N, and (4) continuity properties of CbT,
the c.e. degrees under the Turing reduction
a computable function.
(1) Numberings and effectively closed sets. We show that certain families (or classes
of families) of effectively closed sets such as the decidable, homogeneous, thin, small
or the entire family of effectively closed sets, or string verifiable families possess, or do
not possess, (injective) computable or effective numberings. This works builds upon the
seminal work by Friedberg [46], who constructed an injective numbering of the c.e. sets.
(2) Randomness of closed sets. In the space of closed sets, we give a probability
measure and define a version of the MartinL6f Test for randomness. We show that
random closed sets are never effectively closed, but are, on the other hand, alv i perfect,
have measure zero, and have box dimension log12 Every random closed set contains
random and nonrandom elements, but no nc.e. elements. We also explore alternate
notions for randomness, such as the problem of compressibility of trees. Finally, we
consider the problem of when a randomly chosen closed set meets a closed Q; this is the
study of capacities.
(3) Randomness of continuous functions. As in (2), we give a probability measure
and define a version of the MartinL6f Test for randomness. We show that the image of
a random continuous function is alvii noninjective and perfect, but not necessarily
surjective. Furthermore, computable elements map to random elements. Also, random
closed sets arise as inverse images of 0", but not, in general, as images. The former
motivates a study of pseudodistance functions. Finally, we consider our results in the
context of nrandomness.
(4) Continuity properties in CbT. We show in CbT that for any b / 0, 0', there is an
a > b such that for any x, b A x 0 iff a A x = 0. We prove this by first showing that
that the Seetapun local noncappability theorem in the c.e. Turing degrees [84] also holds
in CbT. This theorem demonstrates that every b / 0, 0' is noncappable with any nontrivial
degree below some a > b (i.e. if x < a and x A b = 0 then x = 0).
CHAPTER 1
INTRODUCTION
This thesis ia an accumulation of my work as a graduate student at the University
of Florida. Much of this work is joint and published, or to be published. The citations
are listed at the beginning of the appropriate chapters. In this chapter we introduce
various notions and expound upon these in later chapters. Each of C'! lpters 25 contains
a distinct topic from computability theory.
1.1 General Overview
Computability theory is a field of mathematical logic; the subject captures the precise
notion of an algorithmic process towards the study of decidable/undecidable problems
in mathematics and nature. Its most notable historical contribution to mathematics is
the disruption of Hilbert's Program by G6del's Incompleteness Theorem. More recent
work has shown that other mathematical problems are unsolvable, in the sense that
no computer algorithm can solve every instantiation of the problem. Examples include
Hilbert's Tenth Problem (to decide whether a given Diophantine equation has solutions),
the word problem for groups (to decide whether a given product of generators and their
inverses is the identity element of a group defined by a finite set of equations between
such products), and the homeomorphy problem (to decide whether the topological spaces
defined by a given pair of simplicial complexes are homeomorphic) [32]. Maturation
of computability, however, through applications and techniques, has broadened its
interactions with other fields, most notably computer science. Each of C'! Ilpters 25 is
a witness to this broadening.
In C'! lpter 2, we focus on the study of effectively closed sets of binary reals. Thought
of as a set representing the solutions to some problem, effectively closed sets characterize
many structures in mathematics and computer science. In algebra, for example, they
represent the prime ideals of a computable enumerable Boolean algebra or a commutative
ring with identity [47]. In graph theory, they represent the set of solutions to ri ,nr:
problems with computable graphs, such as Hamiltonian circuits or vertex partitions [20].
In computer science, effectively closed sets arise in the study of nonmonotonic logics and
Lclanguages [25, 68, 70]. Given the wide variety of applications, the work in ('!i ipter 2
focuses on methods of enumerating various families of effectively closed sets. The idea
is to provide, based on desired properties, complete listings of the objects in this case,
effectively closed sets representing the set of solutions to problems of a certain type.
We show that there is an injective computable enumeration of the entire class of sets, of
certain families of string verifiable classes, and of the decidable and homogeneous classes.
We also show that no computable enumeration exists for thin, perfect thin, small, very
small, or nondecidable classes.
In ('!i lpters 3 and 4, we extend notions related to effective randomness for binary
reals, to closed sets (C'!i plter 3) and continuous functions (C'! plter 4); various global
properties are obtained. A binary real is effectively random if it is impossible for a
computer to find regularity or patterns in it. For a closed set, representing some set
of solutions, this means that it is difficult for a computer to precisely obtain or locally
describe this set of solutions, given the lack of pattern. In both chapters, we obtain
socalled basis and antibasis theorems. For instance, every random closed set contains
random and nonrandom elements, but omits various elements of computabilitytheoretic
interest, such as the properties of being fc.e. (f a polynomial), 1generic, or of incomplete
c.e. degree. We also show that concepts from both chapters are closed related; random
closed sets arise as inverse images of random continuous functions mapping to 0", but
not, in general, as images. Methods employ, ,1 in all of this work range from techniques
in computability and effective randomness, to techniques related to the study of effective
Hausdorff dimension and classical probability.
Finally, in connection with the roots of computability theory, in ('! Ilpter 5 we focus
on the classification of information content by means of Turing degree theory. Mathemati
cal structures or objects are often encoded as sets of natural numbers. Reduction methods,
such as the Turing reducibility, allow the information content of these sets to be classified
into equivalence classes, called degrees. In Chapter 5, the focus is on the bounded Turing
reducibility [16]; we show that capping, the operation which takes the meet of a given
noncomputable incomplete c.e. degree with another noncomputable incomplete c.e. degree
such that the resulting meet is the degree 0 of the computable sets, is continuous. That is,
for any b / 0, O', there is an a > b such that for any x, b A x = 0 iff a A x = 0. As this
is a threequantifier statement in the c.e. bTdegrees, this result gives insight into the three
quanifier theory of the same, whose decidability/undecidability is currently unknown. As
an aside, in recent work by Soare, the bounded Turing reducibility with the identity use
has led to applications of degree theory to differential geometry [72, 87].
The rest of this chapter is devoted to introducing to basic definitions, terminology,
and notations that will be used throughout this entire work. Section 1.2 covers basic no
tions and notations for topics in classical computability (e.g. computable sets, computably
enumerable sets, partial computable functions, Turing degrees). Section 1.3 covers the ba
sics of closed sets in : As algorithmic randomness is a topic covered only in ('!i Ilters 3
and 4, we postpone a general introduction of this topic and provide it in Section 3.2.
1.2 Classical Computability
We generally follow the notation of Soare [86] for notions that arise from classical
computability. For an indepth treatment of the basic foundations of the subject, we refer
the reader there.
A set A is computable if there is a computer program, or equivalently a computable
function, which halts and gives the correct answer on every query to the membership
predicate for A. It is computably enumerable (c.e.) if the computer program is required
to halt only on queries for elements in A; this gives rise to the standard example of a c.e.
set, namely the halting problem the set of Turing machines, officially encoded as a set of
natural numbers, that halt when given their own binary input. Computably enumerable
sets are the domains, therefore, of socalled partial computable functions; we index the
partial computable {0, 1}valued functions as {(,}1E,. Partial computable functionals that
take natural number (m) and real (x) inputs are indexed as te; we will write K1(m) for
the result of applying 4~, to m and x.
Other related notations are standard: 0e,s denotes that portion e, defined by stage
s, and e(x) I means that e, is defined on x (and T means undefined). We also index
the primitive recursive functions, a smaller class of total functions, as {i7} ,Ew. (*,*)
2  wo is typically a computable bijection such that (0, 0) = 0. A and '(A) denote the
complement and power set of A, respectively. z = x E y is the coding together of two reals
x and y, so that z(2n) = x(n) and z(2n + 1) = y(n) for all n.
In computability, a reduction is a binary relation on subsets of N that captures a
relationship between the information content of two sets. The Turing reduction
main reduction used in computability theory. A is Turing reducible to B, written A
if membership in A can be determined by a computer algorithm that has full access to
the membership predicate for B. Intuitively, the information content of A is viewed as
computable, or recoverable, from B. Furthermore B is viewed as an oracle, in terms of
information content, for determining membership in A.
Various restrictions on how much oracle information is allowed to be used in deter
mining the membership of a single element have given rise to different kinds of reductions.
For example, the bounded Turing reduction
determining membership of x in A use at most f(x) amount of B, where f, called the use,
is bounded by a computable function. The identity bounded Turing reductions insists that
the use f be bounded by the identity function.
Reductions often give rise to equivalence classes, called degrees, where two sets are
equivalent if they are mutually reducibile. The Turing degrees that contain c.e. sets are
called c.e. Turing degrees. We denote C and CbT as the structures of the c.e. degrees under
the Turing reductions and the bounded Turing reductions respectively. The study of the
Turing degrees has been one of the 1i i i"r themes in computability research; these degrees
capture the structure of the undecidable problems in arithmetic and nature.
1.3 Closed Sets in Computability
We generally follow the notation of Cenzer [19] for closed sets: For a finite string
a E aW, let 7o = n. We let 0 denote the empty string, which has length 0. A word (a) of
length 1 is may be identified with the symbol a. For two strings ao, T, i that r extends a
and write a c T if lal < Ir and a(i) = r(i) for i < al. Similarly a C: x for x E 2" means
that a(i) = (i) for i < aol. Let a7 denote the concatenation of a and r. Given a finite
string a, let I(a), or alternatively [a], be the interval of all infinite sequences extending a,
i.e. I(ao) = {x E 2" : C x}. Each such interval is a clopen set and the clopen sets are just
finite unions of intervals.
A subset T of u<" is a tree if it is closed under initial segments. The set [T] of infinite
paths through T is defined by x E [T] => (Vn)x [n E T, where x [n = ((0),... ,x(n 1)).
We i that a tree T C w< and set [T] are clopen if there is a nonempty finite S C c<'
so that T = 0 or T = {a : a C or 7 C o for some 7 E S}. P C w" is closed if and only
if P = [T] for some tree T. Now a nonempty closed set P may be identified with a unique
tree Tp {oa : P n I(a) / 0}. Tp has the distinct property of having no dead ends; that is,
if a E Tp then either a^0 E Tp or cral E Tp.
P is an effectively closed set, or IIo class, if P = [T] for some computable tree T Other
definitions are equivalent; in particular P is a fIo class if and only if P = [T] for some
primitive recursive tree T and also if and only if P [T] for some II tree T. Note that
if P is a IIo class, then Tp is a II set, but not in general computable. P is said to be a
decidable Io class if Tp is computable. P is said to be a strong HIo class, if Tp is a IIo set,
or equivalently if P [T] for some A tree; P is said to be a strong A class if Tp is A.
Thus any fII class is also a strong A class. Any decidable fIo class contains a computable
element (in particular the leftmost and rightmost paths) and similarly any strong A class
contains a A element. On the other hand, there exist fII classes with no computable
elements and strong Io classes with no A elements. A IIo class is said to be special if it
does not contain a computable member.
A c.e. open set is defined to be the complement of an effectively closed set. That is, if
P = [T], then ww P = U(<,)T I(a). There is a natural effective enumeration Po, P1,...
of the II classes and thus an enumeration of the c.e. open sets. Thus we can ;v that a
sequence So, Si,... of c.e. open sets is effective if there is a computable function, f, such
that S, = 2'
Pf(,) for all n. For any c.e. set W, we define the c.e. open set generated by
W to be
O(W) U{I(o): () s w}.
Also let
O(W) [n = {x n : x e O(W) and (Vj < n) x(j) < n}.
For a detailed development of II classes, see [20] or [24].
CHAPTER 2
EFFECTIVELY CLOSED SETS AND ENUMERATIONS
The following chapter is joint work with Douglas Cenzer and has been submitted
as an article entitled Effectively Closed Sets and Enumerations [13]. A preliminary
version of this research was originally presented at the Third International Conference
of Computability and Complexity in Analysis in Gainesville, Florida in 2006 by P.
Brodhead. This preliminary work was published in the referred conference proceedings as
Enumerations of IH Classes: A il,.',ii.L i and Decidable Classes (P. Brodhead) in hard
copy and in Springer Electronic Notes in Theoretical Computer Science, Elsevier Science
167 (2007), 289301 [12].
Portions of this work were also presented by P. Brodhead at the 2005 SACNAS
Conference (October 2005, Denver, CO), 7th Annual Graduate Student Conferences in
Logic (April 2006, Madison, WI), the 2007 Association for Symbolic Logic Annual Meeting
(\! I'ch 2007, Gainesville, FL), the 2nd New York Graduate Student Conference in Logic
(\! I'ch 2007, New York, NY), and the 8th Annual Graduate Student Conference in Logic
(April 2007, Chicago, IL). Due to inclimate weather, R. Miller presented in place of P.
Brodhead at the New York conference.
2.1 Introduction
The general theory of numberings was initiated in the mid1950s by Kolmogorov, and
continued under the direction of Mal'tsev and Ershov [44]. A numbering, or enumeration,
of a collection C of objects is a surjective map F : w C. In one of the earliest results,
Friedberg constructed an injective computable numbering i of the Et or computably
enumerable (c.e.) sets such that the relation "n E c (e)" is itself E. More generally, we
will iv that a numbering i of collection of objects with complexity C (such as nc.e.,
E, or U11) is effective if the relation "x cE (e)" has complexity C. We will also consider
enumerations where the relation "x E Q(e)" has a different complexity than C. (For
example, there is a c.e., but not computable, numbering of the computable sets.)
A numbering p is acceptable with respect to a numbering v, denoted v < p, iff there
is a total computable function f such that v = o f. If p is acceptable with respect
to all effective numberings, then p is said to be acceptable. The ordering < gives rise to
an equivalence relation , and two numberings in the same equivalence class are called
equivalent. Furthermore, the structure (C) of all numberings of C modulo forms an
upper semilattice under <. Here injective numberings occur only in the minimal elements
and acceptable numberings occur only in the greatest element. In this chapter, we study
effective numberings of families of effectively closed sets (i.e. II classes).
Enumerations of HI classes were given by Lempp [61] and Cenzer and Remmel
[23, 24], where index sets for various families of HI classes were analyzed. For a given
enumeration &(e) = P, of the II classes and a property R of sets, {e : R(P)} is said to
be an index set. For example, {e : P, has a computable member} is a E complete set. See
[20] for many more results on index sets.
Certain types of HI classes are of particular interest. Let P be a HI class. We will 
that P is thin if for every HI subclass Q of P, there is clopen set U such that Q U n P.
We w that P is homogenous if, given distinct a, r Tp of the same length,
a'i E Tp < rTi Tp.
For P C {0, 1}", P is homogeneous if and only if P is the class of separating sets S(A, B)
for two disjoint c.e. sets A, B, that is,
S(A,B)= {C C : A C C and B n C 0}.
P is small if there is no computable function Q such that, for all n, card(Tp n ;()) > n.
Let Qp(n) be the least k such that card(Tp n wk) > n; then P is very small if the function
Qp dominates every computable function g that is, Qp(n) > g(n) for all but finitely
many n.
A numbering e v> [T,] of HI classes is called a tree numbering and written e v> Te.
Numberings based on primitive recursive trees and on Hn trees are both studied in the
literature (see [23, 24, 20]). If the set {(e, a) : a e T} is computable, then the numbering
(e) = [T,] is said to be a computable numbering.
We begin our study with the family of II classes in Section 2.2. In Sections 2.2.1
2.2.3, several commonly used numberings are studied and shown to be equivalent via a
computable permutation. In Section 2.2.4, we give give a Friedberg numbering of the II
classes; this motivates a study of a general class of families of II classes, called string veri
fiable families in Section 2.3. In Section 2.4, we consider named families of II classes. We
obtain positive results for homogeneous can decidable classes in Sections 2.4.1 and 2.4.2.
We obtain negative results for thin, perfect thin, small, very small, and nondecidable
classes in Sections 2.4.3 and 2.4.4.
2.2 The Family of II Classes
2.2.1 Numberings in the Literature
In this section, we present several different computable numberings of II classes
that have appeared in the literature. We also present an effective, but not computable,
numbering. In each case we demonstrate that each provides a complete numbering of the
II classes.
Numbering 1: Primitive Recursive Functions [23]
For each e, let 7 be the eth primitive recursive function from u to U and let
7 G U, (VT C a) 7Te((T)) 1.
Then Ue is a (uniformly) primitive recursive tree for all e and if {a : 7((a)) = 1} is
any primitive recursive tree, then Ue is that tree. Therefore the sequence Uo, U1,...
contains all primitive recursive trees and hence the mapping li(e) = [Ue] is a computable
numbering of the II classes.
Numbering 2: Computably Enumerable Sets [20]
Let
2(e) O(We).
This is an effective numbering since the relation "x e Q2(e)" is II This can actually be
improved to a computable numbering, as follows.
For each e, recall that We,s is the set of elements enumerated into the eth c.e. set We
by stage s and let
a G Se (VT 7() { W,,\.
Then Se is a (uniformly) primitive recursive tree for all e. Let P = [T] be a II class,
where T is a computable tree. It follows that for some e,
a c T ^ (a) i We.
Then P = [Se]. It follows that the sequence [So], [Si],... contains all nIT classes and hence
the mapping Q(e) [Se] is a computable numbering of the II classes. It is easy to see
that in fact [Se] = 2(e).
Numbering 3: Universal IIt Relation [52, p.73]
There is a universal II relation U C w x 2" such that if Q(x) is a II0 class, then there
is an e e w such that Q {x : U(e, x)}. U is defined in terms of the Kleen Tpredicate, so
that essentially
U(e,x) = V (0) .
Define (' (e) = {x : U(e, x)} to obtain an effective numbering.
To see that this is a computable numbering, let
a e Re 4) 1(0) T
so that ,' (e) = [Re] and the trees Re are uniformly primitive recursive.
Numbering 4: The Halting Problem [53]
Consider the mapping given by
64(6) {X: 4)x(e)(e)T}.
This is a computable numbering, since 4(6) = [T,], where
a Te 4 ),(e) T.
For any computable tree T, choose a so that 1{(n) converges if and only if a c T. Then
a e T V (a)l
so that [T] = 4 ().
Numbering 5: Total Computable Functions
Here we will consider an effective, but not computable numbering i based on
computable trees. This numbering will be used in connection with string verifiable
families of classes in Section 2.3.
Let (e) = [Te], where
This enumeration is uniformly HI, but is not computable, since the relation 0e(m) I is
c.e. noncomputable. Clearly each (e) is a II class. If e, is total and T is a tree such
that, for all a, we have a c T 0 ,e((a)) = 1, then Te = T and is a nI class. Hence
this enumeration has the crucial property that, for every computable tree T, there exists e
such that T =Te.
2.2.2 Equivalence of the Numberings
In this section we show that the computable numberings of section 2.2.1 are mutually
equivalent via a computable permutation. Each of these is equivalent to the effective
enumeration of section 2.2.1 via a Apermutation. We need the following proposition.
Proposition 2.2.1. (a) For each pair i,j with 1 < i < 5 and 1 < j < 4, there is a
computable function f such that bj = o f.
(b) For each j ( 5, there is a A function f such that . = o f.
Proof. (Q1 < V2): Use the S m n Theorem to define f so that
Wf(e) = n : 7e(n) / e}.
Then a E U, a E Sf(,).
( ',_< .): Define f so that, for all m,
S(e) (m) = (least n) (x n) e We).
Then 2 (e) (f(e)).
( <'. 4): Define f so that KI(nu) + (0) for all n. Then x e (e) < x e
4 (f (e)).
( <4 Q i): Recall that 4)4(e) [{a : (e)T}].
Define the primitive recursive function g so that for each e,
'1, if I) (e)T;
TAf(e)(r7))
0, otherwise.
Then 1(e) = 4(9(e)).
(Q1 < .): Define the primitive recursive function f such that, for each e,
( ) if (VT C 7a)7((T)) 1,
0, otherwise.
Then 1l(e) (f(e)).
The rest of the proof follows by composition. O
Theorem 2.2.2. For any computable numbering o which is computably equivalent to )2,
there is a computable permutation p such that '_. o p.
Proof. The proof is a modification of an argument due to Jockusch [86, p. 25]. Let
S= '_.. By assumption, there are computable functions f and g such that )(f(e)) = (e)
and c(g(e)) = (e). Since the numbering ,'_. is based on an enumeration of the partial
computable functions, we can ensure by padding that f is injective. To modify g into an
injective function gl, it is sufficient to effectively compute from each e an infinite set Se of
indices such that p(g(i)) = c(e) for all i e Se. We proceed as follows. Let A and B be
computably inseparable c.e. sets and define computable functions k and such that, for all
e and m:
(k(em)) (e), if m B,
0, if m B
and
((e, T)) = (e), if me A,
2", if m A
That is, we build a tree for &(k(e, m)) which exactly equals the tree for Q(e) for strings of
length s until m e B,+i, in which case no strings of length s + 1 are put into ((k(e, m)).
To build the tree for (f(e, m)), we put in all strings of length s until m e A,+1, in which
case we include only the strings of length s + 1 which are in )(e).
Now let Ce ={k(e, m) : m e A} and De { (e, m) : m A} and let Se g(Ce U De).
Then for j = g(i) e Se, it follows from the definition that b(i) = (e) and therefore
p(g(i)) = 9(g(e)). We will prove in two cases that either g(Ce) is infinite or g(De) is
infinite.
Case I: Suppose that Q(e) / 0 and suppose by way of contradiction that g(Ce)
is finite. Then S {= m : g(k(e, m)) e g(Ce)} is a computable set. Now A C g(Ce)
by definition. On the other hand, if j = g(k(e,m)) e g(Ce) where m e A, then
((j) = (k(e, m)) = ((e) / 0. But for m e B, p(g(k(e, m)) = (k(e, m)) = 0, so that
S n B = 0. This contradicts the assumption that A and B are computably inseparable.
Case II: Suppose that b(e) = 0. It follows as in Case I that g(D,) is infinite.
Thus we may assume without loss of generality that both f and g are oneto
one. Now define a sequence {e, : n e w} and two partitions of u as follows. Let
eo = 0 and for each n, en+ is the least e such that 9p(e) / 9p(ej) for every i <, n. Let
An = {e : (e) = (en)} and B, = {e : 9p(e) = (en)}. Then w = U. AT = Un B, and each
sequence is pairwise disjoint. Furthermore, f(B,) C AT and g(A,) c B,. The remainder
of the proof follows as in the Myhill Isomorphism Theorem [86, p. 24; Also 5.8, p. 25]. O
A similar argument shows that if p is a A numbering of the II classes, then there is
a A permutation p with p = o p. It follows that each of the computable numberings
S,... Q4 are acceptable, that is, they occur in the greatest element of the semilattice
C(P). In the section 2.2.4 we will see that minimal elements exist in the semilattice that
is, injective numberings.
First, however, we provide an alternate proof of Theorem 2.2.2.
2.2.3 Equivalence of the Numberings (Alternate Proof)
At the Third International Conference of Computability and Complexity in Analysis
in Gainesville, Florida in 2006, I presented the argument for Theorem 2.2.2 as given
above; it is a modification of an argument of Jockusch for the c.e. version. Pleased with
the argument's applicability to IIo classes, I urged Robert Soare, who was present in the
audience, to keep the Jockusch argument for the c.e. version in his new and upcoming
book, Computability Theory and Applications [88]. If the alternate proof using the
recursion theorem for the c.e. version of Theorem 2.2.2 [86, p. 43] could not also be
modified to also prove Theorem 2.2.2, then he said he would. We show below that an
alternate proof is possible, modifying the recursion theorem argument.
Theorem 2.2.3. For any computable numbering o which is computably equivalent to b2,
there is a computable permutation p such that 2 = p o p.
Alternate Proof. We proceed, at first, as before. That is, there are computable func
tions f and g such that b(f(e)) = (e) and c(g(e)) = (e). Our goal is to modify g to
obtain an injective function gl. Instead, however, we define gl differently, using an auxil
iary computable function h obtained by the Recursion Theorem. We ensure h satisfies, for
all distinct i and j, (i) gh(e, i) / gh(e,j) and (ii) (h(e,i) = O.
We now define gl, given h satisfying (i) and (ii). Define gi(0) = gh(fg(0),0). To
define gi(k + 1), note that (i) ensures infinitely many distinct gh(e, i) for each e. Let
ao = 0 and ak+1 be the least integer i such that gh(fg(k + 1), i) > gi(k). Define
gi(k + 1) = h(fg(k + 1),ak+1). To see that for all e, '..,,e) = Q, fix e and note that
' ..,(e) = (fg(e),a) h(fg(e),a,) = fg(e) = ':. = e. To see that gi is injective, note that
the definition ensures that for all k, g (k + 1) > gi(k).
We now define h(e, *) for each e, by induction, ensuring that (i) and (ii) are satisfied.
Define h(e, 0) = e. To define h(e, k + 1), use the Recursion Theorem to obtain an n such
that
() = #e(z) if g(n) / gh(e, i) (Vi < k)
undefined otherwise
Notice that if g(n) = gh(e, i) then since i < k, by induction = h(e,i) = '..I.
Qg(T) = Q, the undefined function. Hence e = OT in all instances. Since {a : fa = ~}
is a productive [86, p. 43], let p be a corresponding productive function. Define Wr(x)
WE U {p(x)} and note that each W,(,r) is a distinct c.e. subset of A. Consequently each
gri(n) = '. ; is a distinct partial computable function. Let nk be the least i such that
r(n) / h(e,j) for all j < k. Define
h(e, k n if g(n) / gh(e, i) (Vi < k)
h(e, k+1)
rf (n) otherwise
To see that (i) holds (gh(e, i) / gh(e,j) for distinct i,j), note that '.,., = Qe /
S, gr(,) for all i > 1. Therefore if g(n) / g(e) then for all j, h(e,j) e {e} U
{r'(n)}i>o. Otherwise h(e,j) e {e} U {r'(n)} i1. In either case, for all distinct i,j,
gh(e, i) / gh(e,j).
To see that (ii) holds (/'..., = oe for all i), note that ,e = h(e, i) e {e} U
{r'(n)}i(o or 1), and r'(n) e {a :a O= ,}. So e = (h(e,i) for all i. E
2.2.4 Injective Computable Numberings
In this section, we modify Friedberg's original argument for injective computable
numbering of the c.e. sets, to provide different numbering results needed for 11 classes. In
Section 2.2.4.1, we present Friedberg's original argument. In Section 2.2.4.2, we provide
an injective computable numbering of diiP ii 1 pairs of c.e. sets; this in needed later in
Section 2.4.1 in order to provide an injective computable numbering of the homogeneous
II classes. Finally, in Section 2.2.4.3, we construct a computable injective numbering of
the II classes in 2"; we also provide other injective numbering results for II classes.
2.2.4.1 Original c.e. sets argument
The following is the original Friedberg argument, with slight modifications in the
notation and presentation.
Theorem 2.2.4 ( [46]). There is an injective computable numbering of all c.e. sets.
Proof. Let {We}ee be the standard numbering of the c.e. sets. We will construct
a sequence of c.e. sets {Ye}ee in stages so that e v4 Y, will be the desired injective
numbering. In the construction we will use the notion of one Yindex i following a W
index e with the idea that in the end i will equal We. At some point, however, we may
decide that i will no longer follow e again and we will i that i is released. If i is never
released from following e then it is said to be a lo(,il follower and otherwise it is 1/. ..;'/.
Once released, an index remains free and is never again the follower of any e. At any
particular stage, any Yindex that is not following any Windex is said to be free. Any
nonzero Yindex that has never followed any Windex is said to be unused.
To ensure that no c.e. set is excluded from the Ysets, we will ensure that each We
is infinitely often given the opportunity to be followed. To do this, at stage s = (ns, eS)
all actions in the construction will be taken with respect to We,. Assume without loss of
generality that Yo = 0.
Construction: There are three possible cases at each stage s.
Case 1: If i follows e & We,s n [0, i 1] = We,,,s [0, i 1] (some e < e,), release i and
go on to stage s + 1.
Case 2: Suppose Case 1 does not occur. If Wes, = i,s1, and either i follows some
e < e, i = 0, i is free and i < e, or i is free and i was previously displaced (see Case 3) by
e and released, then go on to stage s + 1 without taking any action.
Case 3: Suppose that Cases 1 and 2 do not occur. Now ensure that e has a follower. If it
does not, choose the lowest unused i / 0 to follow e. Now let ,, = We,,s.
For each j / i such that Yj,s_ = We,,s (in increasing order of j) put the lowest b not
yet in any W or Y into Yj and release j if it is a follower. We i e displaces j .
Verification: Given e E u, let e be the least k so that Wk = We. We will show:
(i) (we) (i) = W
(ii) i / j Yi and Yj are not the same finite sets
(iii) i 7 j and Yj are not the same infinite sets
Verification of (i). Fix e. First note that although e can only have one follower at any
particular stage, it cannot have an infinite number of them which are each released at
some stage. For example, if s and x are sufficiently large, then for all j < e, Wj,8 n [0, x] /
W8, n [0, x]. Hence release can only occur in Case 1 a finite number of times. Furthermore,
Case 2 ensures that release in Case 3 can only occur for any s when e, < e. Therefore, by
the above, if i > x is follower of e and t > s, then Yt1 = Wt / We,,t. Hence i will not
be released in Case 3. Therefore release can only occur in Case 3 a finite number of times.
Now let s be a stage where e never loses a follower. If Case 3 occurs infinitely often
after stage s for e, then it has a permanent follower i so that Y = W. Therefore assume
Case 3 occurs only finitely often. Since e never loses a follower, Case 1 cannot occur.
Hence Case 2 must occur infinitely often. However there are only a finite number of
i such that the 'If' clause holds with "W,s = Ys_1" in Case 2. (For example, {i :
i = 0, i < e, or i is displaced by e} is a finite set since only a finite number of i are
displaced due to Case 3 occurring only a finite number of times. To see also that {i :
for some s, i follows some a < e and "W,s = Y,s1" holds} is finite, note that by the
definition of e, if a < e then Wa W4. So for sufficiently large t, if i follows a then
W,
of i such that the 'If' clause holds with "W6,s = i,s", implies that there is a single i such
that I,1 = W<,s for infinitely many s. Thus Y = W.
Verification of (ii). First note that at any stage s, if W, = 0 then Case 2 ensures that
Case 3 will not be reached so that ec never receives a follower. Furthermore each k / 0
is chosen at some stage s in Case 3 to follow some e and from the previous comments,
Yk,s / 0. Immediately thereafter, Yk,s is ensured to be distinct from all other Y,s (f / k).
This also continues to be true at all subsequent stages t by Case 3.
Now suppose i / j. Since we are supposing both i and Y are finite, there is some
t such that for all s > t, Y,8 = Y and Y,8 = j. As mentioned above Y,s / Y,s and
therefore i / Yj as required.
Verification of (iii). Assume both Y and Y are infinite and i / j. Now i must
eventually follow some Windex. If i is di. il 1, then after it is released i can acquire a
new member only when i is displaced. However i can be displaced only once by each e < i
and never by any e > i. It follows that if i is released then 1 only acquires a finite number
of elements thereafter, contradicting the fact that it is infinite. This argument shows that
both i and j are never released. We iv that i and j are loyal followers.
Suppose now that i and j loyally follow i' and j', respectively. Then Y = Wi, and
Yj W. Now i' / j' since a single Windex cannot have more than one loyal follower.
Assume without loss of generality that i' < j. If W, = Wj,, then for all sufficiently large
s, Wi,,s n [0,j 1] = Wj,,, n [0,j 1] so that Case 1 releases j, a contradiction. Therefore
Yi = Wi, Wy = Yj as required. O
2.2.4.2 Ordered tuples of disjoint c.e. sets
In this section we modify the argument of Theorem 2.2.4 to obtain an injective
computable numbering of all ordered tuples of disjoint c.e. sets.
Theorem 2.2.5. There is an injective computable numbering of all ordered tuples of
disjoint c.e. sets.
Proof. Let {Se}ee, be the standard numbering of the c.e. sets. Then (e, i) v (Se, Si) is a
computable numbering of all tuples of c.e. sets. Obtain a computable numbering (e, i) i
(We, Wi) of all tuples of dlii iiil c.e. sets as follows. At stage s + 1, if a E Se,8+1 \ Se,s then
put a into We,s+i if a g Wi,,. If b e Si,s+l \ Si,s then put b into Wi,s+l if b g We,s+1. Note
that if originally Se n Si = 0 then S, = We and Si = Wi.
Similar to Theorem 2.2.4 (in construction and terminology), we will construct in
stages a sequence of pairs of c.e. sets (Y, Y)(e,i)ew so that (e, i) ( (Y, ) will be the
desired injective numbering. So that no pair is excluded from the Ysets, we will ensure
that each (e, i) is infinitely often given the opportunity to be followed. To do this, at stage
s = (ns, (e, is)) all actions in the construction will be taken with respect to (We, Wi).
Assume without loss of generality that (Yo, Y) = (0, 0).
Construction: There are three possible cases at each stage s.
Case 1: If (e, i) follows (e,, i) and Wa,s n [0, (e, i) 1] = We,, n [0, (e, i) 1] and
Wb,s n [0, (e, i) 1] = Wi,, n [0, (e, i) 1] for some (a, b) < (e, i,), then release (e, i) and
go on to stage s + 1.
Case 2: Suppose Case 1 does not occur. If (Y(,1, Y,s1) = (We,,s, Wi,,s) and either (c, t)
follows some (a, b) < (e,, i), (e, t) = 0, (e, t) is free and (e, t) < (e,, is), or (e, t) is free and
(e, t) was previously displaced (see Case 3) by (es, is), then go on to stage s + 1 without
taking any action.
Case 3: Suppose Cases 1 and 2 do not occur. Now ensure that (es, is) has a follower.
If it does not, choose the least unused (e, i) / 0 to follow (es, is). Now let (Y,,, ,,s) =
(W5s, WiU s ).
For each (c, t) / (e, i) such that (Y,s_1,Yi,s1) = (W",,s, Wis,,) (in increasing order of
(e, t)) put the lowest b not yet in any W or Y into Y, and release (c, t) if it is a follower.
We v (es, i4) displaces (, t) .
Verification: Given (e, i) e wo, let (e,i) be the least (k, ) so that (Wk, W1) =
(We, Wi). We will show:
(i) V(e,i) 3(k,<) (Yk,Y) = W, Wj)
(ii) [(e, i) / (c, t) & Y = Y] Y, and Y, are not the same finite sets
(iii) [(e, i) (c, t) & Y Y,] Y, and Y, are not the same infinite sets
(iv) V(e,i) Y, n 0
Verification of (i). First note that although (e, i) can only have one follower at any
particular stage, it cannot have an infinite number of them which are each released at
some stage. For example, if s and (x, y) are sufficiently large then for all (k, ) < (e, i),
either Wk,, n [0, (x, y)] / W,,, n [0, (x, y)] or W1,, n [0, (x, y)] / Wi,, n [0, (x, y)]. Hence
release can only occur in Case 1 a finite number of times. Furthermore, Case 2 ensures
that release in Case 3 can only occur for any s when (e, is) < (e,i). Therefore, by the
above, if (a, b) > (x, y) is follower of (e,i) and t > s, then Ya,t1 = We,t1 / W,t or
Yb,t = Wi,t1 / Wi,t. Hence (a, b) will not be released in Case 3. Therefore release can
only occur in Case 3 a finite number of times.
Let s be a stage where (e, i) never loses a follower. If Case 3 occurs infinitely often
after stage s for (e,i), then it has a permanent follower (e, i) so that (Ye, Y) = (W, Wi).
Therefore assume Case 3 occurs only finitely often. Since (e, i) never loses a follower, Case
1 cannot occur. Hence Case 2 must occur infinitely often.
There are only a finite number of (c, t) such that the 'If' clause holds with the
equality A(c, t, e,i, s) given by "(Y ,s,,s1) = (We,s, Wi,)" in Case 2. For example,
{(c, t) : (c, t) = 0, (c, t) < (e,i), or (c, t) is displaced by (e,i)} is a finite set since only a
finite number of (c, t) are displaced due to Case 3 occurring only a finite number of times.
To see also that { (, t) : for some s, (c, t) follows some (a, b) < (e,i) and A(c, t, e, i, s) holds}
is finite, note that by the definition of (e,i), if (a, b) < (e,i), then either WE / We or
Wb / Wi. So for sufficiently large t, if (c, t) follows (a, b) then (Y ,t, ,,t ) (W,t, Wi,t)
and so A(c, t, e, i, t) does not hold.
Now Case 2 occurring infinitely often, together with only finite number of (c, L)
such that the 'If' clause holds with A, implies that there is a single (c, t) such that
(Y,,_ 1,Y, ,_1) (We,s, Wi,8) for infinitely many s. Thus (, Y,) (We, W).
Verification of (ii). First note that at any stage s, if (We, Ws) = (0, 0) then Case 2 en
sures that Case 3 will not be reached so that (es, is) never receives a follower. Furthermore
each (a, b) / (0, 0) is chosen at some stage s in Case 3 to follow some (e, i) and from the
previous comments, (Y,s,, Yb,) / (0, 0). Immediately thereafter, (Ys, Yb,s) is ensured to
be distinct from all other (Yk,s,Y ,s) ((a, b) / (k,f)). This also continues to be true at all
subsequent stages t by Case 3.
Now suppose (e, i) / (e, t), = Y,, and both Y, and Y, are finite. There are two
cases. Suppose first that both Yi and Y, are finite. Now there is some t such that for all
s > t, Yk,s Yk (k e {e, i, e, t}). As shown above (Y,,, Y,) / (Y,s,,Y,) and therefore
(Y, Y) / (Y, Y,). Since Y, Y, it follows that Y / Y,.
Suppose now that Yi and Y, are both infinite. Now (e, i) must eventually follow
some (e', i'). If (e, i) is ever released then it is free. Thereafter only the first coordinate
of (Y,, Y) acquires members so that Yi is finite, a contradiction. A similar argument
holds for (c, t). Therefore both (e, i) and (e, L) are never released and are Ic;,.l followers.
Suppose that they follow (e', i') and (c', t'), respectively. Then (Y, Y) = (We,, Wi) and
(Y,Y) = (We,, W,). Note that (', i') / (c', L') since a single Windex cannot have
more than one loyal follower. Assume without loss of generality that (e', i') < (e', L').
Suppose now that We = We,. By assumption, Wi, Y= =, W,, Therefore, for all
sufficiently large s, We,s n [0, (c, t) 1] = We,,s n [0, (c, t) 1] and Wi,, n [0, (c, t) 1]
W,,, n [0, (c, t) 1], so that Case 1 releases (c, t) a contradiction. Therefore We' / We so
that also Y, We/ We = Ye, as required.
Verification of (iii). Suppose that (e, i) / (e, t), Y= Y and both Ye and Y, are infinite.
Now (e, i) must eventually follow some (e', i'). If (e, i) is di1 ,l 1l, then after it is released
Ye can acquire a new member only when (e, i) is displaced. However (e, i) can only be
displaced once by each (c, d) < (e, i) and never by any (c, d) > (e, i). It follows that if (e, i)
is released then Ye only acquires a finite number of elements thereafter, contradicting the
fact that it is infinite. This argument shows that both (e, i) and (e, t) are never released
and are therefore loyal followers. Now apply the same argument given in the later part of
the verification of (ii) to get that Ye / Y,.
Verification of (iv). If (e, i) is a loyal follower of some (e', '), then (Ye, Y) = (We, Wi).
Therefore since (e, i) (We, Wi) is an enumeration of disjoint sets, it follows that
Ye and Yi are disjoint. Otherwise suppose that (e, i) is released at some stage s. Then
Ye,s n Y, 0 and thereafter only Ye can acquire new elements not already included in Y,s.
Therefore disjointness is preserved and Ye n = 0. O
2.2.4.3 Results for effectively closed sets
In this section, we modify the Friedberg argument of Section 2.2.4.1 to construct a
computable injective numbering of the II classes in 2". An alternative proof was sketched
by Raichev [79]. Afterwards we provide other injective numbering results for II classes.
Theorem 2.2.6. There is an injective computable numbering of all II classes in 2".
Proof. Let {W }ee be the computable enumeration of the nonempty c.e. subsets of
2". We will construct a computable numbering {Y : e E ow} in stages Ye,, of a family
of c.e. subsets of 2" so that {O(Y,)},E is an injective numbering of the E classes; our
construction and terminology will be similar to Theorem 2.2.4.
It is important to note that an injective numbering of the c.e. subsets of 2<' will not
automatically yield an injective numbering of the E classes, since each E class will equal
O(W) for many different c.e. sets W. However, if O(V) / O(W) for two c.e. sets V and
W, then there must be some interval I(a) which is included in, , O(V) but not included
in O(W) and hence some stage s such that O(Vs) [ s / O(Ws) [ s at stage s and at any
later stage.
To ensure that no c.e. set is excluded from the Ysets, we will ensure that each We is
infinitely often given the opportunity to be followed. To do this, at stage s = (n, es) all
actions in the construction will be taken with respect to W,. At each stage s, we initiate
at most one new Yi, so that after stage s, we have sets Yo,Y1,...,Yk, for some k, < s. Fix
Yo = 0 and Y1 {0} so that O(Yo) = 0 and O(YI) = 2", respectively.
Construction: There are three possible cases at each stage s.
Case 1: If i follows e, and there exists e < e such that O(We,s) [(i1) = 0(W,,s) [(i1),
then release i and go on to stage s + 1.
Case 2: Suppose that Case 1 does not occur. If O(We,,s) = (Esi), and either i follows
some e < e, i = 0, i = 1, or i is free and either i < es or i was previously displaced (see
Case 3) by es, then go on to stage s + 1 without taking any action.
Case 3: Suppose that Cases 1 and 2 do not occur. Now ensure e has a follower. If it
does not, choose the least unused i / 0, 1 to follow e. Now let i, = We,,,.
If Yj,s_, for some j / i satisfies O(Yj,s_) = O(W,,,s), then put some cj e 2
in what follows, into Yj so that O(Yj,) / O(We,,s).
Let E = {j E c : j / i & O(Yj,_) = O(We,,s)} be the set of indices of equivalent
Yopen sets and suppose that E, = {ci < 62 < ... < IlEs}. Now define Str(k, s) = {a
2k :a O(W,,s) [ k}. Let f(s) be the least k such that IStr(k, s) > IEs,. Then f(s) is
the least level of O(We,,s) where there is enough room to give each equivalent Yj,s an
additional string to distinguish O(Yj,_i) from O(We,s). (Notice that O(We,s) / 2" by
Case 2.) Suppose that Str(f(s), s) = {oi < 2 ... < i Str(s(s),s) }. Now put ac into Yj
and release cj if it is a follower. We w that ce is displaced at stage s.
Verification: Given e c L, let e be the least k such that [O(Wk) = O(We)]. We will
show:
(i) (Ve)(3i) y We;
(ii) i / j implies that O(Yi) and O(Yj) are not equal when both are clopen.
(iii) i / j implies that O(Yi) and O(Yj) are not equal when both are not clopen.
Verification of (i). Fix e. First note that although e can have different followers at
different stages, it cannot have an infinite number of di1 i Ll followers. That is, if s and x
are sufficiently large, then by the definition of e, for all j < e, O(Wj,s) [ x / O(W,s) [ x.
Hence release can only occur in Case 1 a finite number of times. Furthermore, Case 2
ensures that release in Case 3 can only occur for any s when e < e. Therefore, by the
above, if i > x is follower of e and t > s, then O(Y,t1) = O(W,t) / O(W,,t). Hence i
will not be released in Case 3. Therefore release can only occur in Case 3 a finite number
of times.
Now let s be a stage after which e never loses a follower. If Case 3 occurs infinitely
often after stage s for e, then it has a permanent follower i so that O(Yi) = 0(W).
Therefore assume Case 3 occurs only finitely often. Since e never loses a follower, Case
1 cannot occur. Thus Case 2 must occur infinitely often. However there are only a finite
number of i such that the hypothesis of (ii) holds with O(W( ,s) = 0(Y,s_). To see
this, consider the three cases. First, each e < e has only finitely many followers by the
argument above; second, there are only finitely many i < e; and third, only a finite
number of i are displaced by es, due to Case 3 occurring only a finite number of times.
This contradiction shows that e has a permanent follower, as desired.
Verification of (ii). Suppose that U = 0(Y) = O(Yj) (/ 0, 2") is clopen and let
U = O(W6). It follows from compactness, that there is some finite s such that, for all
t > s, O(Yt) [ t = O(Y,t) [t = O(Y). It follows from the verification of (i) above that
there is a stage t > s such that Case 3 applies to e. But then at least one of 0(Y), O(Yj)
must change at stage t. This contradiction verifies (ii).
Verification of (iii). Assume both O(Yi) and O(Yj) are not clopen and i $ j. It follows
that O(Yi) must change infinitely often, since of course O(Y,s) is clopen for each s, and
similarly for O(Yj). Now i must eventually follow some Windex. If i is ever released, then
it is free. Thereafter Y acquires members in Case 3 at stage s only when We,,s = Yi,s.
This implies that Case 2 does not apply at stage s and thus e < i. But each e < i can
only displace i once, again by the hypothesis of Case 2. Thus if i is a di1 .v I1 follower,
then in fact O(Yi) is clopen. Thus we may assume that i is a loyal follower of e and j is a
loyal follower of e'. Then O(We) = O(We') but e / e', since each e can have at most one
loyal follower. Without loss of generality suppose e < e'.
Since O(We) = O(W,), there will be a stage s large enough so that O(We) [ (i 1)
O(We,) [ (i 1). Then since i follows e < e', i will be released at stage s, contradicting the
assumption that i is a loyal follower.
This verification completes the proof. O
The problem of finding an injective enumeration of the II classes in w" remains. For
classes in 2", we have the following generalization of Theorem 2.2.6. It will be useful later.
Let C be the family of clopen subsets of 2".
Theorem 2.2.7. For any family F of II classes in 2" which has a computable numbering,
there is a 11 computable numbering of C U F.
Proof. Let the computable enumeration P, be given. We may assume that C c F by
simply enumerating the clopen sets as {Q2e : e < w} and letting Q2e+1 = P,. Then the
proof of Theorem 2.2.6 produces a 11 computable enumeration of F as desired. O
We have the following corollary from the proof of Theorem 2.2.6.
Corollary 2.2.8. There is an effective numbering of the II classes based on the total
computable functions
Proof. Modify each c.e. set in the standard numbering to enumerate an element only
as long as it is larger than any previously enumerated element. Applying Friedberg's
argument to this class of c.e. sets yields an effective injective numbering e v4 Ce of the
computable sets [91]. Furthermore each Ce still enumerates its elements in increasing
order.
Now suppose {Xe}eew is a corresponding set of characteristic functions. One charac
terization of a nII class P is that P = w" \ O(W) for some computable set W [20]. As a
result, e \ OCe) \ 0(({n : Xe(n) = 1}) is an alternative effective numbering
based on total computable functions (replacing noneffective Numbering 2). O
It is known, for fixed n > 0, that there is a effective injective numbering of the nc.e.
sets [50].
Conjecture 2.2.9. For each n, there is a numbering e v> Ne of nc.e. sets such that there
is an injective computable numbering e v Lw" \ O(Ne) of all closed sets of this form.
For n = 1 the conjecture is given by Theorem 2.2.6.
We next show that Theorem 2.2.6 is not obtainable by any computable procedure
that uniformly selects the minimal index of every II class.
Theorem 2.2.10. There is no computable choice function for indices of FII classes. (i.e. a
computable function f such that f(e) is an index of PF and Pi = PF = f(i) = f(e))
Proof. Suppose that f exists. Let ao, al,... be an enumeration of a noncomputable c.e.
set A. Define a computable function g and trees T,(e) so that if al = n, then
E Tg(e) eg {ao,... ,a,an}.
Then
0 if eeA
S 2" otherwise
For any a c A,e A <+ f(g(e)) = f(g(a)), making A computable. O
It is still possible, however, that some interesting proper family of II classes may be
enumerated by selecting minimal indices from the enumeration of all II classes.
2.3 String Verifiable Families of II Classes
In this section we examine families of classes which are deemed to be string ;' I
able (e.g. decidable or homogeneous classes) or strongly string ;. ,:/7,,'.1: (e.g. strongly
decidable or strongly homogeneous classes). Any string verifiable family has an effective
numbering and any strongly string verifiable family has a computable numbering.
2.3.1 Definition and Examples
First we will define the notions of string verifiable and strongly string verifiable. We
also give some examples.
Notation 2.3.1. Let Fo, F1,... be a computable enumeration of the finite subsets of 2<,
that is, for any n, a c F, b,((ao)) = 1, where bn is the binary expression for the
natural number n. Let E denote the family of finite sequences of positive integers of even
length. Let Po = [To], P = [Ti],... be some computable enumeration of the HI classes in
2".
Definition 2.3.2 (String Verifiability). (i) A string function is a computable function
f : 2
(ii) A family 'H of trees (or, more generally, of subsets of 2<") is string verifiable if there
is a string function h : 2<  E so that for all T, TE GH if and only if the following
condition is satisfied for all a c T, where h(ao) = (mi, m2,..., M2n) and Di = F,m for
i = 1,..., n: There exists i < n such that D2i+l C T and T n D2i+2 = 0 that is,
[T] c S(D2i+1, D2i+2) (the family of separating sets of D2i+1 and D2i+2).
Remark 2.3.3. Note that the family of trees itself is string verifiable among the family of
all subsets of 2<", via the function h(a) =(a, b), where Fa = {r : T E a} and Fb 0.
Example 2.3.1. (a) The Homogeneous Trees. A tree T is said to be homogeneous if
(Va7, r T)[7 = I => (Vi)(,i i T ^i E T)1.
Define the string verification function h as follows. Let A1, A2, A3, A4 enumerate
P({0, 1}<") and let B1,..., B21 1 enumerate the strings of length a. Let h(a)
enumerate in order the set of m2(j,k)+l and m2(j,k)+2 for 1 < j < 4 and 1 < k < 21'1,
where
F2j, = {7 i i Aj, e BkJ
and
F,n.+2 = {r : } U { i B : i Aj, re Bk.
That is, h(a) verifies that T is homogeneous by selecting the unique set Bk {= T
la & 7 E T} and the unique set Aj such that for 7 E Bk, Ti T, e= i E Aj.
(b) The Extendible Trees. Recall that a closed set P is decidable if the P = [T] for some
computable tree T without dead ends. For the purpose of string verification, let us
w that a tree T is extendible if T has no dead ends. This means that that for any
a T, either a0 E T or a 1 e T. In general, Ext(T) = {a I(a) n [T] / 0}
is the set of extendible nodes of T and T is extendible if and only if T = Ext(T).
Thus we let h(a) = (mi, m2, m3, m4), such that Frn {1, 0}, F = Fm4 0 and
Fm3 = {^1}. That is, h verifies that T has no dead ends by either showing that
aO T if F,, C T or that al E T if F3 C T.
Definition 2.3.4. (a) A 11 class P satisfies a finite set of relations 'i C P(2<") (i < n)
if there is a computable tree T such that P = [T] and 7H(T) for each i < n.
(b) A II class P strongly satisfies a finite set of relations Hit C P(2<") (i < n) if there is
a primitive recursive tree T such that P = [T] and 'H(T) for each i < n).
Definition 2.3.5. A family F of classes is [strongly] string verifiable (s.v.) if there is
some finite set of string verifiable relations so that: P E F if P [strongly] satisfies these
relations.
Remark 2.3.6. Note that any string verifiable family of trees contains the empty tree,
so that any string verifiable family of II classes contains the empty class. If P 0, then
P = [T] if and only if T is finite, so any tree T with P = [T] is primitive recursive. So any
strongly string verifiable family also contains the empty class.
2.3.2 Computable and Effective Numberings
We now demonstrate that strongly string verifiable and string verifiable families
possess computable and effective numberings, respectively.
Theorem 2.3.7. (a) Any strongly stringverifiable family of II classes has a com
putable numbering.
(b) Any stringverifiable family of HI classes has an effective numbering.
Proof. Suppose F is a [strongly] stringverifiable family of II classes satisfying string
verifiable (tree) relations 'Ho, ...,'H 9, with corresponding string functions ho, hi,..., h,.
For part (a), let the standard computable enumeration of the HI classes in 2" be given
by P, = [T,], where the sequence T, is uniformly primitive recursive (for example, the
numbering ,'_. given in section 2.2.1). We will define a uniformly computable sequence
S, of trees such that the sequence Q, = [S,] enumerates exactly the family of II classes
strongly satisfying 7o, '1, ..., K,.
For any a e {0, 1}', we determine whether a S, as follows. First check that
a E T,. If so, for each rT E {0, 1} and each i < m, compute hi() = (D, D2,..., D2j)
and determine whether there exists i < j such that D2j+1 C Te and D2j+2 n T, = 0. This
process is computable since each De is a canonical finite set. If the answer is yes, for every
' E {0, 1}', then a E Se and otherwise, a r Se. It is clear that if Te satisfies all of the
relations 7Ho, '1, ..., Hm,, then T, = S,. It follows that every II class in F occurs in the
enumeration Q = [Se]. On the other hand, if T, fails any of the relations, then S, is a
finite set and Q = 0. By assumption, Q, E F in this case as well, so that the sequence
{Q, : e < u} enumerates exactly the family F, as desired.
For part (b), let PE = (e) = T, the uniformly HI enumeration which has the
property that every computable tree occurs in the list {T, : e < u}. We need to do the
string verification in a HI fashion and in particular to check that D2j+2 n T, = 0, which
appears to be E. However, we can simply check that, if p E D2j+2 and e(p) 1, then
e(p) 0. Then the sequence S, is uniformly HI and, if T, has characteristic function ,
and satisfies the string relations, it follows that S, Te. E
We now provide some additional examples of string and strongly string verifiable
classes.
Definition 2.3.8. (i) A fIo class P is strongly decidable if there is a primitive recursive
tree T with no dead ends such that P = [T].
(ii) A TII class P is strongly homogeneous if there is a homogeneous primitive recursive
tree T with no dead ends such that P = [T].
From definition 2.3.8, we immediately have the following corollary to Theorem 2.3.7
Corollary 2.3.9. (a) The family of decidable Hi classes in 2" has an effective num
bering and the family of strongly decidable II classes in 2" has a computable
numbering.
(b) The family of homogeneous FII classes in 2" has an effective numbering and the
family of strongly homogeneous II classes in 2" has a computable numbering.
In the following section we will provide results which demonstrate that decidable
classes in fact have in fact a 11 computable numbering (see Corollary 2.4.2.1). We provide
an alternate explicit proof in section 2.4.2.1. For homogeneous classes, a different approach
is needed to demonstrate that they possess a 11 computable numbering and we do this in
section 2.4.1.
2.3.3 Families Containing the Clopen Classes
In this section we consider stringverifiable families of classes that contain all clopen
classes. We first improve Theorem 2.3.7 to obtain a computable numbering of any string
verifiable family which includes the clopen sets.
Theorem 2.3.10. If F is any stringverifiable family of FII classes, then there is a
computable numbering of C U F.
Proof. We modify the proof of Theorem 2.3.7 so that when the stringverifiable relations
fail, we extend all nodes rather than making them dead ends. Once again, the construc
tion is based on the enumeration e, of the partial computable functions. The construction
is in stages, where at stage s we will have
ne,s = max{n : (Va e {0, 1}")Qe,8((a)) 1},
J,s = {a E {0, 1}ne" : e,s((a)) = I},
and
Qe,s = J es [Se,1,.
Then Qe = n, Qe,,s [Se] will be the desired numbering. To ensure that this numbering is
computable, we will determine whether a E Se at stage a .
For this argument, we assume that Qe(0) = 1 for all e.
Construction. At stage 0 we have ne,o = 0, Je,o = Se,o = {0} and Qe,o = 2w.
At stage s + 1, we check to see whether Qe,s,+((a)) 1 for all a E {0, 1}"'+1. If not,
then ne,s+ = ne,s, Je,s+1 Js and Se,s+1 = Se, U {ai : a S,s, i = 0, 1}. If so, then we
check to see that Qes+1 is the characteristic function of a tree on {0, 1}r"0,+1 and we verify
the string relations up to {0, l}"je+1. If this verification fails, then again ne,s,+ = ne,s and
Je,s+ Je,s. In this case, verification will also fail at all future stages, so that Qe = Qe,s is
a clopen set.
If the tree and stringverifications succeed, then ne,s + = ne,s + 1, so that Je,s+1 c
{0, 1}t"I,+1 and Qe,s+1 change as indicated above. In this case,
Se,s+ = Se,s U {a {0, 1}s+1 : a [(ne,s+l) E Je,s+l}.
If Qe is the characteristic function of the computable tree Te, and if P, = [Te] E F,
then it follows from the construction that Q = Pe, so that Qe E F and furthermore, any
II class P, E F will thereby occur in the numbering. Otherwise, the construction will
make Qe a clopen set. O
Corollary 2.3.11. For any string verifiable family F of II classes, there a 11 computable
numbering of C U F.
Proof. Let F be a string verifiable family. Then there is a computable numbering
of C U F by Theorem 2.3.10. It then follows from Theorem 2.2.7 that there is a 11
computable numbering of C U F. O
Corollary 2.3.12. There a 11 computable numbering of any string verifiable family of
II classes containing all clopen classes.
Corollary 2.3.13. There a 11 computable numbering of the decidable II classes.
2.4 Named Families of II Classes
It this section, we otain numbering results for various named families that commonly
occur in the literature. In the first two subsections (2.4.1, 2.4.2), we expand upon the
results from section 2.3 and we obtain positive numbering results for the homogeneous
and decidable classes. In the two subsections (2.4.3, 2.4.4) that follow these, we obtain
negative results for the thin, perfect thin, small, very small, and nondecidable classes.
2.4.1 Homogeneous Classes
Homogeneous II classes are a stringverifiable family of II classes. Consequently,
by Corollary 2.3.9, they possess an effective numbering. Corollary 2.3.12 falls short
of demonstrating that they possess a computable numbering, as clopen sets are not
necessarily homogeneous. We provide the necessary argument in Theorem 2.4.3 and
show, in fact, that a computable injective numbering exists. We first provide an alternate
characterization of the homogeneous classes; they may be viewed separating classes of c.e.
sets.
Definition 2.4.1. The separating class S(A, B) of two sets A, B C w is given by
S(A,B) {C C u : A C and B n C 0.
In what follows, S,(A, B) will denote the set of characteristic functions fc of C e
S(A, B).
Theorem 2.4.2. P C 2" is a nonempty homogeneous II class iff P = S,(A, B) for some
disjoint c.e. sets A and B.
Proof. () Suppose that P is a nonempty homogeneous class with homogeneous tree
Tp. We will show that P = S,(A,B) where A {= n : 0'0 g Tp} and B = {n : 0'1 t
Tp}.
To see that A and B are c.e., suppose that P = [T] with T computable. Then for
each i {0,1},, O' if Tp iff (3s > (n+ 1)) T n {a: lal = s and 0' i C a} =0.
To see that A and B are disjoint, suppose that n C A. Since P is not empty, there is
some x c P so that x[ n c Tp. Now 0`0 T Tp so that (x [ n) 0 Tp. Hence x(n) / 0
so that x(n) = 1. Therefore (x [ n) l E Tp. Since P is homogeneous, 0"o 1 Tp so that
n B. Hence AnB =0.
To see that S,(A, B) c P, take fc e Sc(A, B) for some C c S(A,B). Then,
nEC a ngB 0"^ leTp f (n+ 1) (fc [n)l eTp&
ngC a ngA 0"^ 0OeTp fcr(n+1) (fcrn) 0Tp
Since for arbitrary n, (fc [ (n + 1)) c Tp, it follows that fc E P.
To see that P C S,(A, B), suppose that x E P and let C = {n : x(n) = 1}. It is clear
that x is the characteristic function fc of C. It suffices to show that C e S(A, B) (so that
fc e Sc(A,B)). Suppose first that n e A. Then 0"0 ^ Tp so that (fc [ n)^0 i Tp.
Since x = fc C Tp, it follows that fc(n) = 1. Hence n E C and A c C. Now suppose
n E B. Then 0" 1 i Tp, so (fc [ n)l 1 Tp. As before, fc(n) = 0 so that n i C. Hence
CnB 0.
() Suppose A and B are disjoint c.e. sets and {A,}sE and {B,}s,, are stage
enumerations of A and B, respectively. Define P = n,[T,] where T, C 2<' is given by
a T, iff (i e A, (i) = 1) A (i B, (i) = 0). Then P is a HI class and
P = S,(A,B). Note that Tp = {a : (Vi < Ia)[(a(i) 0 A i A) V (a(i) =1 A i B)]}
is homogeneous. Furthermore, since A and B are disjoint, Sc(A, B) is nonempty. Hence
S,(A, B) is a nonempty homogeneous class. D
Theorem 2.4.3. There is an injective computable numbering of the homogeneous HI
classes.
Proof. We may obtain an injective numbering of all ordered tuples (A,, B,) of c.e. sets
by Theorem 2.2.4.2. Then by Theorem 2.4.2, P, = S(Ae, Be) is an injective numbering of
the homogeneous classes. Furthermore, S(A{, B,) = [T,] where
Ta The (Vn < a)[(n e Ae,e n a(n) = 1) & (n B,s (nc) = 0)]
This shows that the numbering is computable. O
2.4.2 Decidable Classes
In this section we provide an alternate proof of the existence of injective computable
numberings of the decidable classes (see Corolllary 2.3.13 for the first proof). We also
show that in any computable numbering Q of the decidable classes via trees, we can pro
vide a computable numbering Q of all the trees without dead ends that occur, along with
all clopen classes. Finally, we show that some decidable class in class in the numbering Q
must necessarily have, as a tree, dead ends throughout every occurance in the numbering.
2.4.2.1 An injective computable numbering (Alternate proof)
Since decidability is stringverifiable and every clopen set is decidable, it follows from
Corollary 2.3.12 that the decidable classes have a 11 computable numbering.
This result could not be obtained by using the standard numbering of the FII classes
and modifying each tree as it becomes known that is has a dead end. (For example,
simply extend each such node with, wi, all ones.) This is because, as a consequence to
the following theorem, P being decidable is insufficient to ensure that the unique tree
Tp without dead ends shows up in a computable tree numbering. The following is an
alternate, explicit proof.
Theorem 2.4.4. There is a 11 computable numbering of all decidable classes in 2".
Proof. Let e v We be the injective effective numbering of the computable sets as given
in the proof of Corollary 2.2.8. We will define an effective correspondence between these
sets and the nonempty computable trees without dead ends. We will do this through a
series of three onetoone correspondences namely the correspondences between (1) the
subsets of uw and 2', (2) 2" and 3', and (3) 3' and the nonempty trees without dead ends.
In a stage construction we will then define at stage (e, s) a tree Te,s based on We,s and the
correspondences. Letting T, = nsT,, we will obtain an injective computable numbering
e v4 [T,] of all nonempty decidable HII classes. Furthermore [T,] will correspond to We for
each e. Finally, by appending the empty class to the enumeration we obtain the desired
result. We now define the correspondences.
The onetoone correspondence between the subsets S of w and 2" is given as follows.
Each S corresponds to xs C 2" given by xs(i) = 1 iff i E S.
The onetoone correspondence between 2" and 3" is given as follows. Let x e 2"
and define ao 1i, al = 01, and a2 = 00. Then x corresponds to the unique sequence
(fX(i))iEW 3" where f, : 3 and x = af,(o)af,(1)af,(2) ..
The onetoone correspondence between the set of all nonempty trees T C 2<
without dead ends and 3' is given as follows. Let T be a tree without dead ends and let
ao = 0, oa, ... be an enumeration of the elements of T in order, first by length and then
lexicographically. We define g = gr e 3" by recursion as follows. For each n, define
g(n) = 2 if an^0 and ajn are both in T, g(n) = 1 if an 0 T and aj^l E T and
g(n) = 0 if a,~^0 E T and a,^l T. For each such g we now let T, denote the unique tree
without dead ends corresponding to g.
At stage (e, s), let Te,s be the clopen tree corresponding to We,s. Then, for fixed e,
since the elements of We are enumerated into the set in increasing order, for all s we have
that Te,s+i C Te,s and each We corresponds to Te = ,sTe,s. In fact, the finite sets precisely
correspond to the clopen trees. Now, e v4 We injectively numbers all computable sets.
Therefore {0} U {[Te]}eew is an 'injective computable numbering' of the decidable II
classes. E
2.4.2.2 Trees without dead ends: A numbering result
A class is decidable iff it is the set of infinite paths through computable trees without
dead ends. Given any computable numbering of the II classes via trees, this motivates
capturing, through some computable enumeration, those trees in the numbering that
have no dead ends. The following theorem demonstrates that this is possible injectively,
provided that all clopen trees are also included.
Theorem 2.4.5. Suppose e T> Te is a computable numbering of computable trees in 2<.
Then there is an injective computable numbering of trees consisting precisely of all the Ti
that have no dead ends along with all clopen trees.
Proof. Assume without loss of generality that {Tje}c, contains all clopen trees. We will
construct in stages, as in the terminology of Theorem 2.2.6, a sequence of follower trees
Si. At stage i we will ensure that we have i + 1 trees So, Si,..., Si, constructed up to
level 2', following trees T(so,k ),... T(s,ki) (ki c {m, n}) which are each pairwise distinct
at level 2i. Also, at stage i, initially some of the Si will have the status of being marked
(ki = m) in which case Si will continue to follow T(s,,m) forever. If not, then Si is not
marked (ki = n) and we determine for each i, if Si should be marked. If an Si needs to be
marked then we determine a tree T(s,,m) that it shall hereafter follow. Otherwise each Si
continues to follow T(ss,T) and the stage is complete.
Construction. Stage 0. Find the first tree T, such that T, n {0, 1}20 / 0, denote this
tree as T(so,), and define So = T(so,n) n {0, 1}20
Stage j+1. So,... S have already been constructed up to level 2i and are already
following trees T(So,k)), T(sj,kj). We perform the following two actions at this stage: (1)
Construct So,... ,Sj up to level 2J+l by determining the trees T(so,k+i), .. T(si,k+l,) they
shall follow, and (2) Construct a new tree Sj+I up to level 2j+.
Action (1). Let Uj+I = {(Si, kj) : kj = n and T(s,,kj) has dead ends at level 2J+1}. All
Si such that (Si, kj) g Uj+1 keep their status as marked or unmarked, so kj = kj+l, and
continue to follow T(Si,kj+i). Those Si such that (Si, kj) E Uj+~ will hereafter be marked
and will now follow the tree T(s\,m) = { : T C a or a C r for some r T(s,n) of length
2J}. Note that each marked Si follows a clopen tree T(Sr,m).
Action (2). Let (Sj+1, n) be the least i such that T, is distinct from all T(s,,kj+1)
(i < j) at level 2j+1 and such that T, has no dead ends. Define Sj+1 = T(s+ ,n) n {0 1} j+1
Verification. We now verify that: (i) For each i, limjT(S,,kj) = Si = T,, for some
T,, without dead ends, (ii) For all i, if Ti has no dead ends then there is a c such that
Ti = S, and (iii) i : j implies that Si / Sj.
VerY. ,,i."'. of (i). For all j, kj = n or kj = m. Fix i. By Action (2), at stage i,
(Si, ki) = (Si, n). By Action (1), ke = k+I = n for all > i if Si is never marked. If Si is
marked at stage r > i, then for all s > r, k, = k,+i = m. In either case limjo~Ikj so that
limj(Si, kj) converges to (Si, n) or (S, m). If it converges to (S, m) then Si never diverges
from following the clopen tree T(s,,m). Otherwise Si is never marked and continually
follows T(s,n). Since it is never marked it means that T(s,n) never has dead ends up to
level 2', for all r > i. So T(s,,n) is a tree without dead ends. Either way limjT(Si,kj)l= TR
for some tree T,, without dead ends. Now for all n, Si n {0, 1}<" T(S,,k,) n {0, 1} t and
T(S,k.) C T(sr,k(+,). Therefore Si limjT(s,,k) T.
Verl./ ,'I. i, of (ii). Let Ti be a tree without dead ends. There are two cases. If there
is a stage j and a c such that Ti = T(s,,m) at stage j, then by the construction Ti = Sc. If
not, let i equal the least k such that Tk Ti. Let j be large enough so that T differs from
T, at level 2i for all e < i. If at stage j there already exists a c such that T7 = T(sc,n) then
clearly Ti Sc. Otherwise, by Action (2), some tree Sc follows T7 by no later than stage
j+i.
V1er 'flr. i', of (iii). By Action (2), Si is distinct from all Sj (j < i) at level 2' and
from all Sj (j > i) at level 2j. So S, / Sj if i / j. O
2.4.2.3 Trees with dead ends: A necessity
In any computable numbering Q of Io classes via trees, some decidable class must
necessarily have, as a tree, dead ends throughout every occurance in the numbering (see
Corollary 2.4.7). Two different proofs of this fact may be obtained from Theorems 2.4.6
and 2.4.8 below.
Theorem 2.4.6. In any computable numbering of computable trees in 2<' there is a
computable tree without dead ends outside the image of the numbering.
Proof. Let {T, : e < w} be a uniformly computable sequence of trees. Now use a
diagonalization argument to construct a tree T such that for all n, T n {0, 1}"+1 /
T, n {0, 1}+1, as follows. At stage 0 let T n {0, 1}o {0}. At stage n + 1 we are given
T n {0, 1}" / 0. Therefore there are at least 2 subtrees of {0, 1}"+1 without dead ends
extending T n {0, 1}. Define T n {0, 1}"+1 to be an extension which is different from
T, n {0, 1})+1. O
Corollary 2.4.7. For any computable numbering P, = [T,] of the 10 classes in 2", there
is a decidable II class P such that P / [T,] for any Te without dead ends.
Proof. Let P = [T] where T is the computable tree without dead ends provided by
Theorem 2.4.6. Suppose that P = [T,] for some e. Since T has no dead ends, it follows
that T = Tp and if T, also had no dead ends, then T, = Tp = T. But by the construction,
T n {0, 1}e+1 / T, n {0, 1}e+1, so that T / T,. O
It follows from this corollary that in the standard numbering, {e : T, has no dead
ends} / {e : P, = [T,] is decidable}. In fact both have distinct complexities. By
Konig's Lemma, Ext(P,) = {a e 2<" : I(a) n P, / 0} is nI. So {e : T has no dead
ends} = {e : T, = Ext(P,)} is II. However, {e : P, is decidable} = {e : = [T] for some
computable T without dead ends} = {e : (3a) ~, is a characteristic function for Ext(P,)} is
E. An alternate proof of Corollary 2.4.7 is as a corollary of the following.
Theorem 2.4.8. For any acceptable numbering i of the 10 classes,
{e : i(e) is decidable} is E complete.
Proof. It suffices to prove this for the standard numbering (b2). We will make use of
the wellknown [86] E completeness of {e : We is computable}. It is easy to see that
{e : Q2(e) is decidable} is ZE. For the completeness, define the uniformly computable trees
Tf(e) so that
(i) 0" e Tf(e) for all n;
(ii) 0'1" E Tf () a n We,,.
It follows that 0O1 e Ext(Tf(e)) U n We, so that if /f(e) is decidable, then We
is computable. On the other hand, Ext(Tf(e)) ={0" : n e } U {O0') : s e t, n i We}, so
that if We is computable, then (f (e)) is decidable. Thus We is computable if and only if
b(f (e)) is decidable. O
Note that in [23], a II class P = [Te] in the standard numbering is said to be
decidable if T, has no dead ends, which we now see is probably not the right approach.
2.4.3 Thin and Perfect Thin Classes
In the literature, a MartinPour El theory is a consistent c.e. propositional theory
with additional 'thinness' conditions. The conditions imposed have varied depending
upon the context and motivation of the authors, but include: (1) few c.e. extensions,
(2) essentially undecidable, and (3) wellgenerated. Some authors have chosen to only
impose (1) [21], while others (1) and (2) [19], [24], and finally others (1), (2), and (3)
[34], [40], [29]. The complete consistent extensions of these theories correspond to thin,
perfect thin (or equivalently, special thin [19]), and homogenous thin classes, respectively.
This section is devoted towards demonstrating the nonexistence of computable numbering
of the first two cases by modifying the classical MartinPour El construction of a perfect
thin class. Recently Solomon [89] also modified this theorem to construct a homogeneous
thin class and therefore we conjecture that no computable numberings exist for these
classes.
2.4.3.1 The MartinPour El Construction
Recall that a II class P is thin if for every II subclass Q C P, there is a clopen set U
such that Q = U P. It is perfect iff it has no isolated points.
A perfect class may be defined by a function g : 2<" 2< such that for all a, , a K
'r implies g(a) C r; let us iv that g is extension preserving. Let G(x) = U g(x [n). Then
G(2") is a perfect class. If g is defined in uniformly computable, extensionpreserving
stages g8 (with corresponding G : 2" 2'), so that gs(a) C g+,1(a), then we have
G(2") = neG(2"), so that G(2") is a nI class.
Theorem 2.4.9 (\! ,itinPourEl). For any computable extensionpreserving function
g : 2<' 2<', there exists a perfect thin II class P C G(2").
Proof. Let {P, = [T,] : e E w} be the standard numbering of the II classes and
{(1 : e E w} be the standard numbering {0, 1}valued partial computable functions.
We will construct a computable tree S, corresponding II class P = [S], and a surjective
homeomorphism F : 2" P. F will be constructed by means of an extensionpreserving
map f : 2<" S, with corresponding map F : 2" 2" defined by F(x) = UJ f(x [ n).
We will define f in stages to obtain uniformly computable, extensionpreserving functions
f, so that f = limsf. To ensure that P is thin, we will meet the following requirement for
each e:
Thin(e): (Va E {0, 1}e+l)(V) [(f(a) E Te A a C r) f(r) E Te]
To see that Thin(e) makes P thin, let U {I(f (a)) : 1a = e + 1 & f(a) e T,} and observe
that if P, C P, then P, Pn U.
Construction. Let fo = g. At stage t + 1, we define ft+l as follows. Look for
e < t + 1, r {0, l}e+1, and Tr a with r
If no such e, a, and r exist, then ft+i = ft. Otherwise take the least such e and the
lexicographically least a and r for that e. For all p E 2'<, let ft+l(a p) = ft(>'p); for
p C a (with p / a) or p incomparable with a, let ft+i(p) ft(p).
Verification. It is easy to see by induction on a that for each a, f,(a) converges
to a limit f(a). Then by induction on e, each requirement Thin(e) is satisfied. To see
that f is injective, suppose towards a contradiction that f(a) = f(r) for a / T. By
the construction, a and r must be comparable. Assume, without loss of generality, that
r = a p (p / 0). By induction it is clear that for all t, ft(a) / ft(a p) = ft(). Let
F,(X) = Unf,(r [n), so that P = nF,(2). Since fo = it follows that P C G(2"). O
2.4.3.2 Nonexistence of computable numberings
We now modify the MartinPour El construction to obtain, as a corollary, non
existence of computable numbering for thin and perfect thin classes.
Theorem 2.4.10. Any computable numbering of FIl classes in 2" of Lebesgue measure
zero omits some perfect thin class from its image.
Proof. Let P, = [T,], where {T, : e E u} is uniformly computable. We will construct
a computable extensionpreserving function g 2<' 2<' such that for all e and
all a e {0, 1}e+l, g(a) T,. Then letting G(x) = U,g(x [ n) we will ensure that
G(2") n P, = 0. Replacing fo by g in Theorem 2.4.9, we obtain a perfect thin class P such
that P n P, 0 (and hence certainly P / P,), for all e.
We define g : 2<" 2< recursively, as follows. Define g(0) = 0. Then for each
a E {0, 1}t, compute the shortest and lexicographically least extension r of g(a) such that
7T Tg. Since [T,] has measure zero, it is nowhere dense and thus such a T alv7v exists.
Then let g(a^i)= r^i for i E {0, 1}. O
Corollary 2.4.11. There is no computable numbering of all thin or of all perfect thin II
classes.
Proof. All thin classes have Lebesgue measure zero [85]. Therefore if e v> P, were a
numbering of (perfect) thin classes then Theorem 2.4.10 would provide a (perfect) thin
class P such that P / P, for all e, a contradiction. O
2.4.4 Small, Very Small, and Nondecidable Classes
Binns defined in [11] the notions of small and very small classes as a means of
guaranteeing incompleteness in the lattice of the Medvedev and Muchnik degrees of
subsets of uw. A nonempty II class P is small if there is no computable function I) such
that for all n, ITp n 2(")l n. Let I(n) be the least k such that ITp n ok' > n. A
nonempty II class P is very small if the function T dominates every computable function
g; that is, I(x) > g(x) for all but finitely many x. Let A be a coinfinite c.e. set, i
A = {ao < al < ...}. Recall that A is i,;' ":',,il. if there is no computable function f
such that f(n) > aT for all n and it is dense simple if n v> an dominates every computable
function. In this section we will use these sets to show that no effective numbering exists
for the small, very small, or decidable classes.
2.4.4.1 Numberings and high/noncomputable sets
En route to demonstrating our theorem, we now proceed show that there are no
effective numberings of the high or of the noncomputable sets. As we shall eventually
characterize small an very small classes in terms the degrees of these sets, these results will
be crucial to our argument.
First, we modify Shoenfield's Thickness Lemma [86, p. 131]. Some definitions are
needed. For B C w, let B[y] {(y, z) B : z e w} and that B is piecewise computable
if B[] is computable for all y. For B C A C u, we iv that B is a thick subset of A if for
all y, Bly \ AM is finite.
Lemma 2.4.12 (Thickness Lemma). For any uniformly c.e. sequence {Wi : i E u} of
noncomputable c.e. sets and any piecewise computable c.e. set B, there is a thick c.e.
subset A of B so that Wn %r A for all n.
Proof. The proof as in [86] is modified to ensure that the length and restraint functions
and the requirements incorporate the pair (i, k) in place of the single argument i to make
the argument go through with each Wi in conjuction with each functional Tk. E
We obtain the following corollary.
Corollary 2.4.13. For any uniformly c.e. sequence {W,1 n E u} of noncomputable c.e.
sets, there is a high c.e. set A such that for all i, Wi ,T A.
Proof. This follows from the modified thickness lemma above by the same argument
found in [86, p. 133]. O
Corollary 2.4.14. (a) There is no uniformly c.e. numbering of all high c.e. sets.
(a) There is no uniformly c.e. numbering of all noncomputable c.e. sets.
In fact, it follows that there is no uniformly c.e. numbering of the high or noncom
putable c.e. degrees.
2.4.4.2 Nonexistence of effective numberings
We now proceed characterize the small and very small classes in term of the noncom
putable degrees and the high degrees, respectively. The degree of a II class P is defined to
be the degree of Tp and is thus alviv a c.e. degree (since Tp is a coc.e. set).
We will use the following two classic results.
(1) \! irtin] Any high degree contains a maximal (and hence dense simple) set [86,
pp. 211217].
(2) [Dekker] Any noncomputable c.e. degree contains a hypersimple set [86, p. 81].
Proposition 2.4.15. A c.e. degree is high if and only if it contains a very small II class
PC 2".
Proof. () Suppose a is high, and let A E a be a maximal set, and let p be the
principal function of u A, so that p dominates every computable function. Now let
PA = {0} U {0(t10 : n A}. Then PA is a io class and for each n, the least k such that
ITp n {0, 1}' > k is precisely p(n) + 1 for n > 0 and hence dominates every computable
function.
() Let a be a c.e. degree and suppose that Tp Ea for some very small P. Then the
function f(n) = (least k)[ITp n {0, 1}k > n], which dominates every computable function,
is computable from Tp. It follows from Martin's Theorem [86, p. 208] that Tp is high.
Proposition 2.4.16. A c.e. degree is noncomputable if and only if it contains an infinite,
small II P C 2".
Proof. () Suppose a is a noncomputable c.e. degree, let A E a be hypersimple, and p
be the principal function of u A, so that p is not dominated by any computable function.
Then the ni class PA as defined in the proof of Proposition 2.4.15 will have degree a and
will be small.
(<) Suppose that P is an infinite II0 class and Tp is computable. Then the function
g(n) = (least k)[ITp n {0, 1} > n] is computable and it follows that P is not small. O
Theorem 2.4.17. There is no effective (i.e. ni ) numbering of all nondecidable, of all
infinite small, or of all very small n0 classes in 2".
Proof. Suppose, towards a contradiction, that {Q, = [T,] : n E u} is an effective
numbering of HI classes such that each Q, is nondecidable. Then W,I {(a) : a
Ext(T,)} is a uniformly c.e. numbering of noncomputable c.e. sets. By Corollary 2.4.13,
there is a high c.e. set A such that for all n, Wn, T A. Therefore A is a high degree that
contains a (very) small class not amongst the Qi, a contradiction. O
CHAPTER 3
RANDOM CLOSED SETS
The following chapter is joint work with George Barpalias, Douglas Cenzer, Seyyed
Dashti, and Rebecca Weber and appears in the Journal for Logic and Computation (no.
17, 2007, pages 10411062) as an article entitled Algorithmic Randomness of Closed
Sets [7]. A preliminary version of this research was originally presented at the Com
putability in Europe Conference in Swansea, Wales in 2006 by D. Cenzer. This prelimi
nary work was published in the referred conference proceedings as Random Closed Sets
(P. Brodhead, D. Cenzer, and S. Dashti) in Proceedings of CIE 2006: Logical Approaches
to Computational Barriers, (A. Beckmann, U. Berger, B. Loewe, and J. Tucker, eds.),
Springer Lecture Notes in Computer Science, Vol. 3988 (2006), pages 5564 [14].
Portions of this work were also presented by P. Brodhead at the Greater Boston
Logic Conference (\! ly 2006, Boston, MA), the 2006 SACNAS Conference (October
2006, Tampa, FL), the AMS Fall 2006 Eastern Sectional Meeting (October 2006, Storrs,
CT), the Conference on Logic, Computability, and Randomness (January 2007, Buenos
Aires, Argentina), and the Workshop on Computability and Randomness (December 2007,
Auckland, New Zealand).
Portions of this work were also presented by D. Cenzer at a randomness workshop
at the American Institute of Mathematics (August 2006, Palo, Alto, CA). R. Weber also
presented portions of this material at a randomness workshop at the University of Chicago
(September 2007, Chicago, IL).
3.1 Overview
The literature abounds with results in algorithmic randomess as pertaining to reals
over a finite alphabet, especially within the last few years. Little is known, or even
developed, however, with respect to randomness for closed sets of binary reals. This
chapter is a first approach in this direction.
In this chapter, we consider a notion of effective (i.e. algorithmic) randomness on
the space C of nonempty closed subsets P of 2N; to accomplish this task, we will need use
the definition and machinery of effective randomness for reals, since, through appropriate
coding of closed sets, we will define a closed set to be random iff its code, as a real, is
random. (In fact, later in Chapter 4, we will approach a definition of randomness for
continuous functions in a similar fashion.) Consequently we begin this chapter with an
introduction to algorithmic randomness, including a brief historical background.
More specifically, this chapter is organized as follows. In Section 3.2, we provide an
introduction to algorithmic randomness for reals over a finite alphabet. In Section 3.3,
we give a probability measure and define a version of the MartinL6f Test for closed sets,
leading to a definition of randomness for closed sets. In Section 3.4, we tackle the question
of which types of elements necessarily belong, or do not b. 1.in. to random closed sets. For
instance, every random closed set contains random and nonrandom elements, but no n
c.e. elements. In Section 3.5, we show that random closed sets have measure zero and box
dimension log2 4. In Sections 3.6 3.7, we explore alternate notions for randomness, such as
the problem of compressibility of trees. Finally, in Section 3.8, we consider the problem of
when a randomly chosen closed set meets a closed Q; this is the study of capacities.
3.2 Effective Randomness of Reals
In this section, we present a basic introduction, including a brief historical back
ground, for randomness of reals over a finite alphabet.
3.2.1 Introduction
The study of algorithmic randomness has been of great interest in recent years. The
basic problem is to quantify the randomness of a single real number. Early in the last
century, von Mises [94] 1i::. 1. I that a random real should obey reasonable statistical
tests, such as having a roughly equal number of zeroes and ones of the first n bits, in the
limit. Thus a random real would be stochastic in modern parlance. If one considers only
computable tests, then there are countably many and one can construct a real satisfying all
tests.
An early approach to randomness was through betting. Effective betting on a random
sequence should not allow one's capital to grow unboundedly. The betting strategies used
are constructive martingales, introduced by Ville [93] and implicit in the work of Levy [65],
which represent fair doubleornothing gambling.
MartinL6f [69] observed that stochastic properties could be viewed as special kinds
of measure zero sets and defined a random real as one which avoids certain effectively
presented measure zero sets; see Section 3.2.4. At the same time Kolmogorov [55] defined
a notion of randomness for finite strings based on the concept of i...iii., ..i.;././ A
stronger notion of prefixfree complexity was developed by Levin [64], Gacs [48] and
C'I 1 i [27] and extended to infinite words.
In the following sections, we formalize the notions of constructive martingale ran
domness, MartinL6f randomness, and prefixfree randomness. After their entry into the
literature, Schnorr later proved [83] that all of these notions are equivlant; this is a funda
mental result in the theory of algorithmic randomness. While these definitions and results
are usually given for binary strings and sequences, they carry over to kary strings and
sequences as well. See, for example, Calude [17, 18], or Section 3.2.4 below, where we do
this for the MartinL6f definition of randomness.
3.2.2 Constructive Martingale Randomness
The betting approach to randomness is formalized as follows.
Definition 3.2.1 (Ville [93]). (i) A martingale is a function m : k< [0, oo) such that
for all a E k<',
k1
Mr(7) = m(17i).
i=0
(ii) A martingale m succeeds on X E kN if
lim sup d(X [n) = oc.
nToo
That is, the betting strategy results in an unbounded amount of money made on the
kary infinite sequence X.
(iii) The success set of m is the set S~ [m] of all sequences on which m succeeds.
That is, a martingale on 2<" is the capital function of a fair doubleornothing betting
strategy. When working on 3<" the strategy is tripleornothing.
Definition 3.2.2. A martingale m is constructive (or effective, or c.e.) if it is lower
semicomputable; that is, if there is a computable function m : k x N Q such that
(i) for all a and t, m(ao, t) < mat(a, t + 1) < m(a), and
(ii) for all o, limtoo rh(o, t) = m(a).
In other words, m(w) is approximated from below by rationals uniformly in w.
Definition 3.2.3 (Constructive Martingale Randomness). A sequence in kV is construc
tive martingale random if no constructive martingale succeeds on it.
Comment 3.2.4 (Nonmonotonic Martingales). Some flexibility may be gained by
also considering nonmonotonic martingales; i.e., martingales which bet on the bits of a
sequence out of order. While for a monotonic martingale only the amount of the next
bet is determined from the bits seen previously, for a nonmonotonic martingale both the
amount and the location of the next bet are determined from the bits seen previously
(the next bit may precede them, follow them, or lie in the middle). These martingales
must obey two rules: the standard fairbetting rule that monotonic martingales obey, and
the rule that they never bet on the same bit twice. We refer the reader to Downey and
Hirschfeldt [37] for the formal definition.
Although a priori allowing nonmonotonic martingales strengthens the notion of
randomness, since more strategies must be defeated, in fact in the c.e. case they are
equivalent. Muchnik, Semenov, and Uspensky [71] (Theorem 8.9) show that MLrandom
sequences defeat all computable nonmonotonic martingales (in fact they show this with
respect to general measures, not just the cointoss measure). The proof does not depend
on the computability of the martingale, however; the martingale is used to define a
MartinLof test which may be enumerated equally well alongside the enumeration of the
martingale. Therefore, as defeating all c.e. nonmonotonic martingales is clearly sufficient
to be MLrandom, the two are equivalent.
3.2.3 Prefixfree Randomness
Prefixfree randomness for reals is defined as follows. A Turing machine M which
takes inputs from A*, where A is a finite alphabet, is called prefixfree if it has prefixfree
domain dom(M); that is, if a 1 r are strings in dom(M), then a must equal r. For any
finite string r, the prefixfree ,.. i/l. iH.;, of 'r with respect to M is
KM() = min{ 7, oO : M(a) = r}.
There is a universal prefixfree function U such that, for any prefixfree M, there is a
constant c such that for all r
Ku(7) KM(r) + c.
We let K(r) = Ku(7) and call it the prefixfree .i'.i ili, of r.
Definition 3.2.5 (Prefixfree Randomness). x E {0, 1}" is called prefixfree random if
there is a constant c such that K(x [n) > n c for all n.
This latter inequality means that the initial segments of x are not compressible.
3.2.4 MartinLif (n)randomness
According to the MartinLof definition of randomness, a random real must avoid
certain effectively presented measure zero sets. Inherent in the definition, therefore, is
the chosen measure being used. Fix an alphabet k= {0, ,..., k 1}. We present the
definition of a general probability measure on k", as well as different named measure
types. Typically, however, we will make use of the alphabets {0, 1} or {0, 1, 2}.
Definition 3.2.6 (General Probability Measures). Let f : {0, 1, 2}*  [0, 1] be a function
such that f(0) = 1 and f(a) = i=0,1,2 f(a^ i) for all a. The fprobability measure vf is
defined so that the vfmeasure of the interval [a] is such that vf([a]) = f(a).
Definition 3.2.7 (Different Measures Types). Let f and vf be as in Definition 3.2.6.
(i) vf is computable if f is computable.
(ii) vf is nonatomic, or continuous, if for all x E 3, vf({x}) = 0.
(iii) f and vf are bounded if (3b, c E (0, l))(Va)(Vi)[b f(a) < f(a i) < c f(a)].
[Note that any bounded measure is also continuous.]
(iv) vf is regular if (3bo, bi, b2 (0, 1))(Va) f(a^i) = be f(a)
Now fix a probability measure p. In the literature, p is most typically the Lebesgue
measure.
Definition 3.2.8. A real x E k" is MartinLif random if for every effective sequence
S, S2,... of c.e. open sets with p(S,) < 2", x i n, S,.
The latter condition is equivalent to the condition we get if we replace 2" with q,
for a computable sequence (qi) of rationals such that limit qi = 0. We can also consider an
extended definition of MartinLif randomness, in terms of E sets.
Definition 3.2.9. (i) A E test is a computable collection {V, : n E 2N} of E classes
such that p(Vk) < 2k;
(ii) A real a is E random or nrandom if and only if it passes all E tests (i.e., if
{V,: n E 2N} is a computable collection of Z classes such that p(Vk) < 2k, then
a nn>oVl).
Thus 1random reals are just MartinLbf random reals. See [36] for details on random
and nrandom reals. Kurtz [59] and Kautz [54] proved the following result. Let .I' denote
the nth jump of 0.
Theorem 3.2.10. Let q be a rational number.
(i) For each E class S we can uniformly compute from q and a E index for S, the
index of a EY' 1) class U D S such that U is an open E class and p(U) P(S) < q.
(ii) For each 11 class T, we can uniformly compute from q and a II index for T, the
index of a "II(') class V D T such that V is a closed 1 class and p(V) p(T) < q.
(iii) For each EO class S, we can uniformly compute from q, and a EY index for S and an
oracle for .I ', the index of a I1n_ class V C S such that V is a closed nI1_ class
and p(S) p(V) < q. Moreover, if p(S) is a real computable from .i 1) then the
index for V can be found computably from 1)
(iv) For each 11 class T, we can uniformly compute from q and 110 index for T and an
oracle for .I ', the index of a YZ_1 class U C T such that U is an open ZE_1 class
and p(T) p(U) < q. Moreover, if p(S) is a real computable from 1), then the
index for U can be found computably from 1
Comment 3.2.11. It follows that a real is n + 1random if and only if it is 1random
relative to .
Theorem 3.2.12 (van Lambalgen [92]). The following are equivalent.
1. A B is nrandom.
2. A is nrandom and B is nArandom.
3. B is nrandom and A is nBrandom.
4. A is nBrandom and B is nArandom.
3.3 MartinLMf Randomness of Closed Sets
In this section we define a measure on the space C of nonempty closed subsets of 2"
and use this to define the notion of randomness for closed sets. We then obtain several
properties of random closed sets.
3.3.1 The HitorMiss Topology on C
The standard (hitormiss) topology on C has as a subbasis, the following two types
of sets, where Q is any closed set: V(Q) = {K : K n Q / 0}; W(Q) = {K : K C Q}. A
basis for the hitormiss topology, then, is formed by taking finite intersections of these.
We now consider a refinement of the subbasis sets and obtain a basis for the Borel
sets. We will use the following notation. T" denotes the set T n {0, 1}", and T4" denotes
the set Tn {0, 1}i
Definition 3.3.1. For any tree S and any n, define
C,(S) = {Q C : S = TQ}
That is, C,(S) is the set of closed sets Q E C that agree with S up to level n.
Claim 3.3.2. A basis for the Borel sets is given by the agreement set A:
A = {C,(S) : S is a tree, n E uc}.
To see this, first note that any closed set [T] is the decreasing intersection of clopen
sets
[T'] : UJ{[] : Ee T}.
Therefore we may rewrite subbasis elements V([T]) and W([T]) as
V([T]) nV([T"]) (by definition) and
W([T])= nW([T"]) (by compactness)
But then,
V([T"]) U{n(S) : Sn T" / 0} and
W([T]) = U{ C(s) : s~ c T}
To see the first equality, for example, note that Q n [T"] / 0 if and only if Q E C,(S) for
some S with S n T" / 0. The latter equality holds similarly.
3.3.2 Toward a Measure
To define p(V(Q)) and p(W(Q)), for some fixed measure p and any closed set Q, it
suffices to define p(V(Q,)) and p(W(Q,)) for clopen sets Q, where Q = nQ,. We would
simply define p(V(Q)) =lim,,(V(Q,)) and p(W(Q)) =lim,,(W(Q,)). However, for any
clopen Q, W(Q) is the complement V(2N \ Q). Hence it furthermore suffices to define
p(V(Q)) for clopen sets to get a measure on C.
From the justication of Claim 3.3.2, the latter task may be accomplished by defining a
measure on all Borel basis elements, namely the agreement sets C,(S). To accomplish this,
in the following section we will encode all closed sets Q E C with a canonical code xQ e C.
Then using the Lebesgue measure on 3N, we will define a measure on the sets C,(S) which,
in fact, defines a measure on all closed sets.
3.3.3 Canonical Coding and Measure
The Canonical Coding. An effective onetoone correspondence between the space
C and the space 3" is defined as follows. Let a closed set Q be given and let T = TQ be the
tree without dead ends such that Q = [T]. Define the canonical code x = xQ E {0, 1, 2}"
for Q as follows. Let A = 0, ol, 2,... enumerate the elements of T in order, first by
length and then lexicographically. We now define x = xQ = XT by recursion as follows. For
each n, x(n) = 2 if a'O and a'l are both in T, x(n) = 1 if ajO T and al E T and
x(n) = 0 if a0 E T and a1 i T. For example, if Q = {0,1}", then xQ = (2,2,...) and
if Q = {y}, then XQ = y. Let Qx denote the unique closed set Q such that xQ = x.
Definition 3.3.3 (The Measure). Define the measure p* on C by
p*(X) = ({xQ Q: Q X}).
where p is the Lebesque measure (i.e. the regular measure Pd with bo = bl = b3 = (see
Definition 3.2.7) on 3.
Informally this means that given a E TQ, there is probability 1 that both a^0 E TQ
and atl TQ and, for i = 0, 1, there is probability that only aji E TQ. In particular,
this means that Q n I(a) / 0 implies that for i = 0, 1, Q n I(a i) / 0 with probability .
Comment 3.3.4. At this stage, we have fixed the uniform measure (i.e. all bi = )
towards defining randomness of closed sets. This allows us to more easily demonstrate the
validity of many results. Later, in Section 3.7.1, we will show that the results hold with
any regular measure. Proposition 3.3.5, however, demonstrates that the defined measure
on C, above, holds for any generalized probability measure Pd (see Definition 3.2.6).
Justification for the Coding. Let us also comment briefly on why some other
natural representations were rejected. Suppose first that we simply enumerate all strings
in {0, 1}* as ao, a ,... and then represent T by its characteristic function so that XT(n)
1 <= a,n T. Then in general a code x might not represent a tree. That is, once we
have (01) T we cannot later decide that (011) E T. Suppose then that we allow the
empty closed set by using codes x E {0, 1, 2, 3}* and modify our original definition as
follows. Let x(n) = i have the same definition as above for i < 2 but let x(n) = 3 mean
that neither aO 0 nor a l is in T. Informally, this would mean that for i = 0, 1, a E T
implies that a'i E T with probability '. The advantage here is that we can now represent
all trees. But this is also a disadvantage, since for a given closed set P, there are many
different trees T with P = [T]. The second problem with this approach is that we would
have [T] = 0 with positive probability. We briefly return to this subject in Section 3.7.2.
Now recall the definition of a general probability measure on 3N(Definition 3.2.6). Let
d : 10, 1,2}* [0,1] be a function such that d(0) = 1 and d(a) = E 0,1,2 d(a i) for all
a. Then pd([a]) is defined to be d(a). We may now define, for any such d, pi exactly as in
Definition 3.3.3. Furthermore p* we be deemed computable if d is computable.
Proposition 3.3.5. For any d, the measure p* is defined on all Borel sets in the hitor
miss topology on C. Furthermore, if d is computable, then p* is computable on the family
of clopen sets.
Proof. As discussed in section 3.3.2, it suffices to show that p*(C,(S)) is defined for all
C,(S) E A. Fix a tree S and suppose that {a C : r E S'} is ordered ao < ... < Jk(s,n),
first by length and then lexicographically. Then it is easy to see that
Q G C,(S) I ,[.,n] (k(S, n) + ) C ExQ
Consequently C,(S) = ncs V([al) is clopen in C. Furthermore,
p (C,(S))= ([x[s ] (k(S, n) + )]).
This also demonstrates the computability of p*. O
Definition 3.3.6 (Random Closed Sets). A closed set Q E C is (\! irtinLbf) random iff
its canonical code XQ is MartinL6f random.
This definition clearly relativizes to any oracle in accordance with the definitions of
relative randomness in the Cantor space. Since random reals exist, it follows that random
closed sets exist. Furthermore, there are Ao random reals, so we have the following.
Theorem 3.3.7. There exists a random closed set Q such that TQ is A. O
Note that if TQ is A, then Q must contain A elements (in particular the leftmost
path). Since there exist strong HII classes with no Ao elements, there are strong HII classes
Q such that TQ is not A. The following lemma will be needed throughout.
Lemma 3.3.8. For any Q C 2" which is either closed or open,
p*({P: P C Q}) < P(Q).
Proof. Let Pc(Q) denote {P : P C Q}. We first prove the result for nonempty
clopen sets U in place of Q by the following induction. Suppose U = U 1es I(a), where
S C {0, 1}. For n = 1, either p(U) =1 p*(Pc(U)) or p(U) = and p*(Pc(Q)) = .
For the induction step, let Si {a : ica E S}, let Ui = Ues, I(a), let ui = p(Ui) and let
= p*(Pc(Ui)), for i = 0, 1. Then considering the three cases in which S includes both
initial branches or just one, we calculate that
P*(Pc(U))
1
+ V + I XI).
3
Thus by induction we have
1
*(P c(U)) t (uo + u1 + uoui).
3
Now
2u0oul U2 + U U< + 1,
and therefore
1 1
P*(Pc(U)) < t(uo + uI + Uou1) < ( o + ui) p(U).
3 2
For a closed set Q, let Q = n U,, where U, is clopen and U,,+ c U, for all n. Then
P C Q if and only if P C U, for all n. Thus
so that
(Pc(Q)) =
Finally, for an open set Q, let Q
sets U,. Then, by compactness,
lim *(p'(UW,)) < lim i(Un) = i(Q).
n*oo n*oo
= UT Un be the union of an increasing sequence of clopen
'Pc(Q)= UJPc(Un),
so that
^(Pc(Q)) = lim P*(Pc(U.)) lim /1(U) P (Q).
This completes the proof of the lemma.
3.3.4 Ghost Coding
We wish now to introduce a second method of coding, the ghost coding. A ghost
code of Q is an infinite ternary string whose terms correspond to all nodes of 2< in
lexicographical order. The terms corresponding to the nodes of Q's tree (the "canonical
nodes") agree with the corresponding terms in the canonical code; the remaining ;!I I
nodes" may hold any values. Ghost codes are nonunique, and every closed set has a
nonrandom ghost code (if the closed set itself is random take the code with ghost nodes
all equal to zero, v). This method of coding is more convenient for some purposes;
for example, we will use it to show that if Qo, Qi are closed sets and Q = {^x : x E
Qo} U { lx : x E Q1}, Q is random if and only if the Qi are random relative to each other.
3.3.5 Coding Equivalance
The utility of the ghost codes rests on the following correspondence. Recall van
Lambalgen's theorem (Theorem 3.2.12).
Theorem 3.3.9. The canonical code of a closed set Q C 2" is random if and only if Q has
some random ghost code. Furthermore, for any y, the canonical code r is yrandom if and
only if Q has a ghost code which is yrandom.
Proof. () Suppose the canonical code of Q is nonrandom. Then there is a c.e. mar
tingale m that succeeds on it. From any initial segment a of a ghost code g for Q, the
subsequence a of exactly the canonical nodes of a is computable. Therefore it is com
putable whether the bit of g after a is canonical or ghost. From m, define the martingale
m' which bets as follows:
m'(a i) ={ r(ai) next bit is a canonical node
m'(a) next bit is a ghost node.
That is, m' holds its money on ghost nodes and bets identically to m on canonical nodes.
It is clear that m' succeeds on the ghost code g and thus g is nonrandom.
() Now suppose the canonical code r for Q is random, and let q be an infinite ternary
string that is random relative to r (and so by Theorem 3.2.12 r E q is random). We claim
the ghost code g obtained by using the bits of r as the canonical nodes and the bits of q in
their original order as the ghost nodes is random. It is clear that g is a ghost code for Q.
Suppose m is a c.e. martingale that bets on g. From m it is straightforward to define
a nonmonotonic martingale m' which mimics m's bets exactly but performs them on r D q,
succeeding whenever m succeeds. As r and q were chosen to be relatively random, this will
show g is random.
As discussed previously, from g n it is computable whether g(n) will be a ghost node
or a canonical node, and which position in g or r it occupies in either case. Therefore,
assuming the bits seen so far may be assembled into an initial segment a of g, m' takes
the values m(a i), i < 3, as its bets on the corresponding bit of r or g, whichever is
appropriate. Having seen that bit, then, it can assemble a (1a1 + 1)length initial segment
of g and repeat the process. As m' makes identical bets to m and has identical outcomes,
since it cannot succeed on r E g, m cannot succeed on g and g is random.
To relativize (), suppose that r is yrandom, so that r E y is random by Van
Lambalgen's Theorem 3.2.12. Then in the proof simply choose q to be random relative
to r E y, and then g will be random relative to y. The other direction relativizes in a
straightforward way. O
3.3.6 Coding and Joins of Closed Sets
The primary purpose of the ghost codes is to remove the dependence on the particular
closed set under discussion when interpreting bits of the code as nodes of the tree. This is
especially useful when subdividing the tree, as in the following definition.
Definition 3.3.10. The tree join of closed sets Po and P1 is the closed set
Q {0x : x e Po} U {'x: x e Pi}.
Given ghost codes ro, rl for the Pi, their tree join ro B rl is the code for Q with the
corresponding ghost node values.
The standard recursiontheoretic join is defined by
ro ri (ro(0),ri(0),ro(l),ri(1),...).
We wish to relate the recursiontheoretic join and the tree join.
Lemma 3.3.11. Given two ghost codes ro, rl, the tree join ro B rl is random if and only if
the recursion theoretic join ro E ri is random.
Proof. It is clear that there is a computable permutation 7 which uniformly maps any
tree join ro E] r to the recursiontheoretic join ro D rl. That is, in ro D rl, the entries of
ro and rl alternate, whereas ro E rl starts with a 2, followed by blocks from ro and ri,
as follows. First ro(0), ri(0), then ro(l), ro(2), ri(1), ri(2), and continuing with pairs of
blocks of size 4, 8 and so on. The result now follows from the VonMisesC '!i. !IWald
Computable Selection Theorem [94]; the theorem states that, for any random sequence x
and any computable 11 function g, the sequence z(n) = x(g(n)) is random. E
We now obtain the following corollary of Theorems 3.2.12 and 3.3.9 and Lemma
3.3.11.
Corollary 3.3.12. Suppose Pi, i = 0, 1, are closed sets with canonical codes ri and let P
be the tree join of Po, P1. Then P is random if and only if ro E rl is random.
Proof. () Suppose that ro E ri is random. Then by Theorem 3.2.12, ro and rl are
mutually relatively random. By Theorem 3.3.9, Po has a ghost code go which is random
relative to ri, and so also viceversa, and then P1 has a ghost code gl which is random
relative to go. Again by 3.2.12, the recursiontheoretic join go g gi is random, so by
Theorem 3.3.11 the tree join go ] gi is also random, and hence P possesses a random ghost
code and is random.
() Suppose now that P is random, and therefore possesses a random ghost code g. The
code g may be thought of as a tree join go ] gi, which is therefore random, and so by
Theorem 3.3.11, go gi is random. By Theorem 3.2.12, the individual codes go, gi are
therefore mutually relatively random. Now by the relative version of Theorem 3.3.9, ro is
random relative to gl. But ri is computable from gl and hence ro is random relative to ri
as well. Similarly, ri is rorandom and thus, again by 3.2.12, ro rl is random. E
3.4 Members of Random Closed Sets
In this section we tackle the question of which types of elements necessarily belong,
or do not belong, to random closed sets. The former is addressed in Section 3.4.1 and the
latter in Section 3.4.2.
3.4.1 Positive Results
We shall see, as a consequence of Theorem 3.4.19, that every closed set is perfect and
contains continuum ,rn ir: elements. In this section, we demonstrate other positive results.
For example, every random closed set contains random and nonrandom elements. Other
examples abound. We begin with the first example.
Theorem 3.4.1. Every random closed set contains a random element.
Proof. Suppose that a closed set Q has no random element and consider the following
MartinL6f test on the space C:
U = P IP C and PCV}
where (Vi) is a universal MartinL6f test on the Cantor space. By Lemma 3.3.8, p*(Ui) <
/ip() < 2' so that (Ui) is a MartinL6f test on C. But Q E nU, so Q is not random. O
As a converse to Theorem 3.4.1 we have the following.
Theorem 3.4.2. For any random r E 2", there exists a random closed set containing r as
a path.
The proof of this theorem was originally given by Joe Miller and Antonio Montalbdn
and has been subsequently improved thanks to the anonymous referee.
Proof. Let r be a random real and let x be the canonical code of an rrandom closed set.
We alter x to the code x' of a closed set guaranteed to contain r but changed as little as
possible to achieve that.
To determine x'(n), assume x' [ n has been defined. If x(n) = 2 or x(n) corresponds
to a node not along r, set x'(n) = x(n). If x(n) c {0, 1} corresponds to r(k), set
x'(n) = r(k).
The closed set defined by x' will clearly contain r. For a contradiction, assume x' is
nonrandom and let m' be a c.e. martingale that succeeds on it. We build a nonmonotonic
martingale m to bet on x E r. On bits of x, m will be a tripleornothing martingale; on r,
it will be doubleornothing.
First note that from initial segments of x and r we may reconstruct an initial segment
of x' computably, and we ahbi know from an initial segment of x' whether the next bit
is along r or not, and which bit of r it is. We will construct m so that after every stage of
betting (which will be one bet by m' and one or two bets by m), the value of m is equal to
the value of m'. At every stage it will be clear we have revealed enough bits of x and r to
reconstruct x' to the needed length.
Suppose inductively m and m' hold equal capital after the stage of betting on the
last node of a E x'. If the bit x'(n) following a is not on r, m bets identically to m';
i.e., m(x(n) = i) = m'(aui) for i < 3. In that case x(n) = x'(n) so our inductive
hypothesis holds. If x'(n) is on r, set m(x(n) = 2) = m'(au2) and for i = 0, 1, set
m(x(n) = i) = [m '(a0) + m'(a 1)]. If x'(n) = 2, then the capital for both m
and m' is m'(a'2), so the inductive hypothesis holds and we proceed to the next stage.
Otherwise m bets on r(k) for the appropriate k, setting m(r(k) i) = m'(a i) for
i = 0, 1. On r(k), the sum of m's capital on each of the two outcomes must average to
the previous capital; as the previous capital was 1 [m'(a.'0) + m'(a' 1)] this clearly holds.
By construction r(k) = x'(n) = i, so both m and m' now have capital m'(a i) and the
inductive hypothesis holds. As m' is c.e., m will also be.
As the values of m' along x' are a subsequence of the values of m along x E r, if m'
succeeds so does m, contradicting our assumption on x E r. Therefore x' is the code of a
random closed set containing the given random path r. E
The previous results might si,' 1 that every element of a random closed set is a
random real. However, it turns out that every random closed set contains a nonrandom
real.
We need the following classic result of Chernoff [28] (a version of Bernoulli's Weak
Law of LIr,,/. Numbers) here and also for another theorem to follow. See [67] for an
exposition.
Lemma 3.4.3 (C('! i i. .11). Let E be an event which we will refer to as 'success'. If E
occurs with probability p, then for any natural numbers n and any E with 0 < E < 1, the
probability that out of n mutually independent trials, the number of successes differs from
pn by > Epn is < 2e2pn/3
Theorem 3.4.4. Every random closed set contains a nonrandom real; in particular, the
leftmost and rightmost paths in a random closed set are not random reals.
Proof. We will show that, for a random closed set Q, the leftmost path is not stochas
tically random, that is, the ..imptotic frequency of 0's is 2. Since an effectively random
real in 2" must have ..ii',iiiil1 ic frequence of 1 for 0's and l's, this will suffice to prove
that the leftmost path is not random. We define a MartinL6f test as follows. Fix a ratio
nal E such that 0 < E < 1. For each n, let S, be the family of closed sets (that is, codes
for closed sets) such that the first n bits of the leftmost path have either < 2(1 E)n, or
> (1 + E)n occurrences of 0. By the definition of our probability measure, we have
(n 2\ m 1 nm
3 E*nj ( ) )
It now follows from C'!I. Ii1 all's Lemma 3.4.3 that
p*(Sn) < 2eE22n/9.
Thus the measures of the test sets S, have effective limit zero. It is easy to see that the
sequence {S,} is computably enumerable. For each n, S, is a clopen set and in fact the
union of the finite family of intervals I(c) in C such that a codes a tree up to level n in
which the leftmost path has either < ( )n, or > (1 + )n occurrences of 0.
Furthermore, S' J= U>~ S, is also a MartinL6f test. It follows that for any random
closed set Q, and any E > 0, there is an n such that for all m > n, the frequency of 0's in
the first m bits of the leftmost path is alv,v within E of j. Thus the leftmost path is not
effectively random. D
Recall that the leftmost and rightmost elements of any strong A closed set are
A. Given Theorems 3.4.1 and 3.4.4, we ask: Does a A random closed set contain a A
random path?
Theorem 3.4.5. Every random strong A closed set contains a random A real.
Proof. Let Q be a random strong A0 class. By Theorem 3.4.1, Q contains a random real
x. Let P be a II class in the Cantor space which contains only randoms and contains x
(this exists since the class of random reals is an effective union of I H classes). Note that
P n Q is a nonempty strong A0 class and it follows that the leftmost path of P n Q is a
A real which must be random since it belongs to P. E
The above theorem does not combine with the low basis theorem to establish the
existence of a low random real in any random strong A class. We can use the low basis
theorem, however, to demonstrate the existence of a low random real in any random closed
set with low canonical code.
Theorem 3.4.6. Every random closed set with low canonical code contains a low random
element.
Proof (KjosHanssen). Let Q be a random closed set with low canonical code. By
Theorem 3.4.1, Q contains a random element x. Therefore x E 2" \ U, for some n and
some open U, from the universal MartinL6f test. So, in particular, Q n 2" \ U, is non
empty. Now Q n 2" \ U, is ni relative to TQ. By the low basis theorem, Q n 2" \ U, is I0
has a member y such that y'
Furthermore, since TQ is low, y is also low. O
It is open whether every random closed set with a A canonical code has a low
random element; we conjecture that the answer is no. In the following section, we will
show that there is a random closed set not containing any A0 path.
Our next result, Theorem 3.4.8, uses a method which was used in [56] to show that
every random real is effectively biimmune. We first define this latter notion.
Definition 3.4.7. (i) A set A is effectively immune if it is infinite and there is a
computable function g(x) such that if W1 c A, W, < g(x).
(ii) A is effectively biimmune if A and A are both effectively immune.
Note, in particular, that an effectively immune set cannot contain an infinite c.e. set.
Theorem 3.4.8. If P is a random closed set then all elements of P are effectively bi
immune.
Proof. Suppose that P is a random closed set and A E P. Let (Ui)ijN be a MartinLif
test (in the space 3N) such that there is a computable function f with the property that
if (Ve)jEN is the eth MartinL6f test (under some effective enumeration of all MartinLif
tests) then for all e, i, V (,i) C U, (a standard construction of a universal test gives one
with this property). Since P is random, there is some k such that (the canonical code of)
P is not in Uk; let U = Uk. It suffices to find a computable function g such that
[A P and We CA] == W, < g(x) (31)
for all sets A and all x (the proof that A is effectively immune is entirely similar). Let B,,
be 0 if IWe I n, and otherwise the class of (canonical codes of) trees which contain a path
containing the first n + 1 elements in the standard enumeration of Wx. Then (Bx,,) is a
uniform double sequence of ZE classes and (by the definition of the probability measure on
3N),
pt*(B1) (2)
So for each x, (Bx,2,) is a MartinL6f test in the space 3" and from x we can calculate
the index of it. Then by using the computable function f mentioned above we get a
computable function g such that for all x, Bx,g(9) C U. This means g satisfies (31). O
It is well known that effectively immune sets can compute a fixed point free function,
so we have the following.
Corollary 3.4.9. The paths through a random tree are of fixed point free degree. That is,
each of them computes some fixed point free function.
It is known that every real can be computed by some random real. It is not known,
however, whether any real can be computed by all the paths of some random closed set.
The next theorem, an observation of Ted Slaman at a randomness workshop in Chicago in
2007, is a step in that direction. First, we need the following definitions, which are, in fact,
equivalent notions.
Definition 3.4.10. (i) A real x is Ktrivial if K(x[n) < K(n) + c for some c.
(ii) A real x is a base for 1randomness if there is some y >r x such that y is 1x
random.
Theorem 3.4.11. Any set which is computable from all paths of a random closed set is
Ktrivial.
Proof. If A is computable from all paths, then A is computable from the leftmost and
rightmost paths. Note that each of these latter paths are each computable from the two
halves of the tree. Furthermore these two halves are relatively random to each other.
Hence each is random relative to anything the other computes. So half 1, for example,
computes A and hence half 2 is random relative to A. On the other hand, half 2 also
computes A. Therefore this is an example of something in the cone above A that is
Arandom. So A is a base for 1randomness. O
Much interesting work has been done on the Ktrivial reals. C'!i il ii showed that if A
is Ktrivial, then A r 0'. Solovay constructed a noncomputable Ktrivial real. Downey,
Hirschfeldt, Nies and Stephan [39] showed that no Ktrivial real is c.e. complete. The
notion of a Ktrivial closed set was introduced in [9]. It was shown in particular that every
Ktrivial class contains a Ktrivial member, but there exist Ktrivial II0 classes with no
computable members.
3.4.2 Negative Results
Random closed sets can never contain nc.e., isolated, or 1generic paths, or paths of
incomplete c.e. degree. We build to these facts and prove others along the way.
Theorem 3.4.12. Random closed sets contain no computable elements.
Proof. For any finite string a of length n, the probability that a closed set Q meets
I(a) is ( )". For a computable real y, the sqeuence {Q : Q n I(y[n) / 0} thus forms a
MartinL6f test in the space C of closed sets, which shows that y does not belong to any
MartinL6f random closed set. That is, for each n, {x : Qx n I(y[n) / 0} is a c.e. open set
and has measure (2)" in {0, 1, 2}1, where Qx is the closed set with code x. O
We prove an even stronger result in Theorem 3.4.17. First, however, recall that a
HI class P is decidable if Tp is decidable. It follows that a nonempty decidable HI class
must contain a computable element (for example, the leftmost path). By Theorem 3.4.12,
it follows that no decidable II class can be random. As every random class contains a
random element (Theorem 3.4.1) and has, as we shall show, measure zero (Theorem 3.5.1),
the following proposition demonstrates that this extends to arbitrary II classes.
Proposition 3.4.13. If P is a II class of measure 0, then P has no random elements.
Proof. Let T be a computable tree such that P [T], and for each n, let P,= U{I():
a E T n {0, 1}"}. Then {Pn},EN is an effective sequence of clopen sets with P = P,
and lim, (P) = p(P) = 0. Furthermore, p(P,) = 2 T n {0, 1}'"; this is a computable
sequence. Thus {P,},}N is a MartinL6f test and P has no random elements. O
Alternatively, we can show that no II class is random through the following stronger
result, combined with an appeal to Theorem 3.5.1, from the next section.
Theorem 3.4.14. Let Q be a II class with measure 0. Then no subset of Q is random.
Proof. Let T be a computable tree (possibly with dead ends) and let Q = [T]. Then
Q = n ,, where U, = [T,]. Since p(Q) = 0, it follows from Lemma 3.3.8 that
lim, p*(Pc(Un)) = 0. But Pc(U,) is a computable sequence of clopen sets in C and
p*(Pc(U,)) is a computable sequence of rationals with limit 0. Thus Pc(U,) is a Martin
Lof test, so that for any random closed set, there exists n such that P Pc(U,) and hence
P is not a subset of U,. O
Corollary 3.4.15. No II class can be random. O
We now provide an even stronger version of Theorem 3.4.12; we need the following
definition.
Definition 3.4.16 (fc.e. reals). For any computable, nondecreasing function f, we v
that a real 3 E {0, 1}" is fc.e. if there exists a computable approximating function Q such
that, for all i E N,
(i) Q(i,0) 0
(ii) lim Q0(i, s) = 3(i);
(iii) {s : Q(i, s + 1) / Q(i, s)} has cardinality < f(i).
The reals which are fc.e. for some computable function f are part of the wellknown
Ershov hierarchy [43, 86].
Theorem 3.4.17. Suppose that f is computable and bounded by a polynomial. Then no
random closed set has any fc.e. paths.
Proof. Let f be as above, 3 an fc.e. real and P a closed set containing 3. Let
be the fapproximating function for f. Also let .i C {0, 1}" be the set of different
Qapproximations to f[n during the stages.
A priori, '.1 is exponential. However, for a fixed n, 3 [n can change at most
Ei<, f(i) times, so 'V is also bounded by a polynomial, i.e. there is k E N such
that for almost all n, '1 < nk. Now let
Sn, U {PI P P C& J(7) /0}. (32)
ofMn
Then (S,) is a uniformly c.e. sequence of open sets in the space C of closed sets of 2" and
for all n, P E S,. Also for almost all n,
*(S,) 4 p ({P  P t C & Pn IJ(o) / 0}) = .V  [ Gk *
(TcEM
Since lim, [nk ()"] 0 there is a computable subsequence of (S,) which is a MartinLof
test and so P is not random. E
For any Ktrivial real A and any unbounded nondecreasing computable function h,
A is hc.e. (Nies [75]). Thus it follows from Theorem 3.4.17 that a random closed set can
have no Ktrivial paths. We observe that Theorem 3.4.17 cannot be extended to wc.e. in
general, because there are leftc.e. (and hence wc.e.) random reals, and by Theorem 3.4.2
each of these belongs to a random closed set.
Also, recall from Corollary 3.4.9 that that paths through a random tree are of fixed
point free degree. It is known that fixed point free degrees cannot be 1generic (see [42] for
a proof) or incomplete c.e., and that if they are A0 they compute a promptly simple set
and no pair of them forms a minimal pair (see [57]). So we have the following.
Corollary 3.4.18. The following hold:
No path of a random tree is 1generic.
No pair of A paths of a random tree can be a minimal pair.
Every A path of a random tree computes a promptly simple set.
No path of a random tree can have incomplete c.e. degree.
Theorem 3.4.19. If Q is a random closed set, then Q has no isolated elements.
Proof. Let Q = [T] and suppose by way of contradiction that Q contains an isolated
path x. Then there is some node a E T such that Q n I(a) = {x}. For each n, let
S', = {P C : {re {0,1}" : P n I(a^ ) O0} = 1}.
That is, P E S, if and only if the tree Tp has exactly one extension of a of length n + 1.
It follows that
Pn I(a) = 1 (Vn)P e S,
Now for each n, S,, is a clopen set in C and again by induction, S, has measure () ". Thus
the sequence So, Si,... is a MartinL6f test. It follows that for some n, Q S,. Thus
there are at least two extensions in TQ of a of length n + lal, contradicting the assumption
that x was the unique element of Q n I(a). O
As mentioned previously, it follows that every random closed set is perfect and hence
contains continuum rn ii: elements.
Next we want to find a random closed set which does not contain a A path. Now
it is easy [20, 24] to construct a strong 1 class P of positive measure which contains no
A elements; of course P must contain a random real since it has measure 1. The difficult
problem is to construct a random strong 1 class with no A elements. We have the
following result in this direction, which yields a random strong A closed set with no A
elements.
Theorem 3.4.20. For any set A there is an Arandom closed set Q such that TQ
but Q has no elements
Proof. It is enough if we prove the claim for A = 0 because the argument relativises to
any oracle A in a straightforward way. For A = 0 we use a finite injury construction over
0' to construct Q with the above properties. In the construction we will 0'approximate
the canonical code of a tree T which has no A paths. To make sure that the tree T is
random we fix a 1 class P of positive measure in the space 3" (where the code for T lies)
which contains only randoms, and we make sure that at every stage our approximation
(as a finite ternary string) to T's canonical code can be extended to a path in P. Then by
compactness the canonical code of our tree will be in P and so the tree will be random.
The changes in the approximations are motivated by the requirements:
Re : if 4' is total then the real it defines is not in [T].
Let a, be a finite string approximation of the canonical code a we are building. We will
have as = s. Strategy Re will come into power after stage e and will restrain a up to
some re > e (the default value is re [0] e). Also it might request some changes in a after
the eth bit. We start with ao = 0 and at stage s + 1, assuming inductively that as I and
[ac] n P / 0 we ask for the least i < s such that Ri requires attention. This happens if
(i) The longest defined initial segment T of 4+0' is larger than ever before;
(ii) there exists a e {0, 1,2}* such that a, [(maxj
and T is not consistent with the finite tree with code a.
If there is no such i then we extend as by one bit such that [ac8+] n P / 0. Otherwise
we let a8s+ = a and ri[s + 1] = s + 1. The construction proceeds in a straightforward way
and we can prove inductively that for every e, Re is satisfied, stops requiring attention and
re reaches a limit. Then the limit a = lim8 a, exists and we also have that a is random
by compactness. The satisfaction of the requirements comes from a measuretheoretic
fact. Consider Re and inductively assume that after stage se no Ri with i < e requires
attention. Then r = maxi
[a [maxi
p([a Fr] n P) > 0
and on the other hand, if 3 = KL' we have seen that
pu{7  7 3' and 7 is the canonical code of a tree which has 3 as a path} = 0.
This means that if at stage s, the requirement Re is not yet satisfied, it will receive
attention at a later stage and get satisfied permanently. E
3.5 Measure and Dimension
In this section we show that random closed set have measure zero (Theorem 3.5.1)
and box dimension log2 1 (Theorem 3.5.2).
3.5.1 Measure
Theorem 3.5.1. If Q is a random closed set, then p(Q) = 0.
Proof. We will show that in the space C of closed sets, the p*probability that a closed
set P has Lebesgue measure 0, is 1. This is proved by showing that for each m, I(P) >
2m with p*probability 0. For each m, let
S,= {P : p(P) > 2}.
We claim that for each m, It*(Sm) = 0. The proof is by induction on m.
For m = 0, we have pI(P) > 1 if and only if P = 2', which is if and only if
xp = (2, 2,...), so that So is a singleton and thus has measure 0.
Now assume by induction that Sm has measure 0. Then the probability that a closed
set P = [T] has measure > 2"1 can be calculated in two parts.
(i) If T does not branch at the first level, i To {(0)} without loss of generality.
Now consider the closed set Po {y : Oy E P}. Then p(P) > 2m1 if and only if
p(Po) > 2m, which has probability 0 by induction, so we can discount this case.
(ii) If T does branch at the first level, let Pi {y : i^y e P} for i = 0, 1. Then
p(P) (p(Po) +(Pi)), so that p(P) > 2m1 implies that at least one of p(P,) > 2 m
(Note that the reverse implication is not alvb true.) Let p = p*(Sm+1). The observations
above imply that
1 2 1
p < (1 (1 p)2) p 2,
3 3 3
and therefore p = 0.
To see that a random closed set Q must have measure 0, fix m and let S = S,. Then
S is the intersection of an effective sequence of clopen sets V, where for P = [T],
P Ve p([T,]) > 2.
Since these sets are uniformly clopen, the sequence me = p*(V) is computable. Since
lime m = 0, it follows that this is a MartinLof test and therefore no random set Q
belongs to [n V. Then in general, no random set can have measure > 2' for any m. E
3.5.2 Dimension
Surprisingly, we can compute the (Kolmogorov) box dimension of a random closed
set, and in fact it turns out that all random closed sets have the same dimension. The
intuition for this comes from the following lemma. For any function F mapping the space
C of closed sets into Z, the expected value of F on C is the integral f F(P) with respect to
the probability measure p*.
Lemma 3.5.2. In the space C of closed sets, the expected cardinality of {a E {0, 1})
QIn(a) / 0} is exactly ( )" for every n, where Q is chosen uniformly at random according
to /t*.
Proof. Let S,= {a {0, 1}" : Q n I(a) / 0}, for a randomly chosen Q from C.
The proof is by induction on n. For n 1, we have two cases. With probability 2,
card(S1) = 1 and with probability card(S) = 2. Thus the expected value is exactly
. For n + 1, there are again two cases. With probability 2, card(S1) = 1, so that the
expected card(S,+1) equals the expected card(S,), which is (4)" by induction. With
probability card(SI) =2, in which case the expected card(S,+1) is twice the expected
card(S,), that is, 2(4)". Thus we have the expected value
2 /4" 1 /4 4Y
card(S, +) = ()+ 2 () (4) +
3 3 3 3
El
The box dimension of a closed set in the Cantor space, if it exists, is given by the
following limit:
F ) log2(card(TQ n {0, 1}))
dimB F(Q) = lim
noo 7
(See [6] for this formulation of the box dimension in {0, 1}".) Now by Lemma 3.5.2, the
expected value of card(TQ n {0, 1}") for a random closed set Q is (4)", which s,. I that
the box dimension of Q should be log2 3.
Lemma 3.5.3. Let Q be a random closed set. Then for any E > 0, there exists a m e N
such that, for all n > m, ( )"(1 ))n < card(TQ n {0, 1}) < ( )"(1 + E).
Proof. For each n, let c,(Q), or just c,, denote card(TQ n {0, 1}"). We will use three
applications of ('!,. il l's Lemma 3.4.3. First we show that there exists m such that for
all n > m, C6n > n. Since the tree TQ n {0, 1}6" 1 has at least 6n nodes, it follows from
('!. i i .l'"s Lemma that the number of branching nodes is less than n with probability
S2/6 Thus c6n < n with probability < 2n/6. Then the probability that C6n < K for any
n > m is less than
00 2m/6
E 2n/6
1 21/6"
n=mn
This provides a computable sequence of clopen sets with measures bounded by a com
putable sequence with limit zero and hence a MartinL6f test. It follows that for any
random closed set Q, there exists mo such that c6n > n0 for all n > mo. Now for n > o0,
there are at least 6n2 nodes in TQ n {0, 1}412n1 {0, 1}6n1, so that again by Chernoff's
Lemma, the probability that < n2 of these are branching nodes is < 2n /6. It follows as
above that there exists mi > 3 such that cl2n > n 2 for all n > mi. Now suppose that
m 12mi and that 12n < m < 12(n + 1) < 16n. Then n > mi, so that
cm > c12n > n2 > (m/16)2
Again by ('!, i Ii ll"'s Lemma, the probability that the number of branching nodes from
TQ n {0, 1}" differs from c by >
that e+i differs from 4cn by > c_ c4. For n > mi, we know that c, > (4) so that
A and c4 and hence 2 /9 < 2n/144. Thus the probability p, that c,+i
differs from c by more than is < 2 n/144. Then the probability that for any n > ml,
c+I differs from c, by more than 3rc, is bounded by
n n1 2m/144
S 2/144 1 2144'
n=m n=m
This again provides a MartinL6f test which shows that for any random closed set Q, there
exists m2 so that for n > m2,
(*)
Now given E, choose m > m2
Then for any k,
Cm(f4)2k(_
3
4 1 ) 4 1
3 ) 3 and i <
so that (1 + 1)2 < 1 + and 1 F < (1 
c)k < Cm( 2k( 2k < Cm+2k
3 c2
< )( 2k(1 + )2k < Cn( )2k( + )k
3 \//n 3
Now let k be large enough so that
(1 E)m+k C ()~1( + E)m+k
Then the desired inequality
( )"(1
)" < c < (4 )"(1 + E)
will hold for even n > m + 2k. For odd n, this inequality will hold by the inequality (*)
above.
Theorem 3.5.4. For any random closed set Q, the box dimension of Q is log2 .*
Proof. Given E > 0, let m be given by Lemma 3.5.3. Then for n > m, we have
4
n log2 nlog2(1
0
so that
4
log2 + log(1
and therefore dim ) li
and therefore dimB(Q) = linm
4
E) < log2(card(TQ n {0,1}") <, nlog2 + nlog2(1 + ),
3
log2(card(TQ n {0, 1})) 4
E < 1092 10n2 + 109g E),
, 1 1. !. n of ,j )) 4g
1)2.
V7'
3.6 PrefixFree Complexity of Closed Sets
In this section, we consider randomness for closed sets in terms of incompressibility
of trees. Of course, Schnorr's theorem tells us that P is random if and only if the code
xp E {0, 1, 2}1 for P is prefixfree random, that is, K3((xp n) > n 0(1). (Schnorr's
theorem for arbitrary finite alphabets is shown in [18].) Here we write K3 to indicate that
we would be using a universal prefixfree function U : {0, 1, 2}* {0,, 12}*. However,
many properties of trees and closed sets depend on the levels Tn = T n {0, 1}" of the tree.
For example, if [T,] = U{I(a) : a c T}, then [T] = n, [T,] and p([T]) = lim", p([T,]).
So we want to consider the compressibility of a tree in terms of K(T,). Now there is
a natural representation of Tn as a string of length 2". That is, list {0, 1}" in lexicographic
order as a1,..., a2. and represent Tn by the string el,..., e2~ where ei = 1 if ai E T and
ei = 0 otherwise. Henceforth we identify T, with this natural representation.
It is interesting to note that the code for Tn will have a shorter length than the
natural representation. For example, if [T] = {y} is a singleton, then x = y and for each
n, the code for Tn is x[n. If x is the code for the full tree {0, 1}*, then x= (2, 2,...) and
the code for Tn is a string of (2" 1) 2's, those labels attached to nodes of length < n. For
the remainder of this section, we will use T, to mean the natural representation and x, to
mean the code.
Question. Is there is a formulation of randomness for closed sets in terms of the
incompressibility of T,?
It seems plausible that P = [T] is random if and only if there is a constant c such
that K(T,) > 2" c for all n. However, we will see that this is not possible for any tree.
On the one hand, in Section 3.6.1 we achieve a lower bound for incompressibility. That
is, we show that if P = [T] is random then there is a constant c such that K(T,) >
(0)" c for all n. On the other hand, in Section 3.6.2, we see that the 2" is too high of
an incompressibility bound since there is some c and some random closed set such that
K(T,) < 2" c for all n.
In a larger sense, we seek a formulation of randomness, in terms of the incompressibil
ity of T,, fo thr th objects such as II classes or II classes. In the following two sections
we consider these questions and achieve some lower and upper bounds for these classes of
objects.
3.6.1 Lower Complexity Bounds
First we give a lower bound for the prefixfree complexity of a random tree.
Theorem 3.6.1. If P is a random closed set and T = Tp, then there is a constant c such
that K(T,) > ()" c for all n.
Proof. Let P = [T] be a random closed set. Let m be given by Lemma 3.5.3, for E =
so that for n > m,
card(TT,)> ( .
It follows that the code x, for T, has length > (7). Since x is random, we know that, for
n K m,
K3(X ) > (i a,
for some constant a. Now we can compute x, from T,, so that
K (T,) 3 K (x) b,
for some constant b. The result now follows.
That is, let U (mapping {0, 1}* to {0, 1}*) be a universal prefixfree Turing machine
and let K(T,) = min{(a : U(a) = T,}. Let M be a prefixfree machine M (mapping
{0, 1}* to {0, 1,2}*) such that M(T,) = x,. Then define V by
V(a) = (U(a)).
Then Kv(x n) < K(T,), so that for some constant e, K3(x,) < K(T,) + e and hence
K(T.) > K3x) > (. e.
E
The standard example of a random real, C' iil ,ii's Q [27], is a c.e. real and therefore
A. Thus there exists a A random tree T and by Theorem 3.6.1, K(T ) > ( c for
some c.
We have a more modest result for II classes. That is, there is an effectively closed set
with not too much compressibility, in the following sense.
Theorem 3.6.2. There is a 1 class P [T] such that K(T,) > n for all n.
Proof. Recall the universal prefixfree machine U and let S {a E Dom(U) : U(a)l
21' }. Then S is a c.e. set and can be enumerated as a1, a2,.... The tree T = QT T8 where
T" is defined at stage s. Initially we have To {0, 1}*. We i that at requires attention
at stage s > t when T = U(Ut) T, for some n (so that r = 2") and n > lat Action is
taken by selecting some path pt E T, of length n and defining T"+l to contain all nodes of
T" which do not extend pt. Then T / T,+1 and furthermore T / T, for any r > s + 1 since
future action will only remove more nodes from T,.
At stage s + 1, look for the least t < s + 1 such that at requires action and take the
action described if there is such a t. Otherwise, let T"+ = T".
Let A be the set of t such that action is ever taken on Ut. Recall from the Kraft
Inequality that t 2 < 1. Since pt > Ut, it follows that EteA 2Pt < 1 as well. Now
/I([T]) =1 >t 21IPt > 0 and therefore [T] is nonempty.
It follows from the construction that for each t, action is taken for at at most once.
Now suppose by way of contradiction that U(a) = T, for some at with la < n. There
must be some stage r > t such that for all s > r, T, = T, and such that action is never
taken on any t' < t after stage r. Then at will require action at stage r + 1 which makes
T+1 / TA, a contradiction. E
There is a stronger result for closed 1 classes. Namely, there is a closed 1 class with
the following stronger incompressibility property.
Theorem 3.6.3. There is a nII class P [T] such that K(T4) > 2 for all .
Proof. We will construct a tree T such that T,2 can not be computed from fewer than 2"
bits. We will assume that U(0) T to take care of the case n = 0. At stage s, we will define
the (nonempty) level T,2 of T, using an oracle for 0'.
We begin with To {0}*.
At stage s > 0, we consider
D, {a e Dom(U) : al < 2"}.
Since U is prefixfree, card(D,) < 22". Now there are at least 222 1 trees of height s2
which extend T(s_1)2 and we can use the oracle to choose some finite extension T' = T,2 of
T(s_1)2 such that, for any a E Ds, U(a) / T' and furthermore, U(a) / Tr for any possible
extension T, with s2 < r. That is, since there are < 22s finite trees which equal U(a) for
some a E Ds, there is some extension T' of T(s_1)2 which differs from all of these at level
s2. We observe that the oracle for 0' is used to determine the set Ds.
At stage s, we have ensured that for any extension T C {0, 1}* of T,2, any a with
lal < 22 and any n > s2, U(a) / Tn. It is immediate that K(T,) > 2 '. E
3.6.2 Upper Complexity Bounds
In Theorem 3.6.1 we achieved a lower bound of (')" for the prefixfree complexity of
a tree Tp of a random closed set P. It seems plausible that we might be able to achieve
a higher bound of 2". If true this would actually provide and immediate characterization
of randomness of closed sets in terms of prefixfree complexity of trees. That is, a closed
would be random iff K(T,) > 2' c for some constant c. (To see this, note that from
x [2" we can compute T, uniformly so that K3(x [2) > K(T,) b for some b.) However
the following theorem provides an upper complexity bound less than 2", refuting such a
possibility.
Theorem 3.6.4. For any tree T C {0, 1}*, there are constants k > 0 and c such that
K(T)) < 2' 2k + c for all f.
Proof. For the full tree {0, 1}*, this is clear so suppose that a T for some a E {0, 1}'.
Then for any level f > m, there are 2'm possible nodes for T which extend a and Te may
be uniformly computed from a and from the characteristic function of T. restricted to the
remaining set of nodes. That is, fix a of length m and define a prefixfree computer M as
follows. The domain of M is strings of the form 01T where I = 2 m 2'm. M outputs
the standard representation of a tree Tf such that no extension of a is in Tf and such that
7 tells us whether strings not extending a are in T. It is clear that M is prefixfree and
we have KM (T+) = + 1 + 2' 2'. Thus K(T ) < + + 1 + 2' 2' + c for some constant
c. Now + 1 < 2'/1 for sufficiently large f and thus by adjusting the constant c, we can
obtain c' so that
K(T<) < 2Y 2'm1 + c'.
The following theorem also refutes the possibility that K(T{) > 2ec is a charac
terization of random closed sets in terms of prefixfree randomness. It shows that closed
sets with small measure, such as random closed sets which have measure zero (see Theo
rem 3.5.1), are more compressible.
Theorem 3.6.5. If p([T]) < 2k, then there exists c such that, for all ,
K(T) < 2k+1 + c.
Proof. Suppose that p([T]) < 2k. Then for some level n, T, has < 2 k nodes
al,...,at. Now for any f > n, T. can be computed from the fixed list a,,..., at and the
list of nodes of Te taken from the at most 2 k extensions of al,..., ot. It follows as in the
proof of Theorem 3.6.4 above that for some constant c and all ,
K (T) < 2 + f + 1 + c.
Thus for large enough so that f + 1 < 2 k, we have
K(TT) < 2k+1 + c,
as desired. O
We conjecture that a bound of (4)" would characterize random closed sets in terms of
prefexfree complexity. It would suffice, then, to show that (4)" is a lower bound and that
this bound implies randomness.
We also still seek upper bounds for II or closed II classes towards establishing
prefixfree complexity characterizations of these classes. It seems plausible that II classes
are more compressible (i.e. necessarily have smaller lower bounds) than random closed sets
and we would like to explore this notion further.
3.7 Other Notions of Randomness for Closed Sets
Other notions of randomness that depend on different probability measures, or the
inclusion of trees with dead ends in the encoding, might also be considered.
3.7.1 Randomness with Regular Probability Measures
For any regular measure v, we can define the notion of a vMartinL6f test and the
resulting notion of a vMartinL6f random (or just vrandom) real. It is easy to see that
vrandom reals exist for any v and hence vrandom closed sets exist. The results on ghost
codes and joins will hold for any regular measure. The corresponding version of Lemma
3.3.8 will hold if v is regular with bo and bl < 1. The proofs of Theorem 3.4.14 and
Corollary 3.4.15, that no subset of a measurezero II class is random, also go through
under this assumption.
Some of the results in this chapter may also be obtained for vf where f(a i) <
for i = 0, 1. For example, with respect to vf, a random closed set will have no isolated
elements and it will alvx contain a random element. For any regular measure, either the
leftmost or the rightmost path will be nonrandom, since either b0 + b2 > or bl + b2 The
proof of Theorem 3.4.19 that every random closed set has measure 0 seems to require, for
vfrandomness, that f(a 2) < 1 for all a.
3.7.2 Randomness with the Inclusion of Trees with Deads Ends
Returning to the notion of randomness which allows trees with dead ends, let b3 now
be the probability that a given node has no extensions and let the probability be regular
as above. Then a simple recursion shows the probability p of a given closed set being
empty satisfies the equation
p = b3 + (bo + b)p + b22.
Solving for p, we obtain
(p 1)(b2p b3) 0.
Thus either p = 1 or p = It follows that if b2 < b3, then p 1, that is, almost every
closed set is empty. Suppose now that b3 < b2 and let p, be the probability that a given
tree T has no paths of length n. Then it can be seen by induction that pi < b for all n.
That is, p b3 < b and then
b2
Hence in this case, the probability that a given closed set is empty is b3 < 1. In this case,
one could presumably develop a notion of a random tree and a random closed set and
explore the properties of random closed sets.
3.8 Random Closed Sets and Effective Capacity
In this section we will consider, given a closed set Q, the probability that a randomly
chosen closed set meets Q. This probability is given by pi(V(Q)), where V(Q) is a sub
basis set for the hitormiss topology on C (as given in section 3.3.1) and pi is a given
probability measure. If we define Td(Q) to be precisely p*(V(Q)), it turns out that Id is
a computablee) capacity, in the sense defined below; furthermore, the converse also also
holds. That is, if T is a computablee) capacity, then there is some probability measure ld
for which T = 7d. We then explore the capacities of random and effectively closed sets,
under the uniform measure (i.e. bi = for all i in Definition 3.2.6). This is joint work with
Douglas Cenzer.
3.8.1 Computable Capacities
Definition 3.8.1. A capacity on C is a function T: C  [0, 1] with T(0) = 0 such that
(i) T is monotone increasing, that is,
Qi C Q2 (Q1) <(Q2)
(ii) For n > 2 and any Qi,... Q E C
T7( Q) < i{( i)'+7T(U Q):0 o$/ I c ,2,..., n}}.
i=1 itI
This is the alternating of infinite order property.
(iii) If Q = nQ, and Q,+1 c Q, for all n, then T(Q) = lim,,,T(Q,).
We will assume, unless otherwise specified, that T(2N) = 1 for a given capacity T.
Definition 3.8.2 (Computable Capacities). A capacity T is computable if it is com
putable on the family of clopen sets.
It follows that the capacity of any II class is upper semicomputable. Finally, the
following notational definition will be used throughout.
Definition 3.8.3 (Td(Q), for Q e C). Suppose Q E C. Define Td(Q) := p*(V(Q)), where
V(Q) is a subbasis set for the hitormiss topology on C (as given in section 3.3.1) and pl
is a given probability measure.
That is, Td(Q) is the probability that a randomly chosen closed set meets Q.
We now show, in the following two theorems, that a computable capacity is alv,
obtainable from, or a consequence of, a computable probability measure pi for some d. The
following theorem, in particular, is wellknown. For details on capacity and random set
variables, see [73].
Theorem 3.8.4. If p* is a computablee) probability measure on C, then 7d is a (com
putable) capacity.
Proof. This is easily verified. Certainly T(0) = 0. The alternating property follows
by basic probability. For (iii), suppose that Q = nQ is a decreasing intersection.
Then by compactness, Q n K / 0 if and only if QT n K / 0 for all n. Furthermore,
V(Q,~+) C V(Q,) for all n. Thus
T(Q) = (V(Q)) = (nV(Q,)) = lim,1(V(Q,)) = lim,mT(Q).
The computability of T is easily verified. That is, for any clopen set I(al) U .. U I(ak)
where each ai E {0, 1}", we compute the probability distribution for all trees of height n
and add the probabilities of those trees which contain one of the ci. O
This result has a converse, due to Choquet. See [73] for the general result.
Theorem 3.8.5. If T is a computable capacity, then there is a computable measure p on
the space of closed sets such that T = d.
Proof. Given the values T(U) for all clopen sets I(al) U ... U I(ok) where each ai E
{0, 1}", there is in fact a unique probability measure Pd on these clopen sets such that
T = d and this can be computed as follows.
Suppose first that T(I(i)) = ai for i < 2 and note that each a, < 1 and ao + a1 > 1
by the alternating property. If T = d, then we must have d((0)) + d((2)) = ao and
d((1)) + d((2)) = ai and also d((0)) + d((1)) + d((2)) = 1, so that d((2)) = ao + al 1,
d((0)) = 1 al and d((1)) = 1 ao. This will imply that T(r) = d(r) when rl = 1. Now
suppose that we have defined d(r) and that r is the code for a finite tree with elements
jo,... Un= a and thus d(r i) is giving the probability that a will have one or both
immediate successors. We proceed as above. Let T(I(ai)) = ai T(I(a)) for i < 2. Then
as above d(Tr2) = d(r) (ao + al 1) and d( i) = d(r) (1 ai) for each i. O
3.8.2 Regular Measures and Capacities of Closed Sets
In this section all results are with respect to pj fixed as the uniform measure (i.e.
the regular measure with bo b b2 = see Definition 3.2.6). With this measure, we
will consider the capacities of random closed sets and effectively closed sets. We iv that
Q E C is pI random if XQ is (\! ,rtinLof ) random with respect to the measure Pd.
Theorem 3.8.6. For the regular measure Pd with bi = if R is a p/random closed set,
then Td(R) = 0.
Proof. Fix d as described above so that d(a i) d() and let p~* = We will
compute the probability, given two closed sets Q and K, that Q n K is nonempty. Let
Q. = JI(a7): a CIO, 1} & Q n J(a) / 0}
and similarly for K,. Then Q n K / 0 if and only if Q, n K, / 0 for all n. Let p, be
the probability that Q, n K, / 0 for two arbitrary closed sets K and Q, relative to our
measure i*. It is immediate that pi = since Q n K1 = 0 only when Qi I(i) and
K1 = 1(1 i). Next we will determine the quadratic function f such that p,+ = f(pn).
There are 9 possible cases for Qi and K1, which break down into 4 distinct cases.
Case I: There are two chances that Qi n K1 = 0.
Case II: There are two chances that Q = K1 = I(i), so that Q,+1 n K,+I / 0 with
probability p,.
Case III: There are four chances where Qi = 2" and K = I(i) or vice versa, so that
once again Q,+1 n K,+I / 0 with probability pn.
Case IV: There is one chance that Qi = K1 = 2N, in which case Q,+1 n Ki / 0
with probability 1 (1 p,)2 = 2p, p2. This is because Q,+1 n K,+ = 0 if and only if
both Q,+1 n I(i) n K,, = 0 for both i = 0 and i =1.
Adding these cases together, we see that
6 1 8 12
Pn+1 = Pn + (2p p2) Pn 9P
9 9 9 9
It follows that the sequence (pn)E, is computable and we will see that the limit is zero.
Let f(p) = _p _ p2. Elementary calculus shows that f has fixed points at p 1 and
p = 0 and that for 0 < p < 1, 0 < f(p) < p. Since po = it follows that the sequence
(p }) is monotonic decreasing. Thus the limit exists and limnp = 0 (since it must be a
fixed point of f).
Thus the probability that Q n K / 0 equals limp, = 0. Next we will obtain a
MartinL6f test to prove our result.
For each m, n E u, let
Am,n = {Q : p*({K : Km n Qm / 0}) > 2n}.
Let Cm be the number of trees of height m without dead ends.
For each Q E Am,, there are 2"Cm possible choices for Km such that Km n Qm / 0
and thus at least Dm,n = ()2"lp*(Am,nT)C2 choices for (K,Q) E C x C such that
Km n Qm / 0 with Q E Am,, (since each pair might be counted twice).
Now define a computable sequence (m,),E,, so that pm, < 22n1. Then
Dr ,,n 2(n+l) p*(Am,n)
C2
It follows that
p*(Am ,n) < 2n+lP p < 2n+12n1 2".
Letting
Sn Ur>nAm,r
it follows that p*(S,) < 2" as well.
Now let R be a random closed set. The sequence (Sn,),, is a computable sequence
of c.e. open sets with measure < 2", so that there is some n such that R S,. Thus for
r > n, p*({K : Km, n Rm,, 0}) < 2r and it follows that
*({K : K nR / 0}) limp*({K : K n Rm, / 0}) = 0.
Thus Td(R) = 0, as desired. D
This result seems to depend on the measure. For different regular measures, the
capacity of a random closed set can have different values.
Theorem 3.8.7. For the regular measure Pd with bi = there is a measure zero IIo class
Q such that Td(Q) > 0.
Proof. First let us compute the capacity of X, = {x : x(n) = 0}. For n = 0, we have
Td(Qo) = Now the probability Td(X,+1) that an arbitrary closed set K meets X,+I may
be calculated in two distinct cases. Let K& be as in the proof of Theorem 3.8.6.
Case I If Ko = 2N, then Td(X,+1) =1 (1 d(X,))2.
Case II If Ko = I((i)) for some i < 2, then Td(Xn+) = T(X,).
It follows that d(Xn+l) = 7(X,) + (2d(X,) (7(X,))2) (X,)) ((X,))2.
Now the function f(p) = p lp2 has the property that f(p) > p for 0 < p < 1 and
f(l) = 1. Since Td(X,+1) = f(l(X,)), it follows that lim,Td(X,) = 1 and is the limit of a
computable sequence.
For any a = (no, n1,... n ), with no < ni < .. < nk, similarly define X, { x : (Vi <
k)x(ni) = 0}. A similar argument to that above shows that lim'Td(X,)/Td(X,) = 1.
Now consider the decreasing sequence ck 2= k+2 ith limit 1. C'! i = no such
that Td(X,) > co and for each k, choose n = nk+1 such that Td(X(no,...,nk,n)) > Ck+1.
This can be done since Ck+1 < Ck. Finally, let Q n= X(no,..., ). Then Td(Q)
limkTTd(X(o,...,nk)) > limkCk .
This result can easily be extended to any bounded measure.
CHAPTER 4
RANDOM CONTINUOUS FUNCTIONS
The following chapter is joint work with George Barpalias, Douglas Cenzer, Jeffrey
B. Remmel, and Rebecca Weber and will appear in the Archive for Mathematical Logic as
an article entitled Al,..., I///.:: Randomness of Continuous Functions [8]. A preliminary
version of this research was originally presented at the Third International Conference
of Computability and Complexity in Analysis in Gainesville, Florida in 2006 by J. B.
Remmel. This preliminary work was published in the referred conference proceedings
as Random Continuous Functions (P. Brodhead, D. Cenzer, and J. B. Remmel) in
Proceedings of CCA 2006 (D. Cenzer, R. Dillhage, T. Grubb and Klaus Weihrauch, eds.),
Information Berichte, FernUniversitit (2006), pages 7689 and in Springer Electronic
Notes in Theoretical Computer Science, Elsevier Science 167 (2007) [15].
Portions of this work were also presented by P. Brodhead at the AMS Fall 2006
Eastern Sectional Meeting (October 2006, Storrs, CT) and the Conference on Logic,
Computability, and Randomness (January 2007, Buenos Aires, Argentina).
4.1 Overview
In C!I Ipter 3, we considered a notion of randomness for closed sets. We do the same
for continuous functions here. An introduction to randomness for reals is provided in
Section 3.2.
This chapter is organized as follows. In Section 4.2, we provide a definition of
randomness for continuous functions and show that it is sound. In Section 4.3, we prove
various results for (images of) random continuous functions perfectness, noninjectivity,
and instances of nonsurjectivity; we also study images of computable elements. In
Section 4.4, we tie random closed sets to random closed functions through images: inverse
images of 0" are random closed sets, but images, in general, are not. Continuing on the
theme of inverse images of 0, in Section 4.5 we consider pseudodistance functions. In
Section 4.6, we briefly consider how the results of Chapters 3 and 4 can be relativized for
nrandomness. Finally, in Section 4.7, we describe some directions for future research.
4.2 Definining Randomness for Continuous Functions
A function F : 2" 2" is continuous iff it has a closed graph. It seems reasonable,
then, to define continuous function f to be random, iff the its graph Gr(F) = {x y
y F(x)} is random. However if [T] is the graph of a function and a E T has even
length, then we must have a^0 E T and a^ l T. This means that the family of closed
sets which are the graphs of functions has measure 0 in the space of closed sets and hence
a random closed set will not be the graph of a function. We need, therefore, a different
method to define randomness for continuous functions. We do this below.
4.2.1 Representing Functions
Given a continuous function F : 2" 2N, we are interested in representing it in such
a way so as to be able to consider a notion of algorithmic randomness. We show below
that any such function F may be represented by infinitely many representing functions of
the form f : {0, 1}* {0, 1, 2*. This will allow us, in the following section, to be able to
represent continuous function as elements of 3', socalled representing sequences, and to
consider a such a function as random if it possesses a random representing sequence.
Information output, the key to representation. For any continuous function
F on 2" and any a {0, 1}*, there is a natural number n and binary string T of length
n such that for all u E I(a), F(u) [n = T. In particular, F(u)(n) = r(n) for every such
u. In general, the length of a may be much larger than n, so we may have to extend a by
several bits to get uniformity of F(u) [(n + 1) within the interval around a's extension.
Representing functions. Taking the above into consideration, we may recursively
represent any continuous function F : 2" 2" by some function f : {0, 1}*  {0, 1, 2}*
as follows. Suppose F is given. Let f(0) = 0. For al = m + 1, having defined f(a [i) = ec
for all i < m, let p = (n,..., nk) be the result of deleting all 2s from (el,..., e,). If for all
u I(a), F(u) [k pj, j E {0, 1}, we may let em+l j. If not we must have em+l = 2;
even if so we allow em+l = 2.
The canonical represention. Notice, from the above, that any continuous F has
infinitely many representing functions f : {0, 1}*  {0, 1, 2}*. The representation which
uses as few 2s as possible we shall call the canonical representation.
4.2.2 Representing Sequences
We want to code the representing function as an element of 3" to discuss its algo
rithmic randomness. To do so, first enumerate {0, 1}* = {0} as ao, oa,..., ordered first
by length and then lexicographically. Thus ao = (0), a1 = (1), 2 = (00), etc. We define
representing sequences below.
Definition 4.2.1. (i) (INF, Rem2) Let INF equal the set of y {0, 1, 2}" such that
{n : y(n) / 2} is infinite and, for y E INF, let Rem2(y) be the result of removing
from x all occurrences of 2.
(ii) (Representing functions) A function f : {0, 1}* {0, 1, 2} represents a function
F : 2N 2N if for all x E 2", the sequence y, defined by y(n) = f(x [n) belongs to
INF and Rem2() F(x).
(iii) (Representing sequences) A sequence r E {0, 1, 2} represents the continuous function
F (written F = F,) if the function fr : {0, 1}* {0, 1,2}, defined by f,(a,) = r(n),
represents F.
(iv) (Labelled 2"trees) Given a representing sequence r, the function f, gives rise to a
labelled 2"tree. We attach, or associate, the value of f,(a) with each node a.
Example 4.2.1 (The 2"tree for the Identity, A Geometric Intrepretation). The identity
function can be represented by placing an e on any node a which ends in e. This can also
be pictured geometrically as representing the graph of F as the intersection of a decreasing
sequence of clopen subsets of the unit square. Initially the choice of f((0)) and f((1))
selects from the 4 quadrants. That is, for example, f((O)) = (0) = f((1)) implies that the
graph of F is included in the bottom half of the square and f((0)) = 0 and f((1)) = (1)
implies that the graph excludes the lower right hand quadrant. Successive values of f
continue to restrict the graph of F in a similar fashion.
4.2.3 A Sound Definition
In this section we define a measure /** on the space of functions F : 2" 2" that
allows us to define the notion of randomness for functions on 2N. In short, a function
is random if it possesses a random representing function, or equivalently, a random
representing sequence. We will show that no canonical representing function can be
random, so that necessarily, the definition of randomness for functions is in this existential
format. It may be, however, that no continuous function has a random representing
function. We show that this is not so. In fact, we show that every random representing
function is continuous. Clearly then, random continuous functions exist and, in fact, A
random continuous functions exist. Therefore the definition is sound and the structure
begins to manifest itself as rich.
The Measure for Randomness. The measure which is used to define random
ness for continuous functions is the Lebesgue measure on the space 3" of representing
sequences. Thus for each new bit of input, there is equal probability 1 that f, gives a new
output of 0 for Fr, gives a new output of 1 for Fr, or gives no new output for Fr. This
measure now induces a measure, p** v, on the space F of continuous functions.
Definition 4.2.2. A function F : 2 2" is random if there is a representing sequence
r E 3 for F that is random with respect to the measure **.
We first show that every random representing function is continuous. The following
lemma is needed.
Lemma 4.2.3. Let E be a finite set and let Q C E" be a II class of measure 0. Then no
element of Q is MartinL6f random.
Proof. Let E {0, 1, 2} without loss of generality. Let Q = [T] where T C {0, 1, 2}* is a
computable tree (possibly with dead ends). For each n, let T = T n {0, 1, 2}" and let
Q. = UJ{I(a ) a c T }.
Let g(n) = p(Q,) = T. Then g(n) is a computable sequence and
lims.g(n) = p(Q) = 0.
This MartinL6f test shows that Q has no random elements. (As observed by Solovay, it
is sufficient to have a computable sequence approaching zero rather than the stricter test
with a sequence of measures g(n) < 2".) O
Theorem 4.2.4. (i) The set of representing functions for total functions has measure
one.
(ii) Every random function is continuous.
Proof. (i) Let r E 3 and suppose that f, does not represent a total function. Then
there is some x E 2" and some r E {0, 1}* such that f,(x n) = r for almost all n.
Without loss of generality we may assume that 7 = 0. Let A be the set of functions
f : {0, 1}* {0, 1}* such that f(a) = 0 for arbitrarily long strings a and let p = p**(A).
Then certainly p < 9, since if r(0) and r(1) are both in {0, 1}, then f, A. Considering
the 9 cases for the initial choices of f((0)) and f((1)), we see that
4 1
S= p + [1 (1 p)2],
9 9
so that p2 + p 0, which implies that p 0. (That is, there are 4 cases in which
If((i)) = 1 for i = 0,1 so that immediately f A, there are 4 cases in which only one
of f((i)) = 0, in which case the remaining function g, defined by g(a) = f(i a) must be
in A, and there is one case in which f((i)) = 0 for i = 0, 1, in which case at least one of
the remaining functions must be in A.) Consequently, the set of representing functions for
total functions has measure one.
(ii) Observe that A is a II class, since fr E A if and only if (Vn)(3a e {0, }t")f,(a) =
0. It follows from Lemma 4.2.3 that no representing function on 2* for a random function
on 2" can be in A. As all functions representing partial functions on 2" occur in A, it
follows that every random function is total. Since the graph of a total function is a closed
set, it follows that random functions are continuous. O
Now the set of MartinL6f random elements of {0, 1, 2}" has measure one and there
exists a A0 MartinL6f real. Hence we have the following.
Theorem 4.2.5. There exists a random continuous function which is A computable.
We also first observe that any continuous function will have a representation which is
not random. In fact, the canonical representation itself can never be random.
Proposition 4.2.6. For any continuous function F, the canonical representation is not
random.
Proof. The idea is that whenever the canonical representation labels a node a with 2,
then the two labels on the successor nodes a0 and a l cannot be both 0, or both 1.
Thus we have the following MartinL6f test. Assume by way of contradiction that r is
random and canonical. Let S, be the set of r E 3" such that r has at least e occurrences
of 2 and such that, for the first e occurrences of 2 in r, the corresponding successor values
are not both 0 or both 1. Since r is random, it must have infinitely many occurrences of 2
and since r is canonical, it must belong to every S,. But each S, is a c.e. open set and has
measure < (7), so that no random sequence can belong to every Se. O
The theorem, in fact, demonstrates the need for the existential part of definition of
random functions. In the following sections we will obtain some additional properties of
random continuous functions.
4.3 Random Continous Functions and Images
4.3.1 Perfect Images, in every instance
In this section we show that all random continuous functions alvii have perfect
images. This is a consequence of the following theorem.
Theorem 4.3.1. If F is a random continuous function, then the image F[2N] has no
isolated elements.
Proof. Let f be the random representing function for F and let Q = F[2]. Suppose by
way of contradiction that Q contains an isolated path y. Then there is some finite r C y
such that y is the unique element of I(r) n Q. Fix a such that f(a) = .
For each n, let S, be the set of all g e F such that for all pi, p2 E {0, 1},
1. g(a pl) is compatible with g(ap2),
2. T z g(a pi), and
3. T E g(a^p2)
Then for any each m < n and each p E {0, 1}", we are restricted to at most 7 of the 9
possible choices for f(p 0) and f(p l). This same scenario applies for all p E {0, 1}1,
so that in general, p(S,) < (7)2" 1
Now for each n, S, is a clopen set in F and thus the sequence So, Si,... is a Martin
Lof test. It follows that for some n, F S,. Thus there are two extensions of a of length
n which have incompatible images, contradicting the assumption that y was the unique
element of Q n I((). E
It follows that the image of a random continuous function is perfect and has contin
uum many elements.
4.3.2 Noninjective Images, in every instance
In this section we show that no random continuous function is injective. We will use
the following theorem en route.
Theorem 4.3.2. For any a E {0, 1}*, the probability that the image of a continuous
function F meets I(a) is alvv > .
Proof. The proof is by induction on lal. Without loss of generality, we assume that
a = 0". For each n > 0, let q, be the probability that F[2N] meets I((0")). Let f be the
representing function for F. For n = 1, there are 9 equally probable choices for the pair
f((0)) and f((1)), breaking down into 4 distinct cases.
Case 1. If f((0)) = (1) = f((1)), then F[2"] does not meet I((0)). This occurs just
once.
Case 2. If f((0)) = (0) or f((I)) (0), then F[2"] meets I((0)). This occurs in 5 of
the 9 choices.
Case 3. If f((i)) 0 and f((1 i)) (1), then F[2"] meets I((0)) if and only if
F() [2"] meets I((0)). This occurs in 2 of the 9 choices, with probability qi.
Case 4. If f((0)) = 0 = f((1)), then F[2"] meets I((0)) if at least one of F() [2"]
meets I((0)). This occurs in 1 of the choices, with probability 1 (1 q,)2. That is, F[2N]
fails to meet I((0)) if both F(o)[2"] and F(o)[2"] fail to meet I((0)).
Putting these cases together, we see that
5 2 1
qi + 9qi+ (2qi q),
99 9
so that qi satisfies the quadratic equation
x2 + 5x 5 = 0.
Thus qi is the unique solution in [0,1] of this equation, that is,
v'45 5
55
2
which is indeed > .75.
Now let q, = q and let q,+1 = p. Once again we consider the 9 initial choices, now
breaking down into 6 distinct cases.
Case 1. If f((0)) = (1) f((I)), then F[2"] does not meet I((0"+1)). This occurs
just once.
Case 2. If f((0)) = (0) f((1)), then F[2"] meets I((0"+)) if and only if at
least one of F(o) and F(1) meets I((0")). This occurs just once, and with probability
1(1 q)2 2qq2.
Case 3. If f((i)) (0) and f((1 i)) (1), then F[2"] meets I((0"n+)) if and only if
F(4)[2"] meets I((0O)). This occurs in 2 of the 9 choices, with probability q.
Case 4. If f((i)) 0 and f((1 i)) (1), then F[2"] meets I((0"+')) if and only if
F() [2"] meets I((0+'1)). This occurs in 2 of the 9 choices, with probability p.
Case 5. If f((0)) = 0 = f((1)), then F[2"] meets I((0"+')) if at least one of F(i)[2"]
meets I((0"+1)). This occurs just once, with probability 1 (1 p)2.
Case 6. If f((i)) 0 and f((1 i)) (0), then F[2"] meets I((0"+')) if at least one
of the following two things happens. Either F(i)[2"] meets I((0+'1)), or F(li)[2N] meets
I((0")). This occurs in 2 of the 9 choices, with probability 1 (1 p)(l q).
Putting these cases together, we see that
2 12 2 2 12
p p p pq + q q,
3 9 9 3 9
so that p = qn,+ satisfies the equation
p2 + 3p + 2pq 6q + q2 = 0.
We note that for p = q, the solutions are p = q = 0 and p = q = This explains the value
Sin the statement of theorem.
Now assume by induction that q > . Suppose by way of contradiction that p < . It
follows that
9 9 3
S+ + q 6q+ q2 > 0.
16 4 2
Simplifying, this implies that 16q2 72q + 45 > 0. But this factors into (4q 3)(4q 15)
and is only > 0 when either q < 4 or q > . Since the latter is impossible, we obtain the
desired contradiction that q < . D
Theorem 4.3.3. No random continuous function is injective.
Proof. Let p be the probability that an arbitary continuous function F is injective. It
follows from Theorem 4.3.2 that there is a chance that F has a zero in I(0) and also
in I(1), so that p < i7. Now in general, if F is injective, then it must be injective when
restricted to I(0) and when restricted to I(1). It follows that p < p2, which happens only
for p = 0 and p = 1, given that 0 < p < 1. Since p ,< it follows that p = 0, as
desired. This can be reformulated as a MartinL6f test as follows. First we observe that
F is injective if and only if, the images of each pair of ldii ;il 1 intervals I(a) and I(r) are
disjoint. Let
D(, 7) = {F : F[I(a)] n F[I(r)] =0.
Then D(a, 7) is uniformly c.e. since F[I(a)] n F[I(r)] 0 if and only if there exists n such
that for all extensions a' of a and r' of 7 of length n, f(a') and f(7') are incompatible.
Now let
Sm, {F: (Va, 7 {0,1}m)(a7 / T 7 F E D(a, T)}.
It follows from the observation above that F is injective if and only if F E Pm S,. The
argument above shows that p**(SI) < T6 and that p**(S,m+) < l**(S,)2 and hence
7
It follows that {Sm : m e u} is a MartinLof test and therefore no random continuous
function may belong to every S, and hence no random continuous function can be
injective. E
4.3.3 Nonsurjective Images, in instances
In this section we show that random continuous functions are not necessarily onto.
Definition 4.3.4 (F,, the restriction of F to I(a)). For any function F on 2" and any
a E {0, 1}*, define the restriction F, of F to I(a) by
F,(x) = F(a^x).
Clearly any such restriction of a random continuous function will be random, but
more can be said. Recall van Lambalgen's theorem, Theorem 3.2.12.
Proposition 4.3.5. F is a random continuous function if and only if the functions F(o)
and F(1) are relatively random.
Proof. Let r represent F. Suppose first that F is random. It follows, as in Corol
lary 3.3.12, that F(o) E F(1) is random and hence F(o) and F(1) are relatively random by van
Lambalgen's theorem.
Next suppose that F(o) and F(1) are relatively random and let ri represent F(i) for
i = 0, 1. Let d be any martingale, which we think of as betting on r. Then for i = 0, 1,
we can define a martingale di with oracle rli as follows. We will give the definition for do
and leave dl for the reader. Given a = ro(0),..., ro(2 + q 2) where 0 < q < 2P, use rl
to compute T = r(0),..., r(2"+1 + q 2) and then define di to bet in the same proportion
as d. That is, di(a j'j)/di(a) = d(> j)/d(r) for j < 3. Thus for any node on the left side
of the labelled tree for F, do is making the same bet on the next label that d would have
made, and similarly for di and the right side.
Since the F(i) are relatively random for i = 0, 1, it follows that di does not succeed
and hence there exist upper bounds Bi for {di(ri [n)},E. But it follows from the above
definitions of di that for any p,
d(r [2P 2) do(ro 2P 1) d(ri [2P 1).
This is because the martingale d alternates using do and dl and the result can be viewed
in each alternation as multiplying the capital by some factor. Then in general, for 0 < q <
2P,
d(r [2+ + q 2) do(ro [2P + q 1) di(r [2 1)
and
d(r [2P+ + 2 + q 2) = do(ro [2" 1) d(r, 2P + q 1).
It follows that Bo B1 is an upper bound for {d(r[k) : k E N}, so that d does not succeed
on r. E
Corollary 4.3.6. A random continuous function is not necessarily onto.
Proof. It follows from Proposition 4.3.5 that, for any r e {0, 1}*, there is a random
continuous function with image C I(r). Thus a random continuous function is not
necessarily onto. O
4.3.4 Images of computable elements
In this section we will show that the image of computable element under a random
continuous function is necessarily noncomputable. In fact, it is random. We need the
following proposition.
Proposition 4.3.7. Suppose A C B are two finite sets of symbols. Given X E BN,
let X E A" be the sequence obtained by deleting all symbols in B A from X. If X is
1random, then X is 1random.
Proof. Given X, X as in the proposition, suppose X is not random and let d be a
constructive martingale on A" that succeeds on X. We will construct a martingale d on
B" that succeeds on X. Essentially, d will keep its capital constant on symbols in B A;
it will bet according to d, repeating its bets after bits which hold symbols from B A.
Define d(A) = d(A), and for a E B* and a the corresponding string of A*,
d(a; x) d(c7d ) xcA
d(a) x B A
The function d is clearly constructive, since d is. To show d is a martingale, consider
the sum
d(,7x) dC rx) d(,7) + Y d(17)
d(axx)d (
xEB x A BA
Sd() Sd( ( + d(a)lB Al d(a)[AI + IB AI].
xECA
It remains to show that d succeeds on X. However, that is clear, as on bits which
are in X but not X, d keeps its capital constant, and on bits from X, it acts exactly as d
would. Therefore since d succeeds on X, d succeeds on X and X is nonrandom. O
It is easy to see that, for any random continuous function F and any computable real
x, F(x) is not computable. This follows from our next result.
Theorem 4.3.8. If F is a random continuous function, then, for any computable real x,
F(x) is not computable.
Proof. Suppose that F is random and let x and y be computable reals. For each n, let
S, = {G: G(x[n) y[n}.
Then So, Si,... is an effective sequence of c.e. open sets in F, and an easy induction
shows that p**(S,) = (2/3)". This is a MartinLof test and it follows that F S, for some
n, so that F(x) / y. O
We now strengthen this result to show that the image of a computable element is
random.
Theorem 4.3.9. If F is a random continuous function, then, for any computable real x,
F(x) is a random real.
Proof. Suppose that F is random with representing function fr, let x be a computable
real and let y = F(x). Define the computable function g so that, for each n,
7g(,) X [rn.
By the VonMises(C'lnl ! Wald Computable Selection Theorem, the subsequence
z(n) = r(g(n)) is random in {0, 1,2}". Now y = F(x) may be computed from z by
removing the 2's. Thus F(x) is random by Proposition 4.3.7. E
We note that Fouche [45] has used a different approach to randomness for continuous
functions connected with Brownian motion, first presented by Asarin and Prokovskiy [5],
and has shown that, under this approach, it is also true that for any random continuous
function F, F(x) is not computable for any computable input x.
It follows that a random function F can never be computably continuous and hence
the graph of F is not a IIt class.
4.4 Random Closed Sets arising from random continuous functions
4.4.1 A Positive Result: Inverse Images of 0'
In this section we prove that for any random continuous function F, the set Z(F)
{x : F(x) = 0} is a random closed set. For any subset S of C, let Zs = {F F : Z(F) e
S}.
Lemma 4.4.1. For any open set S, p**(Zs) < i*(S).
Proof. It suffices to prove the result for intervals S = I(a). We will show by induction
on a, that p**(ZI(,)) ()1I4, whereas of course p*(I(a)) = ()1'1. Recall from Corollary
4.4.3 that 0 E F[2"] with probability exactly 3. For a = 1, there are two distinct cases.
Case I Suppose first that a = (i), where i E {0, 1}. Then F E Zs if and only if F has
a zero in I((i)) and has no zero in I((1 i)). Now F has a zero in I((i)) if f((i)) E {0, 2}
and if the restricted function has a zero, which gives probability 2 3= Thus the
combined probability that F E Zs is .
Case II Suppose next that a = (2). Then F E Zs if and only if F has zeroes in both
I((0)) and I((1)). It follows from the argument in Case I that p* (Zs) = .
Notice that Z{} {F : F has no zeroes} has positive measure 1 but p*({0}) = 0.
Now suppose a = n and let r = a i; suppose by induction that p**(Ziy,)) <
p*(JI()). Interpret T as the code for a (finite) binary tree and let p E {0, 1}* be the
terminal node of that tree such that i indicates the branching of p. Again there are two
cases.
Case I Suppose first that i c {0, 1}. Then F E ZI(T) if and only if F E ZI(,) and
furthermore F has a zero in I((p i)) and has no zero in I((p 1 i)). It follows as above
that p**(ZI(T)) (**(Z( ) ( )"1.
Case II Suppose next that i = 2. Then F E ZI(T) if and only if F has zeroes in both
I(p 0) and I(p^1). It follows as above that p**(ZI(T)) p**(Z(,)) (n)"+1
An arbitrary open set is a dliP iil union of intervals and thus the desired inequality
can be extended to open sets. E
Theorem 4.4.2. For any random continuous function G : 2" 2", the set of zeroes of G
is either empty or is a random closed set.
Proof. Suppose that G is a random continuous function which has at least one zero, and
let So, Si,... be a MartinL6f test in C. Then there is a computable function Q such that
Si = UI(a7(i,,)). We may assume without loss of generality p*(Si) < 2i2 and that each
Si is not clopen and that, for each i, the intervals I(ar(i,)) are pairwise disjoint. We will
define a MartinL6f test So, S',... in the space F and use the fact that G must satisfy
{Si}iE to show that Z(G) satisfies {Si}iE.
Fix an interval I(a) in C and let C, = ZI(). Observe that there is a clopen set
B, C 2N and a corresponding finite set To, ... Tk1 of strings such that B, = Uj
associated with a such that, for any Q E C with code r, rE I(a) if and only if Q C B,
and Q n I(rj) / 0 for all j < k. It follows that C is a difference of IIo classes. That is,
F E C if and only if the following two conditions hold.
(i) For each j, F has a zero in I(r7); by compactness, this is equivalent to wicing that
for any there is an extension T {0, 1} of T such that f(r,) E {0, 2}171, where f is the
function on strings representing F.
(ii) F has no zeroes outside of B. Let 2" B = UTAI(T). By compactness, F has no
zeroes outside of B if and only if
(E)(VT e A)(VT' > T) [T' f > (E(m)(f(T'[m) 1)]. (41)
Note that the measure of Co may be computed uniformly from a given the calculation
from Corollary 4.4.3 that whenever f(a) c {0, 2}1'1, then the probability that F has a
zero in I(a) is exactly . For each o, we will uniformly compute a c.e. open set S, CF
such that C, C B, and such that p**(B,) < 2 p. **(C,). There are two stages in the
construction of B,.
Stage I: Let U be the set of codes a' for partial functions f' such that 41 holds with
f' in place of f, and such that furthermore for every j and f such that f' is defined on all
lengthf extensions 7 of T, there is such a T with f'(p) E {0, 2} Vp _ '. It is clear that for
any F E Co, there exists a' E U with F E I(W') and hence
C' C U {I(J') : e U}.
As usual, we may then uniformly compute from U a set U' such that the intervals I(c') for
a' E U' are pairwise dl ii.il in F and
U{I(uT) : U 7 } {U JI(t') :t' E U'T}.
For each o' E U', let Q(c') C I(a) be the II class in F consisting of those extensions of a'
which actually have zeroes in each I(j). Then in fact we have
C = U Q(') : 7 u'}.
As noted above, we can actually compute the measure p**(Q(c')) uniformly from a'
by expressing Q(c') as an effective decreasing intersection of clopen sets. Thus for
each a', we can compute a clopen set B(a') such that Q(a') C B(a') C I(a') and
p**(B(a') 2 **(Q(a')). Let
B,= U B(a') : U'}.
Then we have C, C B, and p**(B,) < p**(C,).
Finally, for each i, let
S( = U i,n)
Then by Proposition 4.3.7, p**(S ) < 2 p*.(Si) < 2 1 and therefore there exists some
i such that G Sj, since F is random. But this means that Z(G) Si and hence Z(F)
meets the MartinLof test. Thus Z(F) is random, as desired. O
4.4.2 A Negative Result: Images, in general
In general, the image of a random continuous function need not be a random closed
set. To see this, recall the statement of Theorem 4.3.2. That is, given a {0, 1}*, the
probability that the image of a continuous function F meets I(a) is .1'. i. > 3. We
obtain the following corollary.
Corollary 4.4.3. For any y e 2",
(a) p**({F : yeF[2"]}) ;
(b) there exists a random continuous function F with y e F[2N].
Proof. (a) Let p be the probability that y e F[2N]. It follows that for each a {0, 1}",
the probability that y E F[I(a)], given that f(a) is consistent with y, also equals p. It
follows from the proof of Theorem 4.3.2 that p = .
(b) Since the random continuous functions have measure 1 in C(2"), it follows that
some random continuous function has y in the image. O
This allows us to demonstrate our result.
Theorem 4.4.4. The image of a random continuous function need not be a random
closed set.
Proof. It was shown in Theorem 3.4.12 that a random closed set has no computable
members. Let F be a random continuous function with 0 in the image, as given by
Corollary 4.4.3. Then F[2N] is not a random closed set. O
4.5 PseudoDistance Functions
In section 4.4.1 we showed that if A is a random continuous function, then A'(0")
is a random closed set, if it is nonempty. This motivates the study of pseudodistance
functions.
Definition 4.5.1. A : 2" 2" is a pseudodistance function for Q C 2" if A is continuous
and A1(0w) = Q.
Comment 4.5.2 (Background). The name comes from a modification of the distance
function distQ : 2" [0, 1] for a closed set Q. For x E 2", distQ(x) is defined to be
min{d(x, y) : y Q} where d is a metric on 2" given by
0 if x = y;
d(x,y) =
2" if n is the least such that x(n) / y(n).
This may be viewed as a computable mapping from 2" x 2" into 2" by representing 0 as
0" and 2" as 0"'10. From this viewpoint, we may view distQ similarly:
0" if x e Q;
distQ(x) =
0"'10 otherwise, where n is the least such that x n TQ.
Every closed set has a characterization in terms of pseudodistance functions, as
follows.
Theorem 4.5.3. Q C 2" is closed iff there is pseudodistance function for Q.
Proof. First suppose that Q is closed and Q = [T]. Define a map A : 2< 2<" by
recursion as follows, with initial mapping 0 v> 0.
A((a)'0 if a^i E T;
(a) 1 otherwise.
Letting A : 2" 2 be defined so that A(X) is the unique Y E n,[A(X [ n)], we obtain
the required pseudodistance function.
Now suppose that A : 2" 2" is a pseudodistance function for Q. This means that
there is some A : 2< 2<" such that A(X) is the unique Y c n[A(X [ n)]. Define the
required tree T (so that Q = [T]) as follows. Put a E T iff A(a) E 0". O
For effectively closed sets, such a characterization in terms of pseudodistance
functions might not seem as immediate, as an effectively closed set may possess a noncom
putable distance function. Nonetheless, the we have the following theorem.
Theorem 4.5.4. Q C 2" is effectively closed iff there is computable pseudodistance
function for Q.
Proof. The follows as in the proof of Theorem 4.5.3, except that A, in both directions, is
now computable. O
It seems plausible that there is a pseudodistance function characterization for random
closed sets. Looking first to distance functions, every random closed set possess the non
random distance function distQ. (To see why distQ is nonrandom, note that if a TQ,
then d is constant on the interval I(a).) If a characterization exists, as in the case for
effectively closed sets, it appears to be not so immediate. On the other hand, we know, by
Theorem 4.4.2, that if Q is a closed set with a random pseudodistance function, then Q is
random. This leads us to the following conjecture.
Conjecture 4.5.5. Q C 2" is random closed iff there is a random pseudodistance
function for Q.
4.6 nRandomness
Recall that a real is n + 1random if and only if it is 1random relative to .I (see
Remark 3.2.11). Now the analogue of Theorem 3.2.10 also holds for {0, 1, 2} for our
measures p* or /**. Therefore our approach also allows us to define the notion of n
random continuous functions (or nrandom closed sets), as follows. A continuous function
F : 2N 2N (or closed set) is defined to be nrandom if and only if it is Martin Lof
random relative to I' '. One can then easily relativize the results of previous sections to
obtain similar results for nrandom continuous functions.
4.7 Future Work
We close this chapter noting that random Brownian motions as studied by Fouche [45]
are a special case of random continuous functions on the real line. This is another area
of interest for further research. That is, we would like to extend the notion of a random
continuous function to functions on the real unit interval [0, 1] and the real line R by
representing functions again in terms of the images of subintervals. We conjecture that
a random continuous real function cannot be left or right computable and in fact, not
weakly computable. We also conjecture that a random continuous function is nowhere
differentiable.
CHAPTER 5
CONTINUITY OF CAPPING IN CBT
The following chapter is joint work with Angsheng Li and Weilin Li and will appear
in the Annals of Pure and Applied Logic as an article entitled C 'ii<.;,.;/ of Capping in
CbT [16]. For this project, P. Brodhead acknowledges support from the National Science
Foundation (under grant number 0714151 as the principal investigator) to conduct this
joint work in Beijing during the summer of 2007 as part of the East Asia and Pacific
Summer Institutes (EAPSI).
This work was presented by P. Brodhead at the First Joint AMSNZMS Meeting
(December 2007, Wellington, New Zealand).
5.1 Introduction
Given sets A, B C w, we i that A is Turing reducible to B, if there is an oracle
Turing machine i'*, such that A 4= B (denoted by A T B). Furthermore, if the bits
of oracle queries are bounded by a computable function, then using recent nomenclature
from Soare [88] we ;i that A is bounded Turing reducible to B, written A
literature often refers to this as the weak truth table reducibility, written
and a bounded Turing (or bT, for short) degree is the equivalence class of a set under
the Turing reductions and the bounded Turing reductions respectively. A degree is called
computably enumerable (c.e.), if it contains a c.e. set. Let C and CbT be the structures
of the c.e. degrees under the Turing reductions and the bounded Turing reductions
respectively.
During the past decades, the studies of the structures C, CbT focused on that of the
algebraic properties, leading to n1 iP r achievements such as the decidability results of
the Zitheory of C, and the E2theory of CbT (AmbosSpies, P. F iP r, S. Lempp and M.
Lerman [3]), and the undecidability results of the E3theory of C (Lempp, Nies, and
Slaman [63]), and of the E4theory of CbT (Lempp and Nies [62]). This progress brings the
decidability problems of the E2theory of C, and the E3theory of CbT into sharper focus,
for which new ingredients are welcome.
In the recent years, the study of the computably enumerable degrees has focused on
Turing definability in the structure C. For instance, Slaman asked in 1985 if there are any
c.e. degrees that are incomplete and nonzero which are definable in the c.e. degrees C.
This question of Slaman is still open. A natural approach to this problem is to find some
definable substructures of C that have nontrivial minimal/maximal and/or least/greatest
members. As a result, topics such at the continuity of the c.e. degrees, started by Lachlan
in 1967, have renewed interest.
In this chapter, we demonstrate the continuity of capping in CbT. This refutes the
existence of a maximal nonbounding degree. It also brings the question of the E3theory
of CbT into sharper focus, as the statement is one of Escomplexity.
To motivate these ideas further, we begin with a brief history of relevant continuity
results in Section 5.2. This motivates our main result and method of proof, described in
Section 5.2.2. The main substance of the proof involves demonstrating that Theorem 5.2.3
holds, that local noncappability holds in CbT. Sections 5.35.6 are devoted to proving this
theorem.
5.2 Continuity Results
The Turing degrees form an uppersemilattice; that is, each pair of elements a, b
has a least upper bound (or join) a V b. A greatest lower bound a A b may or may not
exist. Given a (bounded) Turing degree a, we iv that a is capable if a (bounded)
Turing degree b / 0 exists such that a A b = 0. We iv that a is cuppable if there is a
degree b / 0' such that a V b = 0'. The study of continuity properties the (bounded)
Turing degrees is with respect to meets and joins, and related notions such as capping and
cupping.
5.2.1 Continuity Results in C
In 1979, Lachlan [60] proved the existence of a nonbounding c.e. degree namely, a
noncomputable c.e. degree with no minimal pair below it; Cooper demonstrated in 1974,
that no high c.e. degree can be nonbounding [30]. Returning to the Slaman question,
Downey, Lempp, and Shore demonstrated in 1993 that Cooper's nonbounding degree
could be made high2 [41], leading to the possibility of a maximal nonbounding c.e. degree.
Such a degree could be used to show the existence of a discontinuity, which could be
used to prove the definibility of a c.e. singleton. (To see the former, note that if b > 0
is nonbounding, then for any c.e. a > b, there is some minimal pair r A s below a. Then
either b A r 0 or b As 0, but neither a A r nor a A s equals 0.) However, Seetapun
refuted this possibility (albeit earlier in 1991), demonstrating the nonexistence of a
maximal nonbounding degree [84]. Welch proved a complimentary result in 1981: there is
no maximal bounding degree, in the sense that for all a / 0', there is are b, c such that
bAc 0 and b,c % a [95].
Continuing with capping results, Harrington and Soare [51] proved in 1989, the
nonexistence of maximal minimal pairs that is, for any nontrivial minimal pair (a, b) of
c.e. Turing degrees a, b, there exists a c.e. Turing degree c > a such that (c, b) is still a
minimal pair. Seetapun [84] showed an even stronger result, the continuity of capping: for
any c.e. Turing degree b / 0, 0', there exists a c.e. Turing degree a > b such that for any
c.e. Turing degree x, if x < a, then a A x = 0 if and only if b A x = 0.
AmbosSpies, Lachlan, and Soare [4] proved the dual case of the Harrington and
Soare's result: for any nontrivial splitting x, y of 0', there exists a c.e. degree a < x such
that a V y = 0'. Cooper and Li [31] showed the dual of the Seetapun theorem, that for any
c.e. Turing degree b / 0, 0', there exists a c.e. Turing degree a < b such that for any c.e.
Turing degree x, x V a = 0' if and only if b V x = 0', answering Lachlan's in i, i subdegree
problem.
5.2.2 Continuity Results in CbT and Main Result
There are few results in the topic of definability in the c.e. bTdegrees, CbT. For
instance, we know nothing about the Slaman problem as described in Section 5.1 or the
characterization of definable ideals in CbT. As a matter of fact, little is known of the
continuity properties of the c.e. bTdegrees. An interesting partial result was given by
Stob [90]: in both C and CbT, there are c.e. degrees ao, al > 0, with ao being the unique
complement of al in the interval [0, ao V all, such that if b < ao V al is capable with
any x < ao (i.e. b A x = 0), then x < a. This can be interpreted as a continuity or
discontinuity result in both C and CbT.
The main result of this chapter is the following theorem, an analogue of Seetapun
continuity result.
Theorem 5.2.1 (Continuity of Capping in CbT). For any c.e. bTdegree b : 0, O', there is
a c.e. bTdegree a > b such that for any c.e. bTdegree x, b A x 0 a A x 0.
For this, it suffices to prove Theorem 5.2.3 below, an analog of the Seetapun local
noncappability theorem for the c.e. Turing degrees [84].
Definition 5.2.2 (Local N. i1 '11' ,1ility). A degree b : 0 is locally noncappable if there
is some a > b such that for all x < a, if x A b = 0 then x = 0.
Theorem 5.2.3 (Local Noncappability in CbT). For any c.e. bTdegree b, if b / 0, 0',
then there is a c.e. bTdegree a > b such that if x < a is noncomputable, then x A b / 0.
Proof of Theorem 5.2.1. Assuming Theorem 5.2.3, we can see Theorem 5.2.1. Given
b, let a be the degree in Theorem 5.2.3. For a fixed x e CbT, by Theorem 5.2.3, we con
sider only the case where x % a. Clearly if x A a = 0, then x A b = 0. Assume a A x : 0.
We can choose a c.e. bTdegree y such that y / 0 and y < a, x. Therefore 0 < y < a and
by Theorem 5.2.3, we have y A b / 0, so that x A b / 0. Theorem 5.2.1 follows. O
As a consequence of Theorem 5.2.1, no maximal nonbounding degrees exist in the
c.e. bTdegrees. To see this, suppose b : 0 is a degree which bTbounds no minimal
pairs in CbT, and let a > b be the degree in Theorem 5.2.1. We claim that there are no
bTminimal pairs below a. Suppose to the contrary that x, y is a minimal pair below a.
Since a Ax = x : 0 and a A y =y 0, we can choose nonzero xl to be below both x
and b, and nonzero yi below both y and b. Then (xi,yi) is a minimal pair below b, a
contradiction. An alternative approach is to use the fact that a maximal nonbounding c.e.
degree is equivalent to a nonbounding degree which is not locally noncappable [49, 84].
Consequently, by Theorem 5.2.3, no maximal nonbounding degree can exist.
Our approach to the proof of theorem 5.2.3 is similar to Seetapun's approach for the
c.e. Turing degrees, but it is nonobvious due to the computable bounds of oracle query
bits in both the conditions and conclusions of requirements. That is, bTreductions are
stronger than Turing reductions. So when we require the reductions being built to be
bTreductions, we must satisfy stronger conditions and, in this sense, the problem becomes
harder to solve. For example, A. Li, W. Li, Y. Pan, and L. Tang [66] have shown that the
solution to the i i" subdegree problem in CbT (i.e. the dual to the continuity problem)
is completely different from the result for C (see Cooper and Li [31]). They show that the
statement of the solution in C fails badly in CbT: there exist c.e. bTdegrees a, b such that
0 < a < 0', and for any c.e. bT degree x, b V x = if and only if x > a.
Our approach might not be the only one. Klaus AmbosSpies proved that for any c.e.
set, its Turing degree is capable in the Turing degrees iff its bTdegree is capable in the
bTdegrees [2]; we thank an anonymous referee for pointing this out. Therefore, another
possible approach might be to prove, if possible, that for any two c.e. sets, their Turing
degrees form a minimal pair in the Turing degrees iff their bTdegrees form a minimal pair
in the bounded Turing degrees. As consequence, our continuity result would immediately
follow from Seetapun's continuity result. We comment that although our continuity proof
might be nonobvious, from the above perspective, oftentimes bTdegrees can be handled
much more easily than Turing degrees [1]. AmbosSpies provides various examples [1]. For
example, density of the c.e. bTdegrees can be proved by a finite injury priority argument,
whereas the same result requires an infinite injury argument for the c.e. Turing degrees.
The rest of this chapter is devoted to proving Theorem 5.2.3, the main result. In
section 5.3, we formulate the conditions of the theorem by requirements; in section 5.4, we
arrange all strategies to satisfy the requirements on the nodes of a tree, or more precisely,
the 1,<.:' i.:/; tree T. In section 5.5, we use the priority tree to describe a stagebystage
construction of the objects we need. Finally, in section 5.6 we verify that the construction
in section 5.5 satisfies all of the requirements, finishing the proof of the theorem.
Our notation and terminology are standard and generally follow Soare [86]. During
the course of a construction, notations such as A, KP are used to denote the current
approximations to these objects, and if we want to specify the values immediately at
the end of stage s, then we denote them by As, I[s] etc. For a partial computable
(p.c., or for simplicity, also a Turing) functional, K iv, the use function is denoted by
the corresponding lower case letter Q. The value of the use function of a converging
computation is the greatest number which is actually used in the computation. For a
Turing functional, if a computation is not defined, then we define its use function = 1.
During the course of a construction, whenever we define a parameter, p iv, as fresh, we
mean that p is defined to be the least natural number which is greater than any number
mentioned so far. In particular, if p is defined afresh at stage s, then p > s.
5.3 Requirements and Strategies
In this section we provide the requirements and strategies for proving Theorem 5.2.3.
We restate it here for convenience.
Theorem 5.2.3 (Local Noncappability in CbT). For any c.e. bTdegree b, if b / 0, 0',
then there is a c.e. bTdegree a > b such that if x < a is noncomputable, then x A b / 0.
5.3.1 The requirements
Given a c.e. set B, we will build a c.e. set A to satisfy the following requirements:
e: A (B) V K bT B
Re : Xe= ,e(A,B) (3 c.e. De)[De
Se,i: De / Ai X V T VB
where e, i E u, {(4e, e, Xe) : e E W} is an effective enumeration of all triples (+, ', X) of
all bounded Turing (bT, for short) reductions (, T, and of all c.e. sets X; {A I i E w} is
an effective enumeration of all partial computation functions A; and K is a fixed creative
set. De for all e, are c.e. sets built by us.
Let a, b, x, d be the bTdegrees of A D B, B, X, D, respectively. By the P
requirements, a > b (unless b was already the degree of 0'), and by the Rrequirements,
if x < a there is a d below both x and b such that d / 0 unless either x = 0 or b = 0.
Therefore the requirements are sufficient to prove the theorem.
Before describing the strategies, we introduce some conventions of the bounded
Turing reductions. We will assume that for any given bounded Turing reduction KP or T,
the use functions 0 and i will be increasing in arguments.
5.3.2 A Pstrategy
A Pstrategy will try to satisfy a Prequirement, P i (we drop the index in the
following discussion). We use a node on a tree, 7y i, to denote a Pstrategy. It aims
to ensure that if A = T(B), then there is a bounded Turing reduction A such that
A(B) = K. Therefore the Pstrategy 7 will try to build a bounded Turing reduction A.
A will be built by an cusequence of ;/. /' k. Each cycle k of 7 will be responsible for
defining A(B; k) as follows: first 7 chooses a fresh witness a(k) and waits for a stage, v
, at which we have I(B; a(k)) =0 = A(a(k)). When this occurs, we define A(B; k)
to be K(k) with use function 6(k) = (a(k)). Since i is partial computable, so is the use
function 6 of A. We will ahi,x assume that whenever B changes below the 6use, 6(k)
v, the corresponding computation A(B; k) becomes undefined simultaneously. Suppose
that at a later stage s > v, k is enumerated into K, and B has not changed since A(B; k)
was last created, then A(B; k) / K(k). In this case, we enumerate a(k) into A so that an
inequality '(B; a(k)) / A(a(k)) is created. The key point is that, if A(B; k) / K(k) is a
permanent inequality, so is '(B; a(k)) 0 / 1= A(a(k)).
The Pstrategy 7 will start cycles k in increasing order of k. Cycle k acts only if the
following conditions occur:
1. For all k' < k, A(B; k') = K(k').
2. Either a(k) T, or A(B; k) T and I(B; a(k)) = A(a(k)), or A(B; k) 1= 0 / 1= K(k).
As we have seen in the above analysis, if there is a permanent inequality between
A(B) and K, there is a corresponding permanent inequality between T(B) and A. Since
A is a bounded Turing reduction (with use bound 6), we have that if A is built infinitely
many times, then A(B) is total, and A(B) = K. Hence K is bTreducible to B. Suppose
this never occurs, then the Pstrategy 7 acts only finitely many times, which will be
denoted by 1. Therefore a Pstrategy 7 has only one possible outcome 1, unless K
5.3.3 An Rstrategy
Before describing the Rstrategy, we introduce a convention of the bounded Tur
ing reduction 4. We assume that for any x and any s, if x enters X at stage s, then
S(A,B;x)[s] = 1.
Given an Rrequirement, R ,i, we define the length function of agreement as usual.
That is to ,: At stage s, the length function of agreement f between 4(A, B) and X is
defined as the largest x such that N(A, B) and X agree on all values y < x:
= (X, (A, B))[s] maxx : (Vy < x)[(A, B;y)[s]] = X(y)[s]}
Stage s is said to be Rexpansionary if the length function of agreement increases; that
is, if for all v < s, [s] > [v]. At Rexpansionary stages, an Rstrategy builds bounded
Turing reductions H(X), _(B) (with use functions 0, (, respectively) so that for the least
undefined x < f:
6(X, x) I= D(x) with 0(x) = x and E(B, x) = D(x) with (x) = 0(x) (51)
where 0 is the use of 4(A, B), and D is a c.e. set, whose elements are enumerated into it
by lower priority Sstrategies associated with R, to satisfy the Rrequirement.
Suppose that a is an Rstrategy. As above, the use functions 0 and of the bounded
Turing reductions 6 and E built by a are the identity function and 0 respectively, so that
both 6 and E are bounded Turing reductions. (Notice that 4(A, B) is a bounded Turing
reduction, so that the use 0 is a partial computable function.)
To satisfy O(X) = D and E(B) = D, the Rstrategy a will impose the following
constraints on all Sstrategies with the same global index as the Rstrategy a:
For any s, and any d, d is allowed to be enumerated into D, at stage s only if both
O,(X; d) and 2E,(B; d) are undefined during stage s.
We assume that for the bounded Turing reductions 6 and E, any computation will
automatically become undefined, whenever the oracle changes below the corresponding
use.
By the building of 0, and E,, and by the constraints of a, we have that if 0, and oa
are built infinitely many times, then both O,(X) and ES(B) are total, and both equal Da.
Hence R is satisfied.
Therefore the key point towards the satisfaction of R is that if there are infinitely
many Rexpansionary stages, then both O, and Eo are built infinitely many times.
We thus define the possible outcomes of the Rstrategy a by
0
to denote infinite and finite expansionary stages respectively.
5.3.4 An Sstrategy
Suppose that we want to satisfy an Srequirement, Se,i v. For simplicity, we use
R and S to denote 7R and Se,i respectively. Suppose that a and 3 are the 7 and S
strategies respectively. Let a^(0) C 3.
3 attempts to find some d such that A(d) 1= 0 with an expectation of enumerating d
into D to create an inequality A(d) = 0 / 1 = D(d). However 3 can enumerate a number
d into Da at a stage, s ,, only if both O,(X; d) and E (B; d) are undefined during
stage s as required by the Rstrategy a. Therefore, 3 will prepare a sequence of possible
candidates c's such that
A(c) 1 0 =D(c),
Oa(X; c) and
(B; c) 1.
For the largest c, we build a partial computable f3 as follows:
for every y < ~a(c), if f3(y) is undefined, then define fy(y) B(y).
define d(3) = c, and set c to be undefined, which allows us to define a larger c.
Suppose that there is an error between f3 and B, in the sense that there is a y such
that f3(y) 1= 0 / 1 = B(y) occurs at a stage, v ~,. Then we open an Agap:
build a partial computable function g to simulate Xa as follows: for every x < d(3),
if g(x) is undefined, define g3(x) = Xa(x),
set f3 to be totally undefined (the fa proves wrong, so it is cancelled),
drop the Arestraint by defining rA(3) = 1, and
create a link (a, /3).
[Notice that at stage v, _,(B; d(3)) is undefined due to the Bchange in the domain
of fa. We regard this as a Bpermission for the enumeration of d(o3) into D,. This B
permission will be kept until the current link (a, 3) is either travelled or cancelled so that,
in either case, the link is removed.]
Suppose that 3 creates a link (a, /3) at stage v. Then the link (a, /3) will be travelled
at the next aexpansionary stage s > v. Now we consider two cases:
Case 1. There is an error between gp and X,.
In this case, there is an x <, d(3) which has entered X, since stage v. Therefore
Oc(X; d(/3)) is currently undefined. Together with the condition that E,(B; d(3)) 1,
found at the stage we created the current link (a, /3), 3 is qualified to enumerate d(3) into
Do. S is satisfied by A(d(3)) =0 / 1= D,(d(3)).
Case 2. Otherwise, we know that gp is correct during the gap. Therefore we preserve
g9 on its domain until 3 opens another Agap. For this, we implement:
for every y < Q(d(3)), if f3(y) is undefined, then define f(y) = B(y), and
define the Arestraint rA(/) of 3 to be Q(d(3)).
[Notice that although we have Arestraint at this stage, X, may change due to a
Bchange below Q(d(3)). The definition of fp at this stage allows us to immediately open
an Agap once such a Bchange occurs, in which case, we do not increase the domain of g3
but resume with the old candidate d(3).]
Therefore the Sstrategy 3 is a gap/cogap strategy. It will build partial computable
functions ft and g3 and will proceed as follows:
1. Define a possible candidate c(/3) as fresh.
2. (Building fa) Wait for a stage v at which
A(c(3)) 1= 0= D(c(3)),
O(X,; c(/3)) 1= 0 D (c(3)), and
E,(B; c(/)) 1 0 = D,(c(3)).
Then for c= c(/),
for every y 4< (c), if fy3(y) T, then define f3(y) = B(y),
define d(3) = c,
set c(3) to be undefined, and go back to step 1.
3. (Creating a link (c,/3)) Let s be the current stage. Suppose that there is a b in the
domain of f, that enters B at stage s. Notice that the domain of f, is precisely
everything 4< Q(d(/3)) = 0(d(/3)), so that E,(B; d(3)) becomes undefined at stage s.
Then:
for every x < d(3) = O0(d(3)), if ga(x) is undefined, define it to be X,(x),
define the Arestraint of 3 by rA(/) 1,
set f3 to be totally undefined, and
create a link (ca, /).
4. (Travelling the link (c, /)) We travel the link (c, /) at the next cexpansionary stage
t > s. There are two cases:
Case 4a. (Successful closure) There is an x such that g9(x) 1= 0 / 1 = Xa(x).
(This x must enter X, since the current link (c, /) was created.) Then:
enumerate d(f) into Da, and stop.
Case 4b (Unsuccessful Closure) Otherwise, then
for every y < Q(d(/)), if fs(y) is undefined, then define fs(y) = B(y),
define an Arestraint of 3 by rA(/) (d(/3)).
The Possible Outcomes
We consider the following cases.
Case 1. Case 4a occurs at some stage t.
In this case, lim d(/) [s] 1= d(/) < u, and A(d(/)) = 0 / 1 = D (d(/)) is created. S
is satisfied.
Case 2. Otherwise, and Case 4b occurs infinitely many times.
Notice that g3 is never set to be totally undefined, and that for a fixed number d,
((d) is a fixed number, so that B changes below Q(d) only finitely many times, and so that
Step 3 occurs with the same d only finitely many times. Since Case 4b occurs infinitely
many times, we have that d(j3)[s] will be unbounded over the course of the construction,
and that whenever Step 3 occurs, we build ga on the initial segment of the current d(3).
Therefore g, is built as a computable function.
For an arbitrarily given x, we prove ga(x) I= X,(x). Let s be the stage at which
ga(x) is created. Suppose that si are all stages s' > s at which Step 3 of 3 occurs, and
that for each si, t, E (si, si+l) is the stage at which the link (a, P) created at stage si is
travelled through Case 4b.
By the choice of si, so = s. Since Case 4b occurs at stage to, and to is cexpansionary,
we have that
(i) g3(x) = Xj[so](x) (a will never be visited at stage so).
(ii) For any s E [so,to], g3(x) = X,[s](x).
(iii) g9(x) = 4(A, B; x)[to] = X,[to (x).
By the Arestraint rA(3)[to], and the convention of 4, we have that for any t E [to, Si),
(iv) g(x) = ,(A,B; x)[t] X,[t](x).
Suppose by induction that for n, we have that
(A) For any s E [s,,t,,], gp(x) = X[s](x).
(B) g(x) = X (x)[t1] = 4 (A, B; x) [t,],
(C) For any t E [tn, s,+l), g3(x) = (A, B; x)[t] = X,[t](x), and
(D) gp(x) = X,[sn+l](x).
By (C), (D) for n, and by the choice of tn+l, (A) holds for (n + 1). By (A) for (n + 1),
and the choice of t,+l, we have (B) for (n + 1). By (B) for (n + 1), by the Arestraint at
stage t,1+, and by the convention of 4, (C) holds for (n + 1). (D) for (n + 1) follows from
(C) for (n + 1) and the assumption that a is not visited at stage s,+l.
[Remark. We will arrange the construction so that an Agap can be opened only at
odd stages, and that no Rstrategies can be visited at these stages. This means that no
X, for any Rstrategy a can receive elements at odd stages, since we assume that X, is
enumerated only at stages at which a is visited.]
Therefore for any s > so, either ga(x) = X,(x)[s] or ga(x) = (4(x)[s]. Since
) (A, B; x) equals Xa(x), we have that ga(x) = Xa(x).
This is a global win for the requirement R, so that we don't consider any other
Srequirements with the same global index with the requirement R.
Case 3. Otherwise, and Step 2 occurs infinitely many times.
Since step 3 occurs only finitely many times, f, is set to be totally undefined only
finitely many times. Let f be the final version of fp. By the assumption that step 2 occurs
infinitely many times, f is built as a computable function.
By the choice of f, for any x, once f(x) is created, we have that f(x) = B(x). B is
computable, contradicting the assumption of the theorem. So we assume that this case
will never occur.
Case 4. Otherwise, then by the strategy, we have that lim, c($3)[s] 1= c(3) < w exists,
c(3) g D,, and A(c(3)) = 0 never occurs. Therefore A(c(3)) / 0 D (c(3)). S is satisfied
again.
We use d, g, and w to denote the possible outcomes of 3 corresponding to case 1,
case 2, and case 4 respectively. To guarantee that the true outcome will be the one on the
leftmost visited path, we define the priority ordering of the possible outcomes as follows:
d
With this ordering, we notice that case 3, in case it happens, will be an outcome
between g and w.
5.4 The Priority Tree
In this section, we will build a priority tree of strategies T C A< with A
{0, 1, d, g, w}. Note that there are infinitely many copies of a fixed computable func
tion Ai in {Ai : i E w}. Therefore, to satisfy a fixed 7R requirement, it suffices to satisfy all
Se,i with i > e. Let P < R denote that the priority ranking of P is higher than that of R.
Also let
Definition 5.4.1. (i) Define a priority ranking of the requirements so that, Ve e w:
'Pe < R, < So,e < Sl,e < < Se,e < P'+1
(ii) The possible outcome of a Pstrategy is only 1.
(iii) The possible outcomes of an Rstrategy are 0
(iv) The possible outcomes of an Sstrategy are d
Definition 5.4.2. Given a node we i that:
(i) P, is satisfied at ( if there is a 'P,strategy 7 C c
(ii) 7R is satisfied at ( if either
there is some R,strategy a such that a (l1) C _, or
there is some S,istrategy f3 (for some i) such that 3^ (g) C
(iii) R, is active at if R, is not satisfied at and there there is an R,strategy a with
a (0) C i, such that there is no S,,i',strategy f3' with a c(0) C P3' c f'^ (g) C for any
Ce < e.
In this case, a is unique, and we i that R, is active at via a.
(iv) Se, is satisfied at if 7R is satisfied at or 7R is active at via a, i, and there is
an S,,istrategy f3 such that a (0) C 3 C 3 .
We now define the priority tree T inductively as follows.
Definition 5.4.3 (Priority Tree). (i) Define the root node 0 to be a Postrategy.
(ii) The immediate successors of a node are the possible outcomes of the corresponding
strategy.
(iii) A node i, will work on the highest priority ranking requirement which is not
satisfied and not active at .
Definition 5.4.4. The index I(i) of a node is the index of the requirement on which
the node acts. For example, if is an Re or P,strategy, we define I(i) = e, and if is an
S,istrategy, we define I() = (e, i).
As usual, the priority tree T will have the following properties.
Proposition 5.4.5. Let f be an infinite path through T. Then for any requirement X,
there is a node 0o C f such that either (i) or (ii) below holds,
(i) X is satisfied at ( for any ( with o c ( C f.
(ii) X is active at f for any with Co c C f.
Proof. Let {Xi : i c w} be the priority ranking of the requirements so that Xi < Xi+1 for
all i. We prove this by induction. Assume that the proposition holds for all Xi with i < n.
Let X = X,+1. Fix (' C f so that the proposition holds for all Xi with i < n. Let 1o be an
Xstrategy such that (' c $o c f. Note that 1o must exist by the priority ranking of the
requirements unless X already satisfies Proposition 5.4.5 with the given '. There are three
cases.
Case 1. o is a Pstrategy.
The only outcome of a Pstrategy is 1. Therefore, necessarily o^(1) C f E [T]. So the
proposition holds with t = o^(l).
Case 2. io is a Restrategy for some e.
If o^(1) C f, then the proposition holds with = o^(1). Otherwise o = o^(0) C f.
Now if Ri (i < e) is satisfied at all ( with 0o C ( C f, then by the priority ranking of the
requirements, there is no Si,estrategy f3 D o^(0). Otherwise, by assumption Ri is active at
all ( with 0o C ( C f. So the outcome of Sie for each such Ri must necessarily be d or w.
Hence 1 = (o^(0)^A C f where A is the concatenated outcomes of all Si,strategies (i < e)
such that Ri is active at all ( with 0o c ( C f.
Now l^(k) C f for some k e {d, w, g}. If k = g then the proposition holds with
( = (^(g). If i = (I^(i) C f (i {d, w}) then by the same reasoning above concerning
the Rj (j < e), Re is active at as long as i C ( C f.
Case 3. io is an S,istrategy.
By the priority ranking of the requirements and by the assumption on the Xi (i < n),
it follows the Re is active at all with 0o C ( C f. Therefore either ,w = o^(w) C f
or d = 0o^(d) C f. By the assumption on Re, the proposition holds with j = k, with
appropriate choice of k so that k C f.
Definition 5.4.6. If 3 is an Se,istrategy for some e, i, then define top(3) to be the
longest R,strategy a such that a^(0) C 3.
5.5 The Construction
Our construction will perform different actions at even and odd stages. At even
stages, strategies on the tree will act to satisfy the requirements.
Suppose that B is enumerated at odd stages only, and that at every odd stage, there
is exactly one element that enters B. Given a Bpermission in the construction, we want
to open Agaps for as many Sstrategies as possible. This allows us to specify a Pstrategy
so that we enumerate its witness into A. So we will ensure that Arestraints drop at odd
stages, and also that A is only enumerated at odd stages.
During the course of the construction, we may initialize a node, ( ,iv, which means
that all the actions taken by ( previously, are cancelled, or set to be totally undefined.
Precisely, if an Rstrategy a is initialized, then both 0, and E are set to be totally
undefined, D, is set to be the empty set 0, and all links associated with a are cancelled.
If an Sstrategy 3 is initialized, then both ga and fa are set to be totally undefined,
parameters d(3) and c(3) are both set to be undefined, and any link associated with 3 is
cancelled. If a Pstrategy 7 is initialized, then A, is set to be totally undefined, and all
witnesses of 7 are cancelled.
Notice that an Sstrategy 3 opens its Agap, exactly at stages at which an error
between fa and B occurs, which gives a Bpermission for its current candidate d(3). Our
problem is to make sure that there are infinitely many stages at which all the Sstrategies
on the true path (or the current approximation of the true path) drop their Arestraints
simultaneously.
Given a node suppose that 31 c /2 C .. C P are all Sstrategies 3 with P^(g) C (.
To guarantee that if is a Pstrategy, then there are infinitely many stages at which for
all i = 1, 2, n, the Arestraints rA (i) of i3 drop to 1 infinitely often, proceed as
follows. Let fi be f, for all i = 1, 2, n.
We will arrange the building of various fa, such that: for any s,
1. for any i, if fi[s] is empty, then the current Arestraint rA(3i) is 1, and
2. for any i < j, if both fi[s] and fj[s] are not empty, then dom(fi[s]) 3 dom(fj[s]),
where dom(f) is the domain of f.
Our construction will ensure that for any s, at the end of stage s, the two properties
above hold for all nodes .
Using these properties, we know that whenever we find an error between f, and B,
the same error occurs for fi for all i < n, allowing us to open Agaps for the Sstrategies
3i simultaneously, except for those /3's which are already in their Agaps. Therefore,
if f3, opens an Agap at stage s, then the current stage s is in the Agap of 3i for all
iE {1,2, ,n}.
We are ready to describe the stagebystage construction.
Definition 5.5.1. (The Construction) The construction will be defined as follows.
Stage s = 0. Initialize every node and set A = 0.
Stage s = 2n + 1. Let b be the number that enters B at stage s.
Run the following procedure:
1. Let 3 be the <minimal and cmaximal Sstrategy, if it exists, such that f3(b) 1.
Suppose that 31 C /02 C '.. C f n1 are all nodes 3' such that 3'^(g) C 3. Let 3 = 3,.
2. Initialize all nodes with 3,^(g)
3. In increasing order of i, for 3i, and ci = top(Qi), if f, / 0, then:
for every x < d(Ao), if g3(x) is undefined, then define g(x) = X,(x),
create a link (a~i, /),
set rA(3,i) 1, and
set fa to be totally undefined.
We w that a Pstrategy 6 requires attention at stage s if:
There is some x such that A6(B; x) J/ K(x) and a (x) g A;
For all 3 with 3P(g) C 6, rA(0) = 1 hold during stage s.
4. If there is a Pstrategy which requires attention at stage s, then:
let 7 be the <least such Pstrategy,
let k be the least x such that A (B; x) J/ K(x) and a (x) g A,
enumerate a((k) into A, and
initialize all nodes with > 7, and go to stage s + 1.
5. Otherwise, then go to stage s + 1.
Stage s = 2n + 2. We first specify the root node to be eligible to act at substage
t = 0. At each substage t, we allow the strategy which is eligible to act at this substage to
take action, and then either close the current stage or specify a new node to be eligible to
act at the next substage of stage s.
Substage t. Let ( be the node which is eligible to act at substage t of stage s. If '
has length s, then initialize all nodes % and close the current stage. Otherwise, there
are three cases corresponding to different types of strategy .
Case 1. = 7 is a Pstrategy. Then run the following:
Program 7: 7 will build a bounded Turing reduction A,, and define witnesses a,(k).
For simplicity, we drop the subscription 7 in the description of the program.
1. If there is an n such that a(n) is defined, and 1(WT,(B), A) / a(n), then let 7^(1) be
eligible to act next (i.e. at substage t + 1 of stage s).
2. Otherwise, let k be the least x such that A(B; x) is undefined. Then:
if a(k) 1, then define A(B; k) = K(k) with 6(k) = Q(a(k)),
otherwise, then define a(k) to be fresh, and
initialize all nodes ( > 7, and go to stage s + 1.
Case 2. ( = a is an Restrategy for some e. Run the following
Program a:
1. If s is not cexpansionary, let a^(1) be eligible to act next.
2. Otherwise, and there is a link (c, 3) which was created and has never been cancelled
or travelled. Let /o be the <least such 3, and let /o be eligible to act at the next
substage.
3. Otherwise, then,
let x be the least y such that either O,(X,; y) or E,(B; y) is undefined,
if 0,(X,; x) T, define e,(X,; x) = D (x) with 0(x) = x,
if ,(B; x) T, then define E,(B; x)[s] : D,(x)[s] with use ((x) := (x), where
is the use of N(A, B), and
let a^(0) be eligible to act next.
Case 3. = 3 is an Sistrategy for some e, i. Let a = top(/3). We perform the
following,
Program 3:
1. If 3 has already been satisfied, as defined in 2a below, then 3^(d) is eligible to act.
2. (Travel a link (a, 3)) If a link (a, 3) was created and it has never been cancelled or
travelled since it was created, then travel the link (a, 3) by cases.
Case 2a. (Successful closure) (Ex) g (x) X/ Xa(x). (Notice that g was correct
at the stage we created the current link (a, 3), so this error must occur during the
Agap of the Sstrategy 3.) Then:
enumerate d(3), the largest confirmed candidate of 3, into Da,
we that 3 is .,I.:f/ .1 at stage s,
initialize all nodes > 3, and go to stage s + 1.
Case 2b. (Unsuccessful closure) Otherwise. Then:
define u =max{((y) : g(y) 1} = (d(3)),
set f r[ (O(d(3)) + 1) B ( (d(3))+ 1),
set rA(0) u+1, and
initialize all nodes to the right of 3^(g), and go to stage s + 1.
In either case, the link (a, 3) is removed.
[Remark. We have that if step 2 of program 3 occurs at stage s, then a is visited at
stage s.]
3. (Building f) If c(3) 1= c, AX(c) I= D,(c) = 0, 0,(X,; c) 1, and E,(B; c) 1 then:
Case 3a. (/3') 3'^(g) C 3 and dom(f3,) C [0, (c)]. Then:
initialize all nodes > P3(w), and go to stage s + 1.
Case 3b. Otherwise, then:
for any x < Qa(c), if fa(x) T, then define fa(x) I= B(x),
set d(/) = c(3); d is said to be confirmed,
cancel c(3), so that c(/) T, and
let ^3(g) be eligible to act next.
4. If c(/) T, then define c(/) as fresh, initialize nodes > P^(w), and go to stage s + 1.
5. Otherwise, let P^(w) be eligible to act at the next substage.
This completes the description of the construction.
5.6 The Verification
In this section, we verify the satisfaction of the requirements. First we investigate
some global properties that hold at the end of an arbitrary stage. These properties ensure
that the construction is implemented properly.
Proposition 5.6.1. Let s be a stage.
(i) There is at most one link that is travelled during stage s.
(ii) If a link (ca, ) is travelled at stage s, then before we travel the link, a is visited and
step 2 of program a occurs at stage s.
(iii) There are no a,i, 2, /1, and /a such that a, C a2 C 01 C /2 and both links (ai, P1)
and (a2, 02) exist at the end of stage s.
Proof. It is easy to see that both (i) and (ii) hold by observing the construction.
For (iii), suppose to the contrary that s is the least stage such that there are
al, a2, 1, and 32 with ac C a2 C 13 C /2, and such that both links (al, P/) and
(a2, 02) exist at the end of stage s. By the minimality of s, exactly one of the two links
(cai, 1), and (a2, 02) is created during stage s. We consider two cases.
Case 1. The link (ac, 3i) is created at stage s.
By the construction, s = 2n + 1 for some n, and ft, is set to be totally undefined at
stage s. We analyze the location of 2. If l^(w) C /32, then by the construction at stage s,
/2 is initialized during stage s, so the link (a2, 32) is removed during stage s, contradicting
the choice of 32. If f3 (g) C i2, then by the definition of the priority tree T, top(/32) I a2,
so that there is no link (c2, 2) which can be created in the construction. If fl^(d) C 0/2,
then the current d(32) must be defined after /3 created its inequality at argument d(/31),
after which no link (ci, 31) can be created, since 31 has satisfied its requirement through
A1, (d(31)) = 0 / 1 D(d(31)). So case 1 does not happen.
Case 2. The link (a2, 32) is created at stage s.
Let si < s be the stage at which the current link (al, 31) was created. By the proof in
case 1, we only need to consider the case of 1 ^(w) C /32. By the construction at stage Si,
32 was initialized during stage si. So f3, is totally undefined at the end of stage si. And
in fact, all nodes > 3i^(w) were initialized at stage il. Therefore /32 cannot be visited at
any stage > si unless the link (ca, 0/3)[si] has been removed. This contradicts the choice of
s.
(iii) holds.
The Proposition follows. E
Proposition 5.6.2. (i) Let 3 be an Sstrategy. Then for any s, t, if f3 is totally
undefined at substage t of stage s, then rA3() = 1 holds at the end of substage t of
stage s.
(ii) Let f be an Sstrategy, and s be a stage. Let s be the greatest stage t < s such
that 3 was initialized. Then for any s < sI < 82 < s, if both f3[sl] and f[s2] are
not empty, then
dom(f/p[si]) C dom(f3[s2])
(iii) Let /3, be Sstrategies with 3'^(g) C 3. For any s, if both f3 and fy, are non
empty at the end of stage s, then dom(f3[s]) C dom(f3,[s]).
Proof. Both (i) and (ii) are easy facts by observing the construction.
For (iii), we prove the proposition by induction on the stages. Suppose that (iii) holds
for all s' < s. Consider a stage s at which f3 is built. By program 3 in the construction,
there are two cases.
Case 1. A link (ca,/3) for a = top(/) is travelled unsuccessfully at stage s. Let Si be
the stage at which the current link (ca,/3) was created. Then dom(f3[s]) = dom(f3[si 1)].
Let b = dom(f/[s]), and let so be the first stage at which f3 was defined on [0, b].
By program /3, case 3b of program 3 occurred at stage so. By the construction, there
was no link that was travelled by the substage at which 3 was visited during stage so.
Therefore, 3' was visited at stage so, and case 3b of program 3' occurred at stage so. By
the assumption in case 3b of program /3, we have that both fy and f, are nonempty
at the end of stage So, and dom(f/,[so]) D dom(f/[so]). By the choice of s, 3' has not
been initialized during stages [so, s], by (ii) if f/,[s] is not empty, then dom(f/,[s]) D
dom(f, [so]), (iii) follows in case 1.
Case 2. Case 3b of program 3 occurs at stage s. As the same as that in case 1, by
observing the construction, we have that for any C /3, is visited at stage s, so that
/3' is visited at stage s, and furthermore, case 3b of program 3' occurs at stage s. By the
assumption of case 3b of program /3, the domain of fo is larger than that of f, at the end
of stage s. (iii) follows in case 2.
The proposition follows. O
Definition 5.6.3. Suppose that K (bT B and B %T 0.
(i) Let 6, be the last node which is eligible to act at stage s.
(ii) Define the true path TP e [T] of the construction by TP =liminf,68.
Hereafter whenever we consider a node on the true path, we will use the notation
I E TP rather than I C TP.
Proposition 5.6.4. (Existence of the true path) Suppose ( e TP. Then there is some
a such that ^ (a) is visited infinitely often and initialized only finitely many times. Hence
^(a) e TP.
Proof. We prove by induction on the length of Suppose by induction that the
proposition holds for all c and e TP. Let so be minimal after which will never be
initialized. By the inductive hypothesis, will be visited infinitely often.
We prove the proposition for by cases.
Case 1. = 6 is a P,strategy for some e.
By program 6, there are two subcases to consider.
Subcase la. There are infinitely many 6expansionary stages.
By the construction, step 2 of program 6 occurs infinitely many times, so that A6 is
built infinitely many times, and that A6(B) is built as a total function. Since K bT B,
there is some m such that As(B; m) J/ K(m) holds permanently. Let n be the least such
m, and let A6(B; n) be created at stage v > so. By the choice of n, Ta(B; as(n))[v] = 0
and it will hold permanently. Let u > v be the stage at which n enters K.
Suppose that Pf c 02 C ... C f3 are all Sstrategies f with P^(g) C 6.
By the choice of so, fQ will never be set to be totally undefined after stage so by
initialization for any j. Therefore, for every j E {1, 2, ... ,}, fa is set to be totally
undefined after stage so only if an error occurs between f, and B.
By inductive hypothesis, case 3b of program /3 occurs infinitely many times, so that
fa3 will be built infinitely many times for all j E {1, 2, ,1}.
In particular, f3, will be built infinitely many times. By the assumption of B %T 0,
we can choose a stage s, > u at which there is a number b such that f/3(b) 1= 0 / 1 =
B(b) occurs.
By Proposition 5.6.2 (iii), for any j E {1, 2, }, if f, is not empty at the
beginning of stage si, then there is an error between f3 and B that occurs exactly at
stage si. We have that for every j E {1, 2, .. ,1}, if fa / 0 at the beginning of stage si,
then 3j opens its Agap during stage sl.
By Proposition 5.6.2 (i), for any j E {1, 2, ,1}, if f, is empty at the beginning of
stage si, rA(/j) = 1 holds at both the beginning and the end of stage si.
By program 6, we have that 6 requires attention at stage si, and we let 6 receive
attention by enumerating its witness as(n) into A.
By the choice of n, for any s > si, we have that Ta(B; as(n)) = 0 / 1 = A(a6(n))
holds during stage s, contrary to there being infinitely many Sexpansionary stages. This
case is impossible.
Subcase lb. Otherwise.
In this case A6 is built only finitely many times. Let si > so be minimal after which
A6 will never be built.
By the choice of si, 6^(1) will never be initialized after stage sl, and by program 6, for
any s > si, if J is visited at stage s, so is 6^(1).
The proposition follows in case 1.
Case 2. = a is an Rstrategy.
Observing program a, we consider two subcases.
Subcase 2a. Step 3 of program a occurs infinitely many times.
Then a^(0) E TP. By choice of so, a^(0) will never be initialized after stage so.
Furthermore a^(0) is visited infinitely often. Therefore a^(0) e TP and the proposition
holds in this case.
Subcase 2b. Otherwise.
Suppose that Step 3 of program a occurs at most finitely many times so that a^(0)
is visited at most a finite number of times. We will show that a^(1) e TP. To show that
a^(1) is initialized finitely often, first note that by the choice of so, only nodes ^ D ac(0)
can initialize a^(1).
Let si > so be minimal after which step 3 of program a will never occur. Then for
any s > si, if a node 3' D a^(0) is visited at stage s, then there is an Rstrategy a' C a,
and a link (a', P') which is travelled at stage s.
Suppose that a C a2 C ... C ac~ are all Jstrategies a' with a'^(0) C a. Let
O = a.
We prove by induction that for each i < n, there is a stage after which there will be
no links (aj, 3j) that can be either created or travelled for all j > i and for all pj D ac(0).
For i = n. Define
bF = max{( (x) [t]  t < si, a (x)[t] [}
For every s > si, define
p.[s] = max{y I fa(y)[s] 1, top(3) = a,}
By the construction, we have that for every s > si,
if a link (ac, P') is travelled for some i < n and some f3' D a,^(0), then there is no f
such that f3 is built during stage s for any f with top(3) = a,. In this case, we have
p,[s] < p,[s I], and
if a link (a~, P3) is travelled at stage s, then p[s] < b,.
Therefore in any case we have that for all s > si, p,[s] < b,. By the construction
at odd stages, a link (a,, ,,) can be created at a stage s > si only if there is an element
b < bn that enters B at stage s. Since bn is a fixed number, B changes below bn only
finitely many times. Therefore there are only finitely many stages at which we create links
(ac, /,). Since once a link is travelled, it is removed immediately, there are only finitely
many stages at which a link (ca, P) is either created or travelled.
Let v, > si be minimal after which there will be no link (c,, 3,) that can be either
created or travelled.
Suppose by induction that [ i is a minimal stage after which there will be no link
(ac, 3j) which is either created or travelled for all j E {i + 1, i + 2, n}, and all
4, D c'^(0).
Define
bi = max{, (x)[t] I B(x) 1, t < I }
For any s > i ,I, define
pi[s] = max{y  f(y)[s] 1, 3 aD a(0), top(3) = a}
By the construction, it is easy to see from an inductive argument that for all s > ,,
if a link (acj, s) is travelled for some j < i, and some 3j D ca^(0), then pi[s] <
i [s 1]; and
if a link (a 03) for some f3i a^ (0) is travelled at stage s, then pi[s] < bi.
This shows that for all s > I ,I pi[s] < bi. By the construction, if a link (a /3) for
some 3,i D ca(0) is created at a stage s > I, t, then there is a number b < bi which enters
B at stage s. Since bi is a fixed number, the creation of links (a ,A) for f D a^ (0) occurs
only finitely many times, so that there is a stage > ., I i, after which there will be no
link of the form (a 03) for any f3i a (0) which can be either created or travelled.
Therefore there is a stage vi ', after which no link (cai, /) can be created or travelled
for any i < n and any 3 D ca^(0). So there are only finitely many stages at which some
node ( D a^ (0) is visited.
Thus ca^(1) is initialized only finitely many times.
By the proof above, there are only finitely many stages at which either Step 2 or Step
3 of program a occurs. a^(1) is visited at almost every stage at which a is visited.
Therefore in Subcase 2b, we have that a^(1) is initialized only finitely many times and
visited infinitely often.
Case 3. = 3 is an Sstrategy.
Let a = top(3).
Subcase 3a. a link (c, 3) is travelled once and successfully closed.
Then 3^(d) is visited infinitely often and only initialized finitely many times. So
^ (d) e TP.
Subcase 3b. (c, P) is travelled infinitely often and unsuccessfully closed p's Agap.
As in Subcase 3a, P^{(g) TP.
Subcase 3c. Otherwise.
By the assumption of this case, fa is built only finitely many times, since if this is not
true, then ft is built as a computable function, and ft = B, contradicting the hypothesis
B T 0.
By program 3, lim, c() [s] 1= c() < uc. By the choice of c(3), c(3) g D,. Let si be
the stage after which neither of the step 1, 2, 3, or 4 of program P occurs, therefore, for
any s > si, if f is visited at stage s, so is f^(w).
Therefore f^(w) is initialized only finitely many times, and visited infinitely often,
0^(w) E TP.
The proposition follows in Case 3. D
Since the true path exists only if both K %bT B, and B %T 0 hold as proved in
Proposition 5.6.4, we alv assume the two conditions from now on.
Proposition 5.6.5 (Possible outcomes along TP). Given ( e TP:
(i) If 6 is a Pestrategy for some e, then 6^(1) e TP and A / Te(B).
(ii) If (= a is an Rstrategy, then
(a) if a^(0) e TP, then D, O(X) = _E(B);
(b) if ac(1) e TP, then ko is partial or 4)(A,B) / X,.
(iii) If = p3 is an Sstrategy, then for a =top(3), we have:
(a) if PT(d) e TP, then limd(3)[s] = d(3) < u and A (d(3)) = 0 / 1 D,(d(3)).
(b) if P/(g) e TP, then g3 is a computable function and g3 = Xa.
(c) if /^(w) E TP, then limc(3)[s] 1= c(3) < u and A (c(3)) / 0 = D(c(3)).
Proof. By Proposition 5.6.4, we can choose so minimal after which ( e TP will never be
initialized.
For (i). By Proposition 5.6.4, 6^(1) E TP. Suppose to the contrary that A = e,(B).
By program 6, step 2 of program 6 occurs infinitely many times. Therefore, A6(B) is
total. Since K %bT B, we can choose the least n such that a permanent inequality
A6(B; n) = 0 / 1 = K(n) appears. Let v > so be the stage at which the computation
A6(B; n) was created. Notice that Te(B; a6(n))[v] = 0 and B will never change below
'. (a6(n)) after stage v.
By the proof in case 1 of Proposition 5.6.4, there is a stage Si at which we can
enumerate a6(n) into A.
By the choice of n and si, 4'(B; a6(n)) = 0 / 1 = A(a6(n)) will be preserved forever.
A contradiction.
Therefore we have that A / e(B).
For (ii). Let a^(0) e TP.
By Proposition 5.6.3, both 0, and E, are built infinitely many times.
By Step 3 of program a, it suffices to prove that the following constraints of a are
satisfied during the course of the construction:
For any d, and any s > so, if d is enumerated into D, at stage s, then both O,(X; d)
and E,(B; d) are undefined during stage s.
By Proposition 5.6.1 (i), there is at most one link which is travelled during stage s.
Let d be enumerated into D, at stage s. Then there is an Sstrategy f such that
top(3) = a and case 2a of program 3 occurs at stage s, and d = d(3)[s].
By Proposition 5.6.1 (ii), a is visited at stage s, and there is a link (a, 3) which
was created at a stage s(> so) < s and travelled at stage s successfully. By the
construction at stage s, there was an error between f3 and B occurred at stage s, since
d = d(3)[s] d(3)[s] dom(fg[s]), we have that E,(B; d(3)) becomes undefined during
stage s. By the link (a, 3) [s], step 3 of program a has never occurred during stages
[s, s]. Therefore, E,(B; d) is undefined during and after stage s.
By the assumption of case 2a of program 3 at stage s, there has been an error
between g3 and X, during stages [s, s], therefore, 0,(X,; d) has become undefined
during stage s.
Therefore, for any number d chosen after stage so, the enumeration of d into D, does
alvi respect the constraints imposed by a. Both @,(Xa) = D, and E,(B) = D, are
satisfied.
Let a^(1) E TP. We prove that Xe / ~e(A, B). Suppose to the contrary that
Xe = L(A, B). By the assumption of this case, step 3 of program a occurs only finitely
many times, and that there are infinitely many aexpansionary stages. Therefore, there
is a stage Si > so after which step 3 of program a will never occur. However there are
infinitely many stages at which we travel a link (a, 3) for some 3.
By the choice of sl, E, is a finite set, let s2 be the stage > sl after which B will never
change below max{(~(x)[Sl] I ,(B; x)[si] 1}.
By the Sstrategies, for any 3 with top(3) = a, if f3 is created after stage s2, then
there will be no link (a, 3) which can be created by using the difference between B and fp.
Therefore we can choose a stage s3 > s2 after which there is no link (a, 3) which
can be created for any 3. By the construction, once we travel a link (a, 3), it is removed.
There is a stage s4 > s3 after which there is no link from a to any Sstrategy 3 which can
be either created or travelled. This contradicts the assumption that step 2 of program a
occurs infinitely many times.
We have that Xe / (A, B).
For (iii)(a). If 3^(d) E TP then there is some stage s where Case 2a of program 3
enumerates d(3) = lim, d(3)[s] into D,. Then at all stages t > s, program 3 enacts Case
1. Furthermore since d(3) is only defined when A(d(3)) = 0, we have that Ap(d(3)) 0 /
1 = D (d(3)).
For (iii)(b). By the choice of so, and by the assumption of this case, 0^(g) is never
initialized after stage so.
By program 3, case 3b of program 3 occurs infinitely many times. By the choice of
so, f3 becomes totally undefined at any stage s > so only if a link (a, P) is created at stage
s, and this link will certainly be travelled unsuccessfully, instead of being initialized.
For a fixed number d, 0,(d) is a fined number, so that B changes below 0,(d) only
finitely many times. Therefor d(3)[s] will be unbounded in the construction. By the
construction at odd stages, if a link (a, 3) is created at stage s, then d(/3)[s] is defined, and
gp is built on the initial segment d(3) [s]. Therefor g, is built as a computable function.
Notice that g, will never be set totally undefined after stage so.
We prove that for any x, if ga(x) is created at a stage v > so, then for any s > v,
g (X) = X [s](x).
Given an x, let v > so be the stage at which ga(x) is created and defined as 0 (if it is
1, then ga(x) = Xa(x) takes already the permanent value).
Suppose that i,, < vl < v2 < are all stages > v at which a link (a, 3) is created,
and let ti be the stage at which the link (a,/3P), ] is travelled. Then vo = v.
By the choice of to, the link (c,/) [,,,] is unsuccessfully travelled at stage to, this
means that for any t E [vo, to], g3(x) = X[t](x). Since to is cexpansionary, we have
g9(x) = X,[to](x) = (A, B; x) [to]. By the Arestraint rA(03)[to], the definition of fa [to],
and by the convention of 4,, we have that for any s E [to, vi), g(x) = 4(A, B; x)[s] =
Xa[s](x) is preserved.
Suppose by inductive hypothesis we have:
1. for any s E [va,tn], g9(x) = X[s](x).
2. gp(x) = X,[t](x) = t.(A,B; x)[t,].
3. for any s E [t., v,u+), g/(x) = I)(A, B; x)[s] = X[s](x).
4. g/(y) = Xa[v,+l].
Since the link (a, P)3[vT+l] is unsuccessfully travelled at stage t1+l, (1) holds for n + 1,
and since t+l is cexpansionary, (2) for (n + 1) holds, and furthermore, by the Arestraint
at stage tn+l, (3) for (n + 1) holds, (4) holds since a will never be visited at odd stages, so
there are no elements which enter X, at odd stages.
This proves that g9(x) = Xa(x).
Since x is arbitrarily given, we have that for almost every x, g9(x) = Xa(x), X, is
computable.
(iii)(b) follows.
For (iii)(c). By the assumption in this case, g9 is built only finitely many times. If
fp is built infinitely many times, then the final version of fp, denoted by f, is built as a
computable function, and f = B. A contradiction. Therefore f3 is built only finitely many
times. Let Si > so be such that f3 will never be built at any stage s > si. Furthermore,
we can choose a stage s2 > si after which none of the steps 1, 2, 3, or 4 of program 3 will
occur. Therefore lim, c(/)[s] = c(3) must be chosen before stage s2. Clearly c(3) g D,.
Since 3 is visited infinitely many times, the only reason that case 3b will never occur after
stage s2 is that AX(c(3)) / 0. (iii)(c) follows. O
Proposition 5.6.6 (Psatisfaction Proposition). For each e, P, is satisfied.
Proof. Given e, let 6 be the P,strategy E TP. By Proposition 5.6.5 (i), necessarily
A / ',(B) so that P, is satisfied. The proposition follows. E
Proposition 5.6.7. For each ee c, if X, 4), (A, B) and X, and B are not computable,
then they do not form a minimal pair.
Proof. By Proposition 5.4.5, let a be the longest Restrategy on the true path TP. By
Proposition 5.6.5 (ii), a^(0) E TP and D, = O,(X,) = E(B). By Proposition 5.6.5 (iii),
for any Sstrategy 3, if top(3) = a and /3 TP then either / (w) e TP or ^ (d) e TP.
In either case A3 / D, so that X, and B do not form a minimal pair. The proposition
follows. O
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BIOGRAPHICAL SKETCH
Paul Steven Brodhead was born in Oak Park, IL in 1980. He moved to Richland
Center, WI in 1989 and continued his schooling there until he graduated from Richland
Center High School in 1997. As a Chancellor's scholar, Paul attended the University of
WisconsinMadison from 1997 until 2000, when he earned a B.S. in mathematics. During
the summer of 2000, Paul participated in the NSFfunded undergraduate mathematics
research experience SIMU, at the University of Puerto Rico Humacao. In 2003 Paul
started graduate school at the University of Florida and earned a master's degree in
mathematics in 2005. During the summer of 2006, Paul was a graduate assistant for an
NSF SEAGEPfunded undergraduate mathematics research experience at the University
of the Virgin Islands. He went to the Chinese A, i1I in'v of Sciences, Institute of Software
during the summer of 2007 as an NSF fellow, participating in the East Asia and Pacific
Summer Institutes. As an NSF SEAGEP fellow during the fall of 2007, Paul went to
Victoria University of Wellington (Wellington, New Zealand). As an invited visitor and
lecturer, Paul went to the University of Hawaii at Manoa in the spring of 2008. Paul
earned his Ph.D. in mathematics from the University of Florida in 2008.
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COMPUTABLEASPECTSOFCLOSEDSETSByPAULBRODHEADADISSERTATIONPRESENTEDTOTHEGRADUATESCHOOLOFTHEUNIVERSITYOFFLORIDAINPARTIALFULFILLMENTOFTHEREQUIREMENTSFORTHEDEGREEOFDOCTOROFPHILOSOPHYUNIVERSITYOFFLORIDA2008 1
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c2008PaulBrodhead 2
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TomyFamily,theArtisansofLife 3
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ACKNOWLEDGMENTSIgivethanks,rstandforemost,tomyPh.D.advisorDouglasCenzer,whomustbenotedinthisgenerationasamongthemostpatientandirrevocablykindofmen.Hissenseofhumoralsotrumpsthecloudiestofdays,betheyeventhecloudsofhurricanes{which,infact,occurredasmystudiesprogressedinFlorida.Thesecharacteristics,combinedwithmydetermination,helpedmegainanunderstanding,broadandindepth,ofcomputabilitythatallowedmetosuccessfullyundertakeanddeveloptheresearchpresentedhere.Theroadtothispointbeganlongago,butreacheditspinnacleatFloridaunderhisdirection.TheroadbeganattheUniversityofWisconsin,inmyrstsemester;IwasrstinspiredtopursuemathematicswhiletakingAlgebraandTrigonometry.ProfessorArnoldJohnsonmadethecourseenjoyableandchallenging,makingitunlikeanymathematicscourseIevertook.HimIthank,forashaveIalwayshadhighgoals,hisinuencewascriticalasitwasatthispointinlifethatIdecidedthatIwastopursueaPh.D.inmathematics.Later,inmylastsemesteratWisconsin,hehelpedmedevelopamathematicalmaturitybeyondexpectations,asheledmeinanundergraduateresearchprojectinhisarea,Algebra.Yet,betweentheseexperiences,manyothersmustbethanked.Duringmyundergraduateyears,manyinspiringteacherspushedthemathematicalenvelope,deliveringtheendeavoroftheages,theenjoyablementalfrayofthebreakingofmathematicalbarriers.Ithank:SteenLempp,ReedSolomon,H.JeromeKeisler,ArnoldMiller,MauryBramson,SteveBauman,SimonHellerstein,DanielRider,DanielShea,andJenniferZiebarth.Inparticular,IthankSteenLemppforintroducingmetocomputabilitytheorywhileIwasinhislinearalgebraclass,asitwasatthismomentthatIdeterminedmyPh.D.specialization.Mymindsetwaseverafterwardsdirectedthisway.Forexample,IthankArnoldMillerforhispatienceinlisteningtomynumerousattemptsinusingcomputabilitytheoryinhisyearlongabstractalgebraclass.IthankH.JeromeKeislerforlisteningtomeinhislogiccourse.IalsothankReedSolomonforhelpingmeevenfurtherinhisgraduatelogiccourseandhissubsequentguidanceinmyrstundergraduateresearchproject,thestudyofHilbert'sTenthProblem.IthankforDaniel 4
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SheaalsotheundergraduateadvisorforacceptingmydestinedcourseinmathematicsasIsatinhisabstractanalysiscourse,vigorouslydrivenunderadepartmentmathematicsscholarship,whenyearspriorIenteredhisocewhilststillinAlgebraandTrigonometry,demandinghelpinsettingupmyundergraduatecurriculumforaneventualPh.D.inmathematics.UpontheclosingofmyWisconsinexperience,excellenceinmathematics,ahigherlevelstill,evolvedattheUniversityofPuertoRicoHumacaoduringtheimmediatesummer.IparticipatedintheNSFfundedundergraduateresearchexperience,theSummerInstituteofMathematicsforUndergraduates.Ithank:ReinhardLaubenbacher,AbdulJarrah,RebeccaGarcia,MalarieCummings,CoraSeidler,IvelisseRubio,andHerbertA.Medina.Herbert,Ivelisse,andReinhard,inparticular,helpedmetoseetheurgencyinthetacklingoftheproblemsathand.Iamgreatlyappreciativeofthishelptothisday.MyresearchrigorandcoremathematicaltalentbloomedingraduateschoolatFlorida.Ithank:DouglasCenzer,BillMitchell,JindrichZapletal,JeanLarson,AlexandreTurull,PaulRobinson,JamesKeesling,JonathanKing,ScottMcCullough,StephenSummers,andShariMoscow.IespeciallythankAlexandreTurull.Ihadtheprivilegeofhavingsixsemesterswithhim;hehelpedmetoseeaproblem,itsvalue,anditscomplexities.Ithankmydissertationcommittee:DouglasCenzer,RickSmith,BillMitchell,MuraliRao,andBeverlySanders.IalsothankBernardMair,RickSmith,MuraliRao,andJuanLiu;myexperienceswiththemconvincedmetomoveinvaluabledirectionsthattookmearoundtheglobeandbeyond.Mycollaborators,nearandabroad,youIthank:DouglasCenzer,AngshengLi,WeilinLi,GeorgeBarmpalias,SeyyedDashti,RebeccaWeber,JereyRemmel,RodDowney,NoamGreenberg,KengMengNg,andBjrnKjosHanssen.ManymanymoredoIalsothank{Ihavecollaboratedwithmanypeople,ifevenforashorttime,andtheircontributionsareappreciated.Ithank:PeterHinman,DenisHirschfeldt,CarlJockusch,RussellMiller,JohannaFranklin,BakhadyrKhoussainov,JanRiemann,TedSlaman, 5
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AndreNies,RobertSoare,SteveSimpson,AntonioMontalban,BarbaraCsima,JoeMiller,VeronicaBecher,andWolfgangMerkle.Ithankthepeopleandinstitutionswhocontibutedtheirsupportthroughfunding:theSoutheastAllianceforGraduateEducationandtheProfessoriate,theNationalScienceFoundation,theUniversityofFlorida,theChineseAcademyofSciences,theAssociationforSymbolicLogic,andothers.Manypeopledeservecreditfarbeyondthebrief,orevenomitted,acknowledgementhere;Iamextremelygratefultoyou.Ofspecialconsiderationismyadvisor.Themagnitudeofhissupportisaboveanymeasurablescale.Hissupportinallmyendeavors,professionallyandbeyond,willsurelycontinuetohelpshapeme.Thisdissertationandtheworktherein,andtheexperiencesassociatedwiththeattainmentthereofareawitnesstothis.Tothemanypeoplewhohelpedmealongthisjourney,whichisactuallyjustabeginning,youIthank. 6
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TABLEOFCONTENTS page ACKNOWLEDGMENTS ................................. 4 ABSTRACT ........................................ 10 1INTRODUCTION .................................. 12 1.1GeneralOverview ................................ 12 1.2ClassicalComputability ............................ 14 1.3ClosedSetsinComputability ......................... 15 2EFFECTIVELYCLOSEDSETSANDENUMERATIONS ............ 18 2.1Introduction ................................... 18 2.2TheFamilyof01Classes ............................ 20 2.2.1NumberingsintheLiterature ...................... 20 2.2.2EquivalenceoftheNumberings ..................... 22 2.2.3EquivalenceoftheNumberingsAlternateProof .......... 25 2.2.4InjectiveComputableNumberings ................... 26 2.2.4.1Originalc.e.setsargument .................. 27 2.2.4.2Orderedtuplesofdisjointc.e.sets .............. 29 2.2.4.3Resultsforeectivelyclosedsets .............. 32 2.3StringVeriableFamiliesof01Classes .................... 37 2.3.1DenitionandExamples ........................ 37 2.3.2ComputableandEectiveNumberings ................ 39 2.3.3FamiliesContainingtheClopenClasses ................ 40 2.4NamedFamiliesof01Classes ......................... 42 2.4.1HomogeneousClasses .......................... 42 2.4.2DecidableClasses ............................ 44 2.4.2.1AninjectivecomputablenumberingAlternateproof ... 44 2.4.2.2Treeswithoutdeadends:Anumberingresult ....... 45 2.4.2.3Treeswithdeadends:Anecessity .............. 47 2.4.3ThinandPerfectThinClasses ..................... 49 2.4.3.1TheMartin{PourElConstruction .............. 49 2.4.3.2Nonexistenceofcomputablenumberings .......... 50 2.4.4Small,VerySmall,andNondecidableClasses ............. 51 2.4.4.1Numberingsandhigh/noncomputablesets ......... 51 2.4.4.2Nonexistenceofeectivenumberings ............ 52 3RANDOMCLOSEDSETS ............................. 54 3.1Overview .................................... 54 3.2EectiveRandomnessofReals ......................... 55 3.2.1Introduction ............................... 55 3.2.2ConstructiveMartingaleRandomness ................. 56 3.2.3PrexfreeRandomness ......................... 57 3.2.4MartinLofnrandomness ....................... 58 7
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3.3MartinLofRandomnessofClosedSets .................... 60 3.3.1TheHitorMissTopologyonC .................... 60 3.3.2TowardaMeasure ............................ 61 3.3.3CanonicalCodingandMeasure .................... 61 3.3.4GhostCoding .............................. 64 3.3.5CodingEquivalance ........................... 65 3.3.6CodingandJoinsofClosedSets .................... 66 3.4MembersofRandomClosedSets ....................... 67 3.4.1PositiveResults ............................. 68 3.4.2NegativeResults ............................. 73 3.5MeasureandDimension ............................ 78 3.5.1Measure ................................. 78 3.5.2Dimension ................................ 79 3.6PrexFreeComplexityofClosedSets ..................... 82 3.6.1LowerComplexityBounds ....................... 83 3.6.2UpperComplexityBounds ....................... 85 3.7OtherNotionsofRandomnessforClosedSets ................ 87 3.7.1RandomnesswithRegularProbabilityMeasures ........... 87 3.7.2RandomnesswiththeInclusionofTreeswithDeadsEnds ...... 87 3.8RandomClosedSetsandEectiveCapacity ................. 88 3.8.1ComputableCapacities ......................... 88 3.8.2RegularMeasuresandCapacitiesofClosedSets ........... 90 4RANDOMCONTINUOUSFUNCTIONS ..................... 94 4.1Overview .................................... 94 4.2DeniningRandomnessforContinuousFunctions .............. 95 4.2.1RepresentingFunctions ......................... 95 4.2.2RepresentingSequences ......................... 96 4.2.3ASoundDenition ........................... 97 4.3RandomContinousFunctionsandImages .................. 99 4.3.1PerfectImages,ineveryinstance .................... 99 4.3.2NoninjectiveImages,ineveryinstance ................ 100 4.3.3NonsurjectiveImages,ininstances .................. 103 4.3.4Imagesofcomputableelements ..................... 104 4.4RandomClosedSetsarisingfromrandomcontinuousfunctions ....... 106 4.4.1APositiveResult:InverseImagesof0! ................ 106 4.4.2ANegativeResult:Images,ingeneral ................. 109 4.5PseudoDistanceFunctions ........................... 110 4.6nRandomness .................................. 111 4.7FutureWork ................................... 112 5CONTINUITYOFCAPPINGINCBT ....................... 113 5.1Introduction ................................... 113 5.2ContinuityResults ............................... 114 5.2.1ContinuityResultsinC ......................... 114 5.2.2ContinuityResultsinCbTandMainResult .............. 115 8
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5.3RequirementsandStrategies .......................... 118 5.3.1Therequirements ............................ 118 5.3.2APstrategy ............................... 119 5.3.3AnRstrategy .............................. 120 5.3.4AnSstrategy .............................. 121 5.4ThePriorityTree ................................ 125 5.5TheConstruction ................................ 128 5.6TheVerication ................................. 132 REFERENCES ....................................... 144 BIOGRAPHICALSKETCH ................................ 150 9
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AbstractofDissertationPresentedtotheGraduateSchooloftheUniversityofFloridainPartialFulllmentoftheRequirementsfortheDegreeofDoctorofPhilosophyCOMPUTABLEASPECTSOFCLOSEDSETSByPaulBrodheadMay2008Chair:DouglasCenzerMajor:MathematicsAclosedsetinf0;1gNmaybeviewedthesetofinnitepathsthroughatree;asetAiscomputableifthereisacomputerprogramwhichhaltsandgivesthecorrectansweroneveryquerytothemembershippredicateforA.Anumbering,orenumeration,isamapfromNontoacountablecollectionofobjects;ifthereisacomputablenumberingontoasetANthenwesaythatAiscomputablyenumerableorc.e..Thesetofinnitepathsthroughacomputable,orequivalentlyacoc.e.,treeiscalledaneectivelyclosedset.Inthiswork,weinvestigate:numberingsfordierentfamiliesofeectivelyclosedsets,notionsofrandomnessfornonemptyclosedsubsetsof2N,notionsofrandomnessforcontinuousfunctionsfrom2Nto2N,andcontinuitypropertiesofCbT,thec.e.degreesundertheTuringreduction6bTthatrequiresthateachusebeboundedbyacomputablefunction.Numberingsandeectivelyclosedsets.Weshowthatcertainfamiliesorclassesoffamiliesofeectivelyclosedsets{suchasthedecidable,homogeneous,thin,smallortheentirefamilyofeectivelyclosedsets,orstringveriablefamilies{possess,ordonotpossess,injectivecomputableoreectivenumberings.ThisworksbuildsupontheseminalworkbyFriedberg[ 46 ],whoconstructedaninjectivenumberingofthec.e.sets.Randomnessofclosedsets.Inthespaceofclosedsets,wegiveaprobabilitymeasureanddeneaversionoftheMartinLofTestforrandomness.Weshowthatrandomclosedsetsarenevereectivelyclosed,butare,ontheotherhand,alwaysperfect,havemeasurezero,andhaveboxdimensionlog24 3.Everyrandomclosedsetcontainsrandomandnonrandomelements,butnonc.e.elements.Wealsoexplorealternate 10
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notionsforrandomness,suchastheproblemofcompressibilityoftrees.Finally,weconsidertheproblemofwhenarandomlychosenclosedsetmeetsaclosedQ;thisisthestudyofcapacities.Randomnessofcontinuousfunctions.Asin,wegiveaprobabilitymeasureanddeneaversionoftheMartinLofTestforrandomness.Weshowthattheimageofarandomcontinuousfunctionisalwaysnoninjectiveandperfect,butnotnecessarilysurjective.Furthermore,computableelementsmaptorandomelements.Also,randomclosedsetsariseasinverseimagesof0!,butnot,ingeneral,asimages.Theformermotivatesastudyofpseudodistancefunctions.Finally,weconsiderourresultsinthecontextofnrandomness.ContinuitypropertiesinCbT.WeshowinCbTthatforanyb6=0;00,thereisana>bsuchthatforanyx,b^x=0ia^x=0.WeprovethisbyrstshowingthatthattheSeetapunlocalnoncappabilitytheoreminthec.e.Turingdegrees[ 84 ]alsoholdsinCbT.Thistheoremdemonstratesthateveryb6=0;00isnoncappablewithanynontrivialdegreebelowsomea>bi.e.ifx
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CHAPTER1INTRODUCTIONThisthesisiaanaccumulationofmyworkasagraduatestudentattheUniversityofFlorida.Muchofthisworkisjointandpublished,ortobepublished.Thecitationsarelistedatthebeginningoftheappropriatechapters.Inthischapterweintroducevariousnotionsandexpoundupontheseinlaterchapters.EachofChapters 2 { 5 containsadistincttopicfromcomputabilitytheory.1.1GeneralOverviewComputabilitytheoryisaeldofmathematicallogic;thesubjectcapturestheprecisenotionofanalgorithmicprocesstowardsthestudyofdecidable/undecidableproblemsinmathematicsandnature.ItsmostnotablehistoricalcontributiontomathematicsisthedisruptionofHilbert'sProgrambyGodel'sIncompletenessTheorem.Morerecentworkhasshownthatothermathematicalproblemsareunsolvable,inthesensethatnocomputeralgorithmcansolveeveryinstantiationoftheproblem.ExamplesincludeHilbert'sTenthProblemtodecidewhetheragivenDiophantineequationhassolutions,thewordproblemforgroupstodecidewhetheragivenproductofgeneratorsandtheirinversesistheidentityelementofagroupdenedbyanitesetofequationsbetweensuchproducts,andthehomeomorphyproblemtodecidewhetherthetopologicalspacesdenedbyagivenpairofsimplicialcomplexesarehomeomorphic[ 32 ].Maturationofcomputability,however,throughapplicationsandtechniques,hasbroadeneditsinteractionswithotherelds,mostnotablycomputerscience.EachofChapters 2 { 5 isawitnesstothisbroadening.InChapter 2 ,wefocusonthestudyofeectivelyclosedsetsofbinaryreals.Thoughtofasasetrepresentingthesolutionstosomeproblem,eectivelyclosedsetscharacterizemanystructuresinmathematicsandcomputerscience.Inalgebra,forexample,theyrepresenttheprimeidealsofacomputableenumerableBooleanalgebraoracommutativeringwithidentity[ 47 ].Ingraphtheory,theyrepresentthesetofsolutionstomanyproblemswithcomputablegraphs,suchasHamiltoniancircuitsorvertexpartitions[ 20 ].Incomputerscience,eectivelyclosedsetsariseinthestudyofnonmonotoniclogicsand 12
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!languages[ 25 68 70 ].Giventhewidevarietyofapplications,theworkinChapter 2 focusesonmethodsofenumeratingvariousfamiliesofeectivelyclosedsets.Theideaistoprovide,basedondesiredproperties,completelistingsoftheobjects{inthiscase,eectivelyclosedsets{representingthesetofsolutionstoproblemsofacertaintype.Weshowthatthereisaninjectivecomputableenumerationoftheentireclassofsets,ofcertainfamiliesofstringveriableclasses,andofthedecidableandhomogeneousclasses.Wealsoshowthatnocomputableenumerationexistsforthin,perfectthin,small,verysmall,ornondecidableclasses.InChapters 3 and 4 ,weextendnotionsrelatedtoeectiverandomnessforbinaryreals,toclosedsetsChapter 3 andcontinuousfunctionsChapter 4 ;variousglobalpropertiesareobtained.Abinaryrealiseectivelyrandomifitisimpossibleforacomputertondregularityorpatternsinit.Foraclosedset,representingsomesetofsolutions,thismeansthatitisdicultforacomputertopreciselyobtainorlocallydescribethissetofsolutions,giventhelackofpattern.Inbothchapters,weobtainsocalledbasisandantibasistheorems.Forinstance,everyrandomclosedsetcontainsrandomandnonrandomelements,butomitsvariouselementsofcomputabilitytheoreticinterest,suchasthepropertiesofbeingfc.e.fapolynomial,1generic,orofincompletec.e.degree.Wealsoshowthatconceptsfrombothchaptersareclosedrelated;randomclosedsetsariseasinverseimagesofrandomcontinuousfunctionsmappingto0!,butnot,ingeneral,asimages.Methodsemployedinallofthisworkrangefromtechniquesincomputabilityandeectiverandomness,totechniquesrelatedtothestudyofeectiveHausdordimensionandclassicalprobability.Finally,inconnectionwiththerootsofcomputabilitytheory,inChapter 5 wefocusontheclassicationofinformationcontentbymeansofTuringdegreetheory.Mathematicalstructuresorobjectsareoftenencodedassetsofnaturalnumbers.Reductionmethods,suchastheTuringreducibility,allowtheinformationcontentofthesesetstobeclassiedintoequivalenceclasses,calleddegrees.InChapter 5 ,thefocusisontheboundedTuringreducibility[ 16 ];weshowthatcapping,theoperationwhichtakesthemeetofagiven 13
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noncomputableincompletec.e.degreewithanothernoncomputableincompletec.e.degreesuchthattheresultingmeetisthedegree0ofthecomputablesets,iscontinuous.Thatis,foranyb6=0;00,thereisana>bsuchthatforanyx,b^x=0ia^x=0.Asthisisathreequantierstatementinthec.e.bTdegrees,thisresultgivesinsightintothethreequaniertheoryofthesame,whosedecidability/undecidabilityiscurrentlyunknown.Asanaside,inrecentworkbySoare,theboundedTuringreducibilitywiththeidentityusehasledtoapplicationsofdegreetheorytodierentialgeometry[ 72 87 ].Therestofthischapterisdevotedtointroducingtobasicdenitions,terminology,andnotationsthatwillbeusedthroughoutthisentirework.Section 1.2 coversbasicnotionsandnotationsfortopicsinclassicalcomputabilitye.g.computablesets,computablyenumerablesets,partialcomputablefunctions,Turingdegrees.Section 1.3 coversthebasicsofclosedsetsinNN.AsalgorithmicrandomnessisatopiccoveredonlyinChapters 3 and 4 ,wepostponeageneralintroductionofthistopicandprovideitinSection 3.2 .1.2ClassicalComputabilityWegenerallyfollowthenotationofSoare[ 86 ]fornotionsthatarisefromclassicalcomputability.Foranindepthtreatmentofthebasicfoundationsofthesubject,wereferthereaderthere.AsetAiscomputableifthereisacomputerprogram,orequivalentlyacomputablefunction,whichhaltsandgivesthecorrectansweroneveryquerytothemembershippredicateforA.Itiscomputablyenumerablec.e.ifthecomputerprogramisrequiredtohaltonlyonqueriesforelementsinA;thisgivesrisetothestandardexampleofac.e.set,namelythehaltingproblem{thesetofTuringmachines,ociallyencodedasasetofnaturalnumbers,thathaltwhengiventheirownbinaryinput.Computablyenumerablesetsarethedomains,therefore,ofsocalledpartialcomputablefunctions;weindexthepartialcomputablef0;1g{valuedfunctionsasfege2!.Partialcomputablefunctionalsthattakenaturalnumbermandrealxinputsareindexedase;wewillwritexemfortheresultofapplyingetomandx. 14
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Otherrelatednotationsarestandard:e;sdenotesthatportionedenedbystages,andex#meansthateisdenedonxand"meansundened.Wealsoindextheprimitiverecursivefunctions,asmallerclassoftotalfunctions,asfege2!.h;i:!2!!istypicallyacomputablebijectionsuchthath0;0i=0. AandPAdenotethecomplementandpowersetofA,respectively.z=xyisthecodingtogetheroftworealsxandy,sothatzn=xnandzn+1=ynforalln.Incomputability,areductionisabinaryrelationonsubsetsofNthatcapturesarelationshipbetweentheinformationcontentoftwosets.TheTuringreduction6Tisthemainreductionusedincomputabilitytheory.AisTuringreducibletoB,writtenA6TB,ifmembershipinAcanbedeterminedbyacomputeralgorithmthathasfullaccesstothemembershippredicateforB.Intuitively,theinformationcontentofAisviewedascomputable,orrecoverable,fromB.FurthermoreBisviewedasanoracle,intermsofinformationcontent,fordeterminingmembershipinA.Variousrestrictionsonhowmuchoracleinformationisallowedtobeusedindeterminingthemembershipofasingleelementhavegivenrisetodierentkindsofreductions.Forexample,theboundedTuringreduction6bTrequiresthatquerytoanoracleBfordeterminingmembershipofxinAuseatmostfxamountofB,wheref,calledtheuse,isboundedbyacomputablefunction.TheidentityboundedTuringreductionsinsiststhattheusefbeboundedbytheidentityfunction.Reductionsoftengiverisetoequivalenceclasses,calleddegrees,wheretwosetsareequivalentiftheyaremutuallyreducibile.TheTuringdegreesthatcontainc.e.setsarecalledc.e.Turingdegrees.WedenoteCandCbTasthestructuresofthec.e.degreesundertheTuringreductionsandtheboundedTuringreductionsrespectively.ThestudyoftheTuringdegreeshasbeenoneofthemajorthemesincomputabilityresearch;thesedegreescapturethestructureoftheundecidableproblemsinarithmeticandnature.1.3ClosedSetsinComputabilityWegenerallyfollowthenotationofCenzer[ 19 ]forclosedsets:Foranitestring2!n,letjj=n.Welet;denotetheemptystring,whichhaslength0.Awordaof 15
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length1ismaybeidentiedwiththesymbola.Fortwostrings;,saythatextendsandwritevifjj6jjandi=ifori
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thatSn=2N)]TJ/F24 11.955 Tf 11.961 0 Td[(Pfnforalln.Foranyc.e.setW,wedenethec.e.opensetgeneratedbyWtobeOW=[fI:hi2Wg:AlsoletOWn=fxn:x2OWand8j
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CHAPTER2EFFECTIVELYCLOSEDSETSANDENUMERATIONSThefollowingchapterisjointworkwithDouglasCenzerandhasbeensubmittedasanarticleentitledEectivelyClosedSetsandEnumerations[ 13 ].ApreliminaryversionofthisresearchwasoriginallypresentedattheThirdInternationalConferenceofComputabilityandComplexityinAnalysisinGainesville,Floridain2006byP.Brodhead.ThispreliminaryworkwaspublishedinthereferredconferenceproceedingsasEnumerationsof01Classes:AcceptabilityandDecidableClassesP.BrodheadinhardcopyandinSpringerElectronicNotesinTheoreticalComputerScience,ElsevierScience16707,289301[ 12 ].PortionsofthisworkwerealsopresentedbyP.Brodheadatthe2005SACNASConferenceOctober2005,Denver,CO,7thAnnualGraduateStudentConferencesinLogicApril2006,Madison,WI,the2007AssociationforSymbolicLogicAnnualMeetingMarch2007,Gainesville,FL,the2ndNewYorkGraduateStudentConferenceinLogicMarch2007,NewYork,NY,andthe8thAnnualGraduateStudentConferenceinLogicApril2007,Chicago,IL.Duetoinclimateweather,R.MillerpresentedinplaceofP.BrodheadattheNewYorkconference.2.1IntroductionThegeneraltheoryofnumberingswasinitiatedinthemid1950sbyKolmogorov,andcontinuedunderthedirectionofMal'tsevandErshov[ 44 ].Anumbering,orenumeration,ofacollectionCofobjectsisasurjectivemapF:!!C.Inoneoftheearliestresults,Friedbergconstuctedaninjectivecomputablenumberingofthe01orcomputablyenumerablec.e.setssuchthattherelationn2e"isitself01.Moregenerally,wewillsaythatanumberingofcollectionofobjectswithcomplexityCsuchasnc.e.,0n;or0niseectiveiftherelationx2e"hascomplexityC.Wewillalsoconsiderenumerationswheretherelationx2e"hasadierentcomplexitythanC.Forexample,thereisac.e.,butnotcomputable,numberingofthecomputablesets.Anumberingisacceptablewithrespecttoanumbering,denoted6,ithereisatotalcomputablefunctionfsuchthat=f.Ifisacceptablewithrespect 18
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toalleectivenumberings,thenissaidtobeacceptable.Theordering6givesrisetoanequivalencerelation,andtwonumberingsinthesameequivalenceclassarecalledequivalent.Furthermore,thestructureLCofallnumberingsofCmoduloformsanuppersemilatticeunder6.Hereinjectivenumberingsoccuronlyintheminimalelementsandacceptablenumberingsoccuronlyinthegreatestelement.Inthischapter,westudyeectivenumberingsoffamiliesofeectivelyclosedsetsi.e.01classes.Enumerationsof01classesweregivenbyLempp[ 61 ]andCenzerandRemmel[ 23 24 ],whereindexsetsforvariousfamiliesof01classeswereanalyzed.Foragivenenumeratione=Peofthe01classesandapropertyRofsets,fe:RPegissaidtobeanindexset.Forexample,fe:Pehasacomputablemembergisa03completeset.See[ 20 ]formanymoreresultsonindexsets.Certaintypesof01classesareofparticularinterest.LetPbea01class.WewillsaythatPisthinifforevery01subclassQofP,thereisclopensetUsuchthatQ=UP.WesaythatPishomogenousif,givendistinct;2TPofthesamelength,_i2TP_i2TP:ForPf0;1g!,PishomogeneousifandonlyifPistheclassofseparatingsetsSA;Bfortwodisjointc.e.setsA;B,thatis,SA;B=fC!:ACandBC=;g:Pissmallifthereisnocomputablefunctionsuchthat,foralln,cardTP!n>n.LetPnbetheleastksuchthatcardTP!k>n;thenPisverysmallifthefunctionPdominateseverycomputablefunctiong{thatis,Pn>gnforallbutnitelymanyn.Anumberinge7![Te]of01classesiscalledatreenumberingandwrittene7!Te.Numberingsbasedonprimitiverecursivetreesandon01treesarebothstudiedintheliteraturesee[ 23 24 20 ].Ifthesetfe;:2Tegiscomputable,thenthenumberinge=[Te]issaidtobeacomputablenumbering. 19
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Webeginourstudywiththefamilyof01classesinSection 2.2 .InSections 2.2.1 { 2.2.3 ,severalcommonlyusednumberingsarestudiedandshowntobeequivalentviaacomputablepermutation.InSection 2.2.4 ,wegivegiveaFriedbergnumberingofthe01classes;thismotivatesastudyofageneralclassoffamiliesof01classes,calledstringveriablefamiliesinSection 2.3 .InSection 2.4 ,weconsidernamedfamiliesof01classes.WeobtainpositiveresultsforhomogeneouscandecidableclassesinSections 2.4.1 and 2.4.2 .Weobtainnegativeresultsforthin,perfectthin,small,verysmall,andnondecidableclassesinSections 2.4.3 and 2.4.4 .2.2TheFamilyof01Classes2.2.1NumberingsintheLiteratureInthissection,wepresentseveraldierentcomputablenumberingsof01classesthathaveappearedintheliterature.Wealsopresentaneective,butnotcomputable,numbering.Ineachcasewedemonstratethateachprovidesacompletenumberingofthe01classes.Numbering1:PrimitiveRecursiveFunctions[ 23 ]Foreache,letebetheethprimitiverecursivefunctionfrom!to!andlet2Ue8vehi=1:ThenUeisauniformlyprimitiverecursivetreeforalleandiff:hi=1gisanyprimitiverecursivetree,thenUeisthattree.ThereforethesequenceU0;U1;:::containsallprimitiverecursivetreesandhencethemapping1e=[Ue]isacomputablenumberingofthe01classes.Numbering2:ComputablyEnumerableSets[ 20 ]Let2e=!!)222(OWe:Thisisaneectivenumberingsincetherelationx22e"is01.Thiscanactuallybeimprovedtoacomputablenumbering,asfollows. 20
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Foreache,recallthatWe;sisthesetofelementsenumeratedintotheethc.e.setWebystagesandlet2Se8vhi=2We;jj:ThenSeisauniformlyprimitiverecursivetreeforalle.LetP=[T]bea01class,whereTisacomputabletree.Itfollowsthatforsomee,2Thi=2We:ThenP=[Se].Itfollowsthatthesequence[S0];[S1];:::containsall01classesandhencethemappinge=[Se]isacomputablenumberingofthe01classes.Itiseasytoseethatinfact[Se]=2e.Numbering3:Universal01Relation[ 52 ,p.73]Thereisauniversal01relationU!2!suchthatifQxisa01class,thenthereisane2!suchthatQ=fx:Ue;xg.UisdenedintermsoftheKleenTpredicate,sothatessentiallyUe;xxe":Dene3e=fx:Ue;xgtoobtainaneectivenumbering.Toseethatthisisacomputablenumbering,let2Ree":sothat3e=[Re]andthetreesReareuniformlyprimitiverecursive.Numbering4:TheHaltingProblem[ 53 ]Considerthemappinggivenby4e=fx:xeee"g:Thisisacomputablenumbering,since4e=[Te],where2Teee": 21
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ForanycomputabletreeT,chooseasothatanconvergesifandonlyif2T.Then2Taa#;sothat[T]=4a.Numbering5:TotalComputableFunctionsHerewewillconsideraneective,butnotcomputablenumberingbasedoncomputabletrees.ThisnumberingwillbeusedinconnectionwithstringveriablefamiliesofclassesinSection 2.3 .Let5e=[Te],where2Te8[ehi#)167(!ehi=1]:Thisenumerationisuniformly01,butisnotcomputable,sincetherelationem#isc.e.noncomputable.Clearlyeach5eisa01class.IfeistotalandTisatreesuchthat,forall,wehave2Tehi=1,thenTe=Tandisa01class.Hencethisenumerationhasthecrucialpropertythat,foreverycomputabletreeT,thereexistsesuchthatT=Te.2.2.2EquivalenceoftheNumberingsInthissectionweshowthatthecomputablenumberingsofsection 2.2.1 aremutuallyequivalentviaacomputablepermutation.Eachoftheseisequivalenttotheeectiveenumerationofsection 2.2.1 viaa03permutation.Weneedthefollowingproposition. Proposition2.2.1. a Foreachpairi;jwith16i65and16j64,thereisacomputablefunctionfsuchthatj=if. b Foreachj65,thereisa03functionfsuchthat5=if.Proof.162:UsetheS)]TJ/F24 11.955 Tf 11.955 0 Td[(m)]TJ/F24 11.955 Tf 11.955 0 Td[(nTheoremtodenefsothatWfe=fn:en6=1g:Then2Ue2Sfe. 22
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263:Denefsothat,forallm,xfem=leastnhxni2We:Then2e=3fe.364:Denefsothatxan=xforalln.Thenx23ex24fe.461:Recallthat4e=[f:ee"g].Denetheprimitiverecursivefunctiongsothatforeache,fehi=8>><>>:1;ifee";0;otherwise.Then1e=4ge.165:Denetheprimitiverecursivefunctionfsuchthat,foreache,fehi=8>><>>:1;if8vehi=1;0;otherwise.Then1e=5fe.Therestoftheprooffollowsbycomposition. Theorem2.2.2. Foranycomputablenumbering'whichiscomputablyequivalentto2,thereisacomputablepermutationpsuchthat2='p.Proof.TheproofisamodicationofanargumentduetoJockusch[ 86 ,p.25].Let=2.Byassumption,therearecomputablefunctionsfandgsuchthatfe='eand'ge=e.Sincethenumbering2isbasedonanenumerationofthepartialcomputablefunctions,wecanensurebypaddingthatfisinjective.Tomodifygintoaninjectivefunctiong1,itissucienttoeectivelycomputefromeacheaninnitesetSeofindicessuchthat'gi='eforalli2Se.Weproceedasfollows.LetAandBbecomputablyinseparablec.e.setsanddenecomputablefunctionskand`suchthat,forall 23
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eandm:ke;m=8>><>>:e;ifm=2B;;;ifm2Band`e;m=8>><>>:e;ifm2A;2!;ifm=2AThatis,webuildatreeforke;mwhichexactlyequalsthetreeforeforstringsoflengthsuntilm2Bs+1,inwhichcasenostringsoflengths+1areputintoke;m.Tobuildthetreefor`e;m,weputinallstringsoflengthsuntilm2As+1,inwhichcaseweincludeonlythestringsoflengths+1whichareine.NowletCe=fke;m:m2AgandDe=f`e;m:m2AgandletSe=gCe[De.Thenforj=gi2Se,itfollowsfromthedenitionthati=eandtherefore'gi='ge.WewillproveintwocasesthateithergCeisinniteorgDeisinnite.CaseI:Supposethate6=;andsupposebywayofcontradictionthatgCeisnite.ThenS=fm:gke;m2gCegisacomputableset.NowAgCebydenition.Ontheotherhand,ifj=gke;m2gCewherem2A,then'j=ke;m=e6=;.Butform2B,'gke;m=ke;m=;,sothatSB=;.ThiscontradictstheassumptionthatAandBarecomputablyinseparable.CaseII:Supposethate=;.ItfollowsasinCaseIthatgDeisinnite.Thuswemayassumewithoutlossofgeneralitythatbothfandgareonetoone.Nowdeneasequencefen:n2!gandtwopartitionsof!asfollows.Lete0=0andforeachn,en+1istheleastesuchthat'e6='eiforeveryi6n.LetAn=fe:e=engandBn=fe:'e=eng.Then!=SnAn=SnBnandeachsequenceispairwisedisjoint.Furthermore,fBnAnandgAnBn.TheremainderoftheprooffollowsasintheMyhillIsomorphismTheorem[ 86 ,p.24;Also5.8,p.25]. Asimilarargumentshowsthatif'isa03numberingofthe01classes,thenthereisa03permutationpwith'=2p.Itfollowsthateachofthecomputablenumberings 24
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1;:::;4areacceptable,thatis,theyoccurinthegreatestelementofthesemilatticeLP.Inthesection 2.2.4 wewillseethatminimalelementsexistinthesemilattice{thatis,injectivenumberings.First,however,weprovideanalternateproofofTheorem 2.2.2 .2.2.3EquivalenceoftheNumberingsAlternateProofAttheThirdInternationalConferenceofComputabilityandComplexityinAnalysisinGainesville,Floridain2006,IpresentedtheargumentforTheorem 2.2.2 asgivenabove;itisamodicationofanargumentofJockuschforthec.e.version.Pleasedwiththeargument'sapplicabilityto01classes,IurgedRobertSoare,whowaspresentintheaudience,tokeeptheJockuschargumentforthec.e.versioninhisnewandupcomingbook,ComputabilityTheoryandApplications[ 88 ].Ifthealternateproofusingtherecursiontheoremforthec.e.versionofTheorem 2.2.2 [ 86 ,p.43]couldnotalsobemodiedtoalsoproveTheorem 2.2.2 ,thenhesaidhewould.Weshowbelowthatanalternateproofispossible,modifyingtherecursiontheoremargument. Theorem2.2.3. Foranycomputablenumbering'whichiscomputablyequivalentto2,thereisacomputablepermutationpsuchthat2='p.AlternateProof.Weproceed,atrst,asbefore.Thatis,therearecomputablefunctionsfandgsuchthatfe='eand'ge=e.Ourgoalistomodifygtoobtainaninjectivefunctiong1.Instead,however,wedeneg1dierently,usinganauxiliarycomputablefunctionhobtainedbytheRecursionTheorem.Weensurehsatises,foralldistinctiandj,ighe;i6=ghe;jandiihe;i=e.Wenowdeneg1,givenhsatisfyingiandii.Deneg1=ghfg;0.Todeneg1k+1,notethatiensuresinnitelymanydistinctghe;iforeache.Leta0=0andak+1betheleastintegerisuchthatghfgk+1;i>g1k.Deneg1k+1=hfgk+1;ak+1.Toseethatforalle,g1e=e,xeandnotethatg1e=ghfge;ae=hfge;ae=fge=ge=e.Toseethatg1isinjective,notethatthedenitionensuresthatforallk,g1k+1>g1k. 25
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Wenowdenehe;foreache,byinduction,ensuringthatiandiiaresatised.Denehe;0=e.Todenehe;k+1,usetheRecursionTheoremtoobtainannsuchthatnz=8><>:ezifgn6=ghe;i8i6kundenedotherwiseNoticethatifgn=ghe;ithensincei6k,byinductione=he;i=ghe;i=gn=n,theundenedfunction.Hencee=ninallinstances.Sincefa:a=ngisaproductive[ 86 ,p.43],letpbeacorrespondingproductivefunction.DeneWrx=Wx[fpxgandnotethateachWrinisadistinctc.e.subsetofA.Consequentlyeachgrin=rinisadistinctpartialcomputablefunction.Letnkbetheleastisuchthatrin6=he;jforallj6k.Denehe;k+1=8><>:nifgn6=ghe;i8i6krnknotherwiseToseethatiholdsghe;i6=ghe;jfordistincti;j,notethatge=e6=rin=grinforalli>1.Thereforeifgn6=gethenforallj,he;j2feg[fringi>0.Otherwisehe;j2feg[fringi>1.Ineithercase,foralldistincti;j,ghe;i6=ghe;j.Toseethatiiholdshe;i=eforalli,notethate=n,he;i2feg[fringi>or1,andrin2fa:a=ng.Soe=he;iforalli. 2.2.4InjectiveComputableNumberingsInthissection,wemodifyFriedberg'soriginalargumentforinjectivecomputablenumberingofthec.e.sets,toprovidedierentnumberingresultsneededfor01classes.InSection 2.2.4.1 ,wepresentFriedberg'soriginalargument.InSection 2.2.4.2 ,weprovideaninjectivecomputablenumberingofdisjointpairsofc.e.sets;thisinneededlaterinSection 2.4.1 inordertoprovideaninjectivecomputablenumberingofthehomogeneous01classes.Finally,inSection 2.2.4.3 ,weconstructacomputableinjectivenumberingofthe01classesin2!;wealsoprovideotherinjectivenumberingresultsfor01classes. 26
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2.2.4.1Originalc.e.setsargumentThefollowingistheoriginalFriedbergargument,withslightmodicationsinthenotationandpresentation. Theorem2.2.4[ 46 ]. Thereisaninjectivecomputablenumberingofallc.e.sets.Proof.LetfWege2!bethestandardnumberingofthec.e.sets.Wewillconstructasequenceofc.e.setsfYege2!instagessothate7!Yewillbethedesiredinjectivenumbering.IntheconstuctionwewillusethenotionofoneYindexifollowingaWindexewiththeideathatintheendYiwillequalWe.Atsomepoint,however,wemaydecidethatiwillnolongerfolloweagainandwewillsaythatiisreleased.Ifiisneverreleasedfromfollowingethenitissaidtobealoyalfollowerandotherwiseitisdisloyal.Oncereleased,anindexremainsfreeandisneveragainthefollowerofanye.Atanyparticularstage,anyYindexthatisnotfollowinganyWindexissaidtobefree.AnynonzeroYindexthathasneverfollowedanyWindexissaidtobeunused.Toensurethatnoc.e.setisexcludedfromtheYsets,wewillensurethateachWeisinnitelyoftengiventheopportunitytobefollowed.Todothis,atstages=hns;esiallactionsintheconstructionwillbetakenwithrespecttoWes.AssumewithoutlossofgeneralitythatY0=;.Construction:Therearethreepossiblecasesateachstages.Case1:Ififollowses&We;s[0;i)]TJ/F15 11.955 Tf 12.044 0 Td[(1]=Wes;s[0;i)]TJ/F15 11.955 Tf 12.044 0 Td[(1]somee
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i 8e9iYi=W^e ii i6=j)167(!YiandYjarenotthesamenitesets iii i6=j)167(!YiandYjarenotthesameinnitesetsVericationofi.Fixe.Firstnotethatalthough^ecanonlyhaveonefolloweratanyparticularstage,itcannothaveaninnitenumberofthemwhichareeachreleasedatsomestage.Forexample,ifsandxaresucientlylarge,thenforallj<^e,Wj;s[0;x]6=W^e;s[0;x].HencereleasecanonlyoccurinCase1anitenumberoftimes.Furthermore,Case2ensuresthatreleaseinCase3canonlyoccurforanyswhenes<^e.Therefore,bytheabove,ifi>xisfollowerof^eandt>s,thenYi;t)]TJ/F23 7.97 Tf 6.586 0 Td[(1=W^e;t)]TJ/F23 7.97 Tf 6.586 0 Td[(16=Wes;t.HenceiwillnotbereleasedinCase3.ThereforereleasecanonlyoccurinCase3anitenumberoftimes.Nowletsbeastagewhere^eneverlosesafollower.IfCase3occursinnitelyoftenafterstagesfor^e,thenithasapermanentfollowerisothatYi=W^e.ThereforeassumeCase3occursonlynitelyoften.Since^eneverlosesafollower,Case1cannotoccur.HenceCase2mustoccurinnitelyoften.Howeverthereareonlyanitenumberofisuchthatthe`If'clauseholdswithW^e;s=Yi;s)]TJ/F23 7.97 Tf 6.586 0 Td[(1"inCase2.Forexample,fi:i=0;i6^e;oriisdisplacedby^egisanitesetsinceonlyanitenumberofiaredisplacedduetoCase3occuringonlyanitenumberoftimes.Toseealsothatfi:forsomes;ifollowssomea<^eandW^e;s=Yi;s)]TJ/F23 7.97 Tf 6.587 0 Td[(1"holdsgisnite,notethatbythedenitionof^e,ifa<^ethenWa6=W^e.Soforsucientlylarget,ififollowsathenW^e;t6=Yi;t)]TJ/F23 7.97 Tf 6.586 0 Td[(1.ThereforeCase2occuringinnitelyoften,togetherwithonlynitenumberofisuchthatthe`If'clauseholdswithW^e;s=Yi;s)]TJ/F23 7.97 Tf 6.586 0 Td[(1",impliesthatthereisasingleisuchthatYi;s)]TJ/F23 7.97 Tf 6.587 0 Td[(1=W^e;sforinnitelymanys.ThusYi=W^e.Vericationofii.Firstnotethatatanystages,ifWes=;thenCase2ensuresthatCase3willnotbereachedsothatesneverreceivesafollower.Furthermoreeachk6=0ischosenatsomestagesinCase3tofollowsomeeandfromthepreviouscomments,Yk;s6=;.Immediatelythereafter,Yk;sisensuredtobedistinctfromallotherY`;s`6=k.ThisalsocontinuestobetrueatallsubsequentstagestbyCase3. 28
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Nowsupposei6=j.SincewearesupposingbothYiandYjarenite,thereissometsuchthatforalls>t,Yi;s=YiandYj;s=Yj.AsmentionedaboveYi;s6=Yj;sandthereforeYi6=Yjasrequired.Vericationofiii.AssumebothYiandYjareinniteandi6=j.NowimusteventuallyfollowsomeWindex.Ifiisdisloyal,thenafteritisreleasedYicanacquireanewmemberonlywheniisdisplaced.Howevericanbedisplacedonlyoncebyeachei.ItfollowsthatifiisreleasedthenYionlyacquiresanitenumberofelementsthereafter,contradictingthefactthatitisinnite.Thisargumentshowsthatbothiandjareneverreleased.Wesaythatiandjareloyalfollowers.Supposenowthatiandjloyallyfollowi0andj0,respectively.ThenYi=Wi0andYj=Wj0.Nowi06=j0sinceasingleWindexcannothavemorethanoneloyalfollower.Assumewithoutlossofgeneralitythati0
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s=hns;hes;isiiallactionsintheconstructionwillbetakenwithrespecttoWes;Wis.AssumewithoutlossofgeneralitythatY0;Y0=;;;.Construction:Therearethreepossiblecasesateachstages.Case1:Ifhe;iifollowshes;isiandWa;s[0;he;ii)]TJ/F15 11.955 Tf 21.148 0 Td[(1]=Wes;s[0;he;ii)]TJ/F15 11.955 Tf 21.148 0 Td[(1]andWb;s[0;he;ii)]TJ/F15 11.955 Tf 19.707 0 Td[(1]=Wis;s[0;he;ii)]TJ/F15 11.955 Tf 19.707 0 Td[(1]forsomeha;bi
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above,ifha;bi>hx;yiisfollowerofhe;iiandt>s,thenYa;t)]TJ/F23 7.97 Tf 6.587 0 Td[(1=We;t)]TJ/F23 7.97 Tf 6.586 0 Td[(16=Wes;torYb;t)]TJ/F23 7.97 Tf 6.586 0 Td[(1=Wi;t)]TJ/F23 7.97 Tf 6.586 0 Td[(16=Wis;t.Henceha;biwillnotbereleasedinCase3.ThereforereleasecanonlyoccurinCase3anitenumberoftimes.Letsbeastagewherehe;iineverlosesafollower.IfCase3occursinnitelyoftenafterstagesforhe;ii,thenithasapermanentfollowerhe;iisothatYe;Yi=We;Wi.ThereforeassumeCase3occursonlynitelyoften.Sincehe;iineverlosesafollower,Case1cannotoccur.HenceCase2mustoccurinnitelyoften.Thereareonlyanitenumberofh;isuchthatthe`If'clauseholdswiththeequality;;e;i;sgivenbyY;s)]TJ/F23 7.97 Tf 6.587 0 Td[(1;Y;s)]TJ/F23 7.97 Tf 6.587 0 Td[(1=We;s;Wi;s"inCase2.Forexample,fh;i:h;i=0;h;i6he;ii;orh;iisdisplacedbyhe;iigisanitesetsinceonlyanitenumberofh;iaredisplacedduetoCase3occuringonlyanitenumberoftimes.Toseealsothatfh;i:forsomes;h;ifollowssomeha;bit,Yk;s=Ykk2fe;i;;g.AsshownaboveYe;s;Yi;s6=Y;s;Y;sandthereforeYe;Yi6=Y;Y.SinceYi=Y,itfollowsthatYe6=Y. 31
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SupposenowthatYiandYarebothinnite.Nowhe;iimusteventuallyfollowsomehe0;i0i.Ifhe;iiiseverreleasedthenitisfree.ThereafteronlytherstcoordinateofYe;YiacquiresmemberssothatYiisnite,acontradiction.Asimilarargumentholdsforh;i.Thereforebothhe;iiandh;iareneverreleasedandareloyalfollowers.Supposethattheyfollowhe0;i0iandh0;0i,respectively.ThenYe;Yi=We0;Wi0andY;Y=W0;W0.Notethathe0;i0i6=h0;0isinceasingleWindexcannothavemorethanoneloyalfollower.Assumewithoutlossofgeneralitythathe0;i0ihe;ii.Itfollowsthatifhe;iiisreleasedthenYeonlyacquiresanitenumberofelementsthereafter,contradictingthefactthatitisinnite.Thisargumentshowsthatbothhe;iiandh;iareneverreleasedandarethereforeloyalfollowers.NowapplythesameargumentgiveninthelaterpartofthevericationofiitogetthatYe6=Y.Vericationofiv.Ifhe;iiisaloyalfollowerofsomehe0;i0i,thenYe;Yi=We0;Wi0.Thereforesincehe;ii7!We;Wiisanenumerationofdisjointsets,itfollowsthatYeandYiaredisjoint.Otherwisesupposethathe;iiisreleasedatsomestages.ThenYe;sYi;s=;andthereafteronlyYecanacquirenewelementsnotalreadyincludedinYi;s.ThereforedisjointnessispreservedandYeYi=;. 2.2.4.3ResultsforeectivelyclosedsetsInthissection,wemodifytheFriedbergargumentofSection 2.2.4.1 toconstructacomputableinjectivenumberingofthe01classesin2!.AnalternativeproofwassketchedbyRaichev[ 79 ].Afterwardsweprovideotherinjectivenumberingresultsfor01classes. 32
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Theorem2.2.6. Thereisaninjectivecomputablenumberingofall01classesin2!.Proof.LetfWege2!bethecomputableenumerationofthenonemptyc.e.subsetsof2
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2k:62OWes;skg.Let`sbetheleastksuchthatjStrk;sj>jEsj.Then`sistheleastlevelofOWes;swherethereisenoughroomtogiveeachequivalentYj;s)]TJ/F23 7.97 Tf 6.587 0 Td[(1anadditionalstringtodistinguishOYj;s)]TJ/F23 7.97 Tf 6.586 0 Td[(1fromOWes;s.NoticethatOWes;s6=2!byCase2.SupposethatStr`s;s=f12:::jStr`s;sjg.NowputjintoYjandreleasejifitisafollower.Wesaythatjisdisplacedatstages.Verication:Givene2!,let^ebetheleastksuchthat[OWk=OWe].Wewillshow: i 8e9iYi=W^e; ii i6=jimpliesthatOYiandOYjarenotequalwhenbothareclopen. iii i6=jimpliesthatOYiandOYjarenotequalwhenbotharenotclopen.Vericationofi.Fixe.Firstnotethatalthough^ecanhavedierentfollowersatdierentstages,itcannothaveaninnitenumberofdisloyalfollowers.Thatis,ifsandxaresucientlylarge,thenbythedenitionof^e,forallj<^e,OWj;sx6=OW^e;sx.HencereleasecanonlyoccurinCase1anitenumberoftimes.Furthermore,Case2ensuresthatreleaseinCase3canonlyoccurforanyswhenes<^e.Therefore,bytheabove,ifi>xisfollowerof^eandt>s,thenOYi;t)]TJ/F23 7.97 Tf 6.587 0 Td[(1=OW^e;t)]TJ/F23 7.97 Tf 6.587 0 Td[(16=OWes;t.HenceiwillnotbereleasedinCase3.ThereforereleasecanonlyoccurinCase3anitenumberoftimes.Nowletsbeastageafterwhich^eneverlosesafollower.IfCase3occursinnitelyoftenafterstagesfor^e,thenithasapermanentfollowerisothatOYi=OW^e.ThereforeassumeCase3occursonlynitelyoften.Since^eneverlosesafollower,Case1cannotoccur.ThusCase2mustoccurinnitelyoften.HoweverthereareonlyanitenumberofisuchthatthehypothesisofiiholdswithOW^e;s=OYi;s)]TJ/F23 7.97 Tf 6.587 0 Td[(1.Toseethis,considerthethreecases.First,eache
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Vericationofii.SupposethatU=OYi=OYj6=;;2!isclopenandletU=OW^e.Itfollowsfromcompactness,thatthereissomenitessuchthat,forallt>s,OYi;tt=OYj;tt=OYi.Itfollowsfromthevericationofiabovethatthereisastaget>ssuchthatCase3appliesto^e.ButthenatleastoneofOYi;OYjmustchangeatstaget.Thiscontradictionveriesii.Vericationofiii.AssumebothOYiandOYjarenotclopenandi6=j.ItfollowsthatOYimustchangeinnitelyoften,sinceofcourseOYi;sisclopenforeachs,andsimilarlyforOYj.NowimusteventuallyfollowsomeWindex.Ifiiseverreleased,thenitisfree.ThereafterYiacquiresmembersinCase3atstagesonlywhenWes;s=Yi;s)]TJ/F23 7.97 Tf 6.586 0 Td[(1.ThisimpliesthatCase2doesnotapplyatstagesandthuses
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Corollary2.2.8. Thereisaneectivenumberingofthe01classesbasedonthetotalcomputablefunctionsProof.Modifyeachc.e.setinthestandardnumberingtoenumerateanelementonlyaslongasitislargerthananypreviouslyenumeratedelement.ApplyingFriedberg'sargumenttothisclassofc.e.setsyieldsaneectiveinjectivenumberinge7!Ceofthecomputablesets[ 91 ].FurthermoreeachCestillenumeratesitselementsinincreasingorder.Nowsupposefege2!isacorrespondingsetofcharacteristicfunctions.Onecharacterizationofa01classPisthatP=!!nOWforsomecomputablesetW[ 20 ].Asaresult,e7!!!nOCe=!!nOfn:en=1gisanalternativeeectivenumberingbasedontotalcomputablefunctionsreplacingnoneectiveNumbering2. Itisknown,forxedn>0,thatthereisaeectiveinjectivenumberingofthenc.e.sets[ 50 ]. Conjecture2.2.9. Foreachn,thereisanumberinge7!Neofnc.e.setssuchthatthereisaninjectivecomputablenumberinge7!!!nONeofallclosedsetsofthisform.Forn=1theconjectureisgivenbyTheorem 2.2.6 .WenextshowthatTheorem 2.2.6 isnotobtainablebyanycomputableprocedurethatuniformlyselectstheminimalindexofevery01class. Theorem2.2.10. Thereisnocomputablechoicefunctionforindicesof01classes.i.e.acomputablefunctionfsuchthatfeisanindexofPeandPi=Pefi=feProof.Supposethatfexists.Leta0;a1;:::beanenumerationofanoncomputablec.e.setA.DeneacomputablefunctiongandtreesTgesothatifjj=n,then2Tge$e62fa0;:::;ang:ThenPge=8><>:;ife2A2!otherwiseForanya2A;e2A$fge=fga,makingAcomputable. 36
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Itisstillpossible,however,thatsomeinterestingproperfamilyof01classesmaybeenumeratedbyselectingminimalindicesfromtheenumerationofall01classes.2.3StringVeriableFamiliesof01ClassesInthissectionweexaminefamiliesofclasseswhicharedeemedtobestringverablee.g.decidableorhomogeneousclassesorstronglystringveriablee.g.stronglydecidableorstronglyhomogeneousclasses.Anystringveriablefamilyhasaneectivenumberingandanystronglystringveriablefamilyhasacomputablenumbering.2.3.1DenitionandExamplesFirstwewilldenethenotionsofstringveriableandstronglystringveriable.Wealsogivesomeexamples. Notation2.3.1. LetF0;F1;:::beacomputableenumerationofthenitesubsetsof2
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Denethestringvericationfunctionhasfollows.LetA1;A2;A3;A4enumeratePf0;1g
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2.3.2ComputableandEectiveNumberingsWenowdemonstratethatstronglystringveriableandstringveriablefamiliespossesscomputableandeectivenumberings,respectively. Theorem2.3.7. a Anystronglystringveriablefamilyof01classeshasacomputablenumbering. b Anystringveriablefamilyof01classeshasaneectivenumbering.Proof.SupposeFisa[strongly]stringveriablefamilyof01classessatisfyingstringveriabletreerelationsH0;H1;:::;Hm,withcorrespondingstringfunctionsh0;h1;:::;hm.Forparta,letthestandardcomputableenumerationofthe01classesin2!begivenbyPe=[Te],wherethesequenceTeisuniformlyprimitiverecursiveforexample,thenumbering2giveninsection 2.2.1 .WewilldeneauniformlycomputablesequenceSeoftreessuchthatthesequenceQe=[Se]enumeratesexactlythefamilyof01classesstronglysatisfyingH0;H1;:::;Hm.Forany2f0;1gn,wedeterminewhether2Seasfollows.Firstcheckthat2Te.Ifso,foreach2f0;1gnandeachi6m,computehi=D1;D2;:::;D2janddeterminewhetherthereexistsi
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Wenowprovidesomeadditionalexamplesofstringandstronglystringveriableclasses. Denition2.3.8. i A01classPisstronglydecidableifthereisaprimitiverecursivetreeTwithnodeadendssuchthatP=[T]. ii A01classPisstronglyhomogeneousifthereisahomogeneousprimitiverecursivetreeTwithnodeadendssuchthatP=[T].Fromdenition 2.3.8 ,weimmediatelyhavethefollowingcorollarytoTheorem 2.3.7 Corollary2.3.9. a Thefamilyofdecidable01classesin2!hasaneectivenumberingandthefamilyofstronglydecidable01classesin2!hasacomputablenumbering. b Thefamilyofhomogeneous01classesin2!hasaneectivenumberingandthefamilyofstronglyhomogeneous01classesin2!hasacomputablenumbering.Inthefollowingsectionwewillprovideresultswhichdemonstratethatdecidableclassesinfacthaveinfacta11computablenumberingseeCorollary 2.4.2.1 .Weprovideanalternateexplicitproofinsection 2.4.2.1 .Forhomogeneousclasses,adierentapproachisneededtodemonstratethattheypossessa11computablenumberingandwedothisinsection 2.4.1 .2.3.3FamiliesContainingtheClopenClassesInthissectionweconsiderstringveriablefamiliesofclassesthatcontainallclopenclasses.WerstimproveTheorem 2.3.7 toobtainacomputablenumberingofanystringveriablefamilywhichincludestheclopensets. Theorem2.3.10. IfFisanystringveriablefamilyof01classes,thenthereisacomputablenumberingofC[F.Proof.WemodifytheproofofTheorem 2.3.7 sothatwhenthestringveriablerelationsfail,weextendallnodesratherthanmakingthemdeadends.Onceagain,theconstructionisbasedontheenumerationeofthepartialcomputablefunctions.Theconstructionisinstages,whereatstageswewillhavene;s=maxfn:82f0;1gne;shi#g; 40
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Je;s=f2f0;1gne;s:e;shi=1g;andQe;s=[Je;s=[Se;s]:ThenQe=TsQe;s=[Se]willbethedesirednumbering.Toensurethatthisnumberingiscomputable,wewilldeterminewhether2Seatstagejj.Forthisargument,weassumethate=1foralle.Construction.Atstage0wehavene;0=0,Je;0=Se;0=f;gandQe;0=2!.Atstages+1,wechecktoseewhethere;s+1hi#forall2f0;1gns+1.Ifnot,thenne;s+1=ne;s,Je;s+1=JsandSe;s+1=Se;s[f_i:2Se;s;i=0;1g.Ifso,thenwechecktoseethate;s+1isthecharacteristicfunctionofatreeonf0;1gne;s+1andweverifythestringrelationsuptof0;1gne;s+1.Ifthisvericationfails,thenagainne;s+1=ne;sandJe;s+1=Je;s.Inthiscase,vericationwillalsofailatallfuturestages,sothatQe=Qe;sisaclopenset.Ifthetreeandstringvericationssucceed,thenne;s+1=ne;s+1,sothatJe;s+1f0;1gne;s+1andQe;s+1changeasindicatedabove.Inthiscase,Se;s+1=Se;s[f2f0;1gs+1:ne;s+12Je;s+1g:IfeisthecharacteristicfunctionofthecomputabletreeTe,andifPe=[Te]2F,thenitfollowsfromtheconstructionthatQe=Pe,sothatQe2Fandfurthermore,any01classPe2Fwilltherebyoccurinthenumbering.Otherwise,theconstructionwillmakeQeaclopenset. Corollary2.3.11. ForanystringveriablefamilyFof01classes,therea11computablenumberingofC[F.Proof.LetFbeastringveriablefamily.ThenthereisacomputablenumberingofC[FbyTheorem 2.3.10 .ItthenfollowsfromTheorem 2.2.7 thatthereisa11computablenumberingofC[F. 41
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Corollary2.3.12. Therea11computablenumberingofanystringveriablefamilyof01classescontainingallclopenclasses. Corollary2.3.13. Therea11computablenumberingofthedecidable01classes.2.4NamedFamiliesof01ClassesItthissection,weotainnumberingresultsforvariousnamedfamiliesthatcommonlyoccurintheliterature.Inthersttwosubsections 2.4.1 2.4.2 ,weexpandupontheresultsfromsection 2.3 andweobtainpositivenumberingresultsforthehomogeneousanddecidableclasses.Inthetwosubsections 2.4.3 2.4.4 thatfollowthese,weobtainnegativeresultsforthethin,perfectthin,small,verysmall,andnondecidableclasses.2.4.1HomogeneousClassesHomogeneous01classesareastringveriablefamilyof01classes.Consequently,byCorollary 2.3.9 ,theypossessaneectivenumbering.Corollary 2.3.12 fallsshortofdemonstratingthattheypossessacomputablenumbering,asclopensetsarenotnecessarilyhomogeneous.WeprovidethenecessaryargumentinTheorem 2.4.3 andshow,infact,thatacomputableinjectivenumberingexists.Werstprovideanalternatecharacterizationofthehomogeneousclasses;theymaybeviewedseparatingclassesofc.e.sets. Denition2.4.1. TheseparatingclassSA;BoftwosetsA;B!isgivenbySA;B=fC!:ACandBC=;g.Inwhatfollows,ScA;BwilldenotethesetofcharacteristicfunctionsfCofC2SA;B. Theorem2.4.2. P2!isanonemptyhomogeneous01classiP=ScA;Bforsomedisjointc.e.setsAandB.Proof.!SupposethatPisanonemptyhomogeneousclasswithhomogeneoustreeTP.WewillshowthatP=ScA;BwhereA=fn:0n_062TPgandB=fn:0n_162TPg.ToseethatAandBarec.e.,supposethatP=[T]withTcomputable.Thenforeachi2f0;1g,0n_i62TPi9s>n+1Tf:jj=sand0n_ivg=;. 42
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ToseethatAandBaredisjoint,supposethatn2A.SincePisnotempty,thereissomex2Psothatxn2TP.Now0n_062TPsothatxn_062TP.Hencexn6=0sothatxn=1.Thereforexn_12TP.SincePishomogeneous,0n_12TPsothatn62B.HenceAB=;.ToseethatScA;BP,takefC2ScA;BforsomeC2SA;B.Then,n2C!n62B!0n_12TP!fCn+1=fCn_12TP&n62C!n62A!0n_02TP!fCn+1=fCn_02TPSinceforarbitraryn,fCn+12TP,itfollowsthatfC2P.ToseethatPScA;B,supposethatx2PandletC=fn:xn=1g.ItisclearthatxisthecharacteristicfunctionfCofC.ItsucestoshowthatC2SA;BsothatfC2SCA;B.Supposerstthatn2A.Then0n_062TPsothatfCn_062TP.Sincex=fC2TP,itfollowsthatfCn=1.Hencen2CandAC.Nowsupposen2B.Then0n_162TP,sofCn_162TP.Asbefore,fCn=0sothatn62C.HenceCB=;.SupposeAandBaredisjointc.e.setsandfAsgs2!andfBsgs2!arestageenumerationsofAandB,respectively.DeneP=s[Ts]whereTs2
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2.4.2DecidableClassesInthissectionweprovideanalternateproofoftheexistenceofinjectivecomputablenumberingsofthedecidableclassesseeCorolllary 2.3.13 fortherstproof.Wealsoshowthatinanycomputablenumberingofthedecidableclassesviatrees,wecanprovideacomputablenumbering^ofallthetreeswithoutdeadendsthatoccur,alongwithallclopenclasses.Finally,weshowthatsomedecidableclassinclassinthenumberingmustnecessarilyhave,asatree,deadendsthroughouteveryoccuranceinthenumbering.2.4.2.1AninjectivecomputablenumberingAlternateproofSincedecidabilityisstringveriableandeveryclopensetisdecidable,itfollowsfromCorollary 2.3.12 thatthedecidableclasseshavea11computablenumbering.Thisresultcouldnotbeobtainedbyusingthestandardnumberingofthe01classesandmodifyingeachtreeasitbecomesknownthatishasadeadend.Forexample,simplyextendeachsuchnodewith,say,allones.Thisisbecause,asaconsequencetothefollowingtheorem,PbeingdecidableisinsucienttoensurethattheuniquetreeTPwithoutdeadendsshowsupinacomputabletreenumbering.Thefollowingisanalternate,explicitproof. Theorem2.4.4. Thereisa11computablenumberingofalldecidableclassesin2!.Proof.Lete7!WebetheinjectiveeectivenumberingofthecomputablesetsasgivenintheproofofCorollary 2.2.8 .Wewilldeneaneectivecorrespondencebetweenthesesetsandthenonemptycomputabletreeswithoutdeadends.Wewilldothisthroughaseriesofthreeonetoonecorrespondences{namelythecorrespondencesbetweenthesubsetsof!and2!,2!and3!,and3!andthenonemptytreeswithoutdeadends.Inastageconstructionwewillthendeneatstagee,satreeTe;sbasedonWe;sandthecorrespondences.LettingTe=sTe;s,wewillobtainaninjectivecomputablenumberinge7![Te]ofallnonemptydecidable01classes.Furthermore[Te]willcorrespondtoWeforeache.Finally,byappendingtheemptyclasstotheenumerationweobtainthedesiredresult.Wenowdenethecorrespondences. 44
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TheonetoonecorrespondencebetweenthesubsetsSof!and2!isgivenasfollows.EachScorrespondstoxS22!givenbyxSi=1ii2S.Theonetoonecorrespondencebetween2!and3!isgivenasfollows.Letx22!anddenea0=1;a1=01,anda2=00.Thenxcorrespondstotheuniquesequencefxii2!23!wherefx:!!3andx=afxafxafx:::TheonetoonecorrespondencebetweenthesetofallnonemptytreesT2
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Proof.AssumewithoutlossofgeneralitythatfTege2!containsallclopentrees.Wewillconstructinstages,asintheterminologyofTheorem 2.2.6 ,asequenceoffollowertreesSi.Atstageiwewillensurethatwehavei+1treesS0;S1;:::;Si,constructeduptolevel2i,followingtreesTS0;ki;:::;TSi;kiki2fm;ngwhichareeachpairwisedistinctatlevel2i.Also,atstagei,initiallysomeoftheSiwillhavethestatusofbeingmarkedki=minwhichcaseSiwillcontinuetofollowTSi;mforever.Ifnot,thenSiisnotmarkedki=nandwedetermineforeachi,ifSishouldbemarked.IfanSineedstobemarkedthenwedetermineatreeTSi;mthatitshallhereafterfollow.OtherwiseeachSicontinuestofollowTSi;nandthestageiscomplete.Construction.Stage0.FindthersttreeTisuchthatTif0;1g206=;,denotethistreeasTS0;n,anddeneS0=TS0;nf0;1g620.Stagej+1.S0;:::;Sjhavealreadybeenconstructeduptolevel2jandarealreadyfollowingtreesTS0;kj;:::;TSj;kj.Weperformthefollowingtwoactionsatthisstage:ConstructS0;:::;Sjuptolevel2j+1bydeterminingthetreesTS0;kj+1,:::,TSj;kj+1theyshallfollow,andConstructanewtreeSj+1uptolevel2j+1.Action.LetUj+1=fSi;kj:kj=nandTSi;kjhasdeadendsatlevel2j+1g.AllSisuchthatSi;kj62Uj+1keeptheirstatusasmarkedorunmarked,sokj=kj+1,andcontinuetofollowTSi;kj+1.ThoseSisuchthatSi;kj2Uj+1willhereafterbemarkedandwillnowfollowthetreeTSi;m=f:vorvforsome2TSi;noflength2jg.NotethateachmarkedSifollowsaclopentreeTSi;m.Action.LetSj+1;nbetheleastisuchthatTiisdistinctfromallTSi;kj+1i6jatlevel2j+1andsuchthatTihasnodeadends.DeneSj+1=TSj+1;nf0;1g6j+1.Verication.Wenowverifythat:iForeachi,limjTSi;kj#=Si=TniforsomeTniwithoutdeadends,iiForalli,ifTihasnodeadendsthenthereisacsuchthatTi=Sc,andiiii6=jimpliesthatSi6=Sj.Vericationofi.Forallj,kj=norkj=m.Fixi.ByAction,atstagei,Si;ki=Si;n.ByAction1,k`=k`+1=nforall`>iifSiisnevermarked.IfSiismarkedatstager>i,thenforalls>r,ks=ks+1=m.Ineithercaselimj>ikj#sothat 46
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limjSi;kjconvergestoSi;norSi;m.IfitconvergestoSi;mthenSineverdivergesfromfollowingtheclopentreeTSi;m.OtherwiseSiisnevermarkedandcontinuallyfollowsTSi;n.SinceitisnevermarkeditmeansthatTSi;nneverhasdeadendsuptolevel2r,forallr>i.SoTSi;nisatreewithoutdeadends.EitherwaylimjTSi;kj#=TniforsometreeTniwithoutdeadends.Nowforalln,Sif0;1g6n=TSi;knf0;1g6nandTSi;knTSi;kn+1.ThereforeSi=limjTSi;kj=Tni.Vericationofii.LetTibeatreewithoutdeadends.Therearetwocases.IfthereisastagejandacsuchthatTi=TSc;matstagej,thenbytheconstrctionTi=Sc.Ifnot,letbiequaltheleastksuchthatTk=Ti.LetjbelargeenoughsothatTbidiersfromTeatlevel2jforalleiatlevel2j.SoSi6=Sjifi6=j. 2.4.2.3Treeswithdeadends:AnecessityInanycomputablenumberingof01classesviatrees,somedecidableclassmustnecessarilyhave,asatree,deadendsthroughouteveryoccuranceinthenumberingseeCorollary 2.4.7 .TwodierentproofsofthisfactmaybeobtainedfromTheorems 2.4.6 and 2.4.8 below. Theorem2.4.6. Inanycomputablenumberingofcomputabletreesin2
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Corollary2.4.7. ForanycomputablenumberingPe=[Te]ofthe01classesin2!,thereisadecidable01classPsuchthatP6=[Te]foranyTewithoutdeadends.Proof.LetP=[T]whereTisthecomputabletreewithoutdeadendsprovidedbyTheorem 2.4.6 .SupposethatP=[Te]forsomee.SinceThasnodeadends,itfollowsthatT=TPandifTealsohadnodeadends,thenTe=TP=T.Butbytheconstruction,Tf0;1ge+16=Tef0;1ge+1,sothatT6=Te. Itfollowsfromthiscorollarythatinthestandardnumbering,fe:Tehasnodeadendsg6=fe:Pe=[Te]isdecidableg.Infactbothhavedistinctcomplexities.ByKonig'sLemma,ExtPe=f22
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2.4.3ThinandPerfectThinClassesIntheliterature,aMartinPourEltheoryisaconsistentc.e.propositionaltheorywithadditional`thinness'conditions.Theconditionsimposedhavevarieddependinguponthecontextandmotivationoftheauthors,butinclude:fewc.e.extensions,essentiallyundecidable,andwellgenerated.Someauthorshavechosentoonlyimpose1[ 21 ],whileothersand2[ 19 ],[ 24 ],andnallyothers,2,and[ 34 ],[ 40 ],[ 29 ].Thecompleteconsistentextensionsofthesetheoriescorrespondtothin,perfectthinorequivalently,specialthin[ 19 ],andhomogenousthinclasses,respectively.ThissectionisdevotedtowardsdemonstratingthenonexistenceofcomputablenumberingsofthersttwocasesbymodifyingtheclassicalMartinPourElconstructionofaperfectthinclass.RecentlySolomon[ 89 ]alsomodiedthistheoremtoconstructahomogeneousthinclassandthereforeweconjecturethatnocomputablenumberingsexistfortheseclasses.2.4.3.1TheMartin{PourElConstructionRecallthata01classPisthinifforevery01subclassQP,thereisaclopensetUsuchthatQ=UP.Itisperfectiithasnoisolatedpoints.Aperfectclassmaybedenedbyafunctiong:2
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Wewilldenefinstagestoobtainuniformlycomputable,extensionpreservingfunctionsfssothatf=limsfs.ToensurethatPisthin,wewillmeetthefollowingrequirementforeache:Thine:82f0;1ge+18[f2Te^v!f2Te]ToseethatThinemakesPthin,letU=fIf:jj=e+1&f2TegandobservethatifPeP,thenPe=PU.Construction.Letf0=g.Atstaget+1,wedeneft+1asfollows.Lookfore
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Wedeneg:2n.LetnbetheleastksuchthatjTP!kj>n.Anonempty01classPisverysmallifthefunctiondominateseverycomputablefunctiong;thatis,x>gxforallbutnitelymanyx.LetAbeacoinnitec.e.set,say A=fa0anforallnanditisdensesimpleifn7!andominateseverycomputablefunction.Inthissectionwewillusethesesetstoshowthatnoeectivenumberingexistsforthesmall,verysmall,ordecidableclasses.2.4.4.1Numberingsandhigh/noncomputablesetsEnroutetodemonsratingourtheorem,wenowproceedshowthattherearenoeectivenumberingsofthehighorofthenoncomputablesets.Asweshalleventuallycharacterizesmallanverysmallclassesintermsthedegreesofthesesets,theseresultswillbecrucialtoourargument.First,wemodifyShoeneld'sThicknessLemma[ 86 ,p.131].Somedenitionsareneeded.ForB!,letB[y]=fhy;zi2B:z2!gandsaythatBispiecewisecomputable 51
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ifB[y]iscomputableforally.ForBA!,wesaythatBisathicksubsetofAifforally,B[y]nA[y]isnite. Lemma2.4.12ThicknessLemma. Foranyuniformlyc.e.sequencefWi:i2!gofnoncomputablec.e.setsandanypiecewisecomputablec.e.setB,thereisathickc.e.subsetAofBsothatWn66TAforalln.Proof.Theproofasin[ 86 ]ismodiedtoensurethatthelengthandrestraintfunctionsandtherequirementsincorporatethepairhi;kiinplaceofthesingleargumentitomaketheargumentgothroughwitheachWiinconjuctionwitheachfunctionalk. Weobtainthefollowingcorollary. Corollary2.4.13. Foranyuniformlyc.e.sequencefWn:n2!gofnoncomputablec.e.sets,thereisahighc.e.setAsuchthatforalli,Wi66TA.Proof.Thisfollowsfromthemodiedthicknesslemmaabovebythesameargumentfoundin[ 86 ,p.133]. Corollary2.4.14. a Thereisnouniformlyc.e.numberingofallhighc.e.sets. a Thereisnouniformlyc.e.numberingofallnoncomputablec.e.sets.Infact,itfollowsthatthereisnouniformlyc.e.numberingofthehighornoncomputablec.e.degrees.2.4.4.2NonexistenceofeectivenumberingsWenowproceedcharacterizethesmallandverysmallclassesintermofthenoncomputabledegreesandthehighdegrees,respectively.Thedegreeofa01classPisdenedtobethedegreeofTPandisthusalwaysac.e.degreesinceTPisacoc.e.set.Wewillusethefollowingtwoclassicresults.[Martin]Anyhighdegreecontainsamaximalandhencedensesimpleset[ 86 ,pp.211{217].[Dekker]Anynoncomputablec.e.degreecontainsahypersimpleset[ 86 ,p.81]. Proposition2.4.15. Ac.e.degreeishighifandonlyifitcontainsaverysmall01classP2!. 52
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Proof.!Supposeaishigh,andletA2abeamaximalset,andletpbetheprincipalfunctionof!)]TJ/F24 11.955 Tf 12.821 0 Td[(A,sothatpdominateseverycomputablefunction.NowletPA=f0!g[f0n10!:n=2Ag.ThenPAisa01classandforeachn,theleastksuchthatjTPf0;1gnj>kispreciselypn+1forn>0andhencedominateseverycomputablefunction.Letabeac.e.degreeandsupposethatTP2aforsomeverysmallP.Thenthefunctionfn=leastk[jTPf0;1gkj>n],whichdominateseverycomputablefunction,iscomputablefromTP.ItfollowsfromMartin'sTheorem[ 86 ,p.208]thatTPishigh. Proposition2.4.16. Ac.e.degreeisnoncomputableifandonlyifitcontainsaninnite,small01P2!.Proof.!Supposeaisanoncomputablec.e.degree,letA2abehypersimple,andpbetheprincipalfunctionof!)]TJ/F24 11.955 Tf 11.489 0 Td[(A,sothatpisnotdominatedbyanycomputablefunction.Thenthe01classPAasdenedintheproofofProposition 2.4.15 willhavedegreeaandwillbesmall.SupposethatPisaninnite01classandTPiscomputable.Thenthefunctiongn=leastk[jTPf0;1gkj>n]iscomputableanditfollowsthatPisnotsmall. Theorem2.4.17. Thereisnoeectivei.e.01numberingofallnondecidable,ofallinnitesmall,orofallverysmall01classesin2!.Proof.Suppose,towardsacontradiction,thatfQn=[Tn]:n2!gisaneectivenumberingof01classessuchthateachQnisnondecidable.ThenWn=fhi:=2ExtTngisauniformlyc.e.numberingofnoncomputablec.e.sets.ByCorollary 2.4.13 ,thereisahighc.e.setAsuchthatforalln,Wn66TA.ThereforeAisahighdegreethatcontainsaverysmallclassnotamongsttheQi,acontradiction. 53
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CHAPTER3RANDOMCLOSEDSETSThefollowingchapterisjointworkwithGeorgeBarpalias,DouglasCenzer,SeyyedDashti,andRebeccaWeberandappearsintheJournalforLogicandComputationno.17,2007,pages1041{1062asanarticleentitledAlgorithmicRandomnessofClosedSets[ 7 ].ApreliminaryversionofthisresearchwasoriginallypresentedattheComputabilityinEuropeConferenceinSwansea,Walesin2006byD.Cenzer.ThispreliminaryworkwaspublishedinthereferredconferenceproceedingsasRandomClosedSetsP.Brodhead,D.Cenzer,andS.DashtiinProceedingsofCIE2006:LogicalApproachestoComputationalBarriers,A.Beckmann,U.Berger,B.Loewe,andJ.Tucker,eds.,SpringerLectureNotesinComputerScience,Vol.39882006,pages5564[ 14 ].PortionsofthisworkwerealsopresentedbyP.BrodheadattheGreaterBostonLogicConferenceMay2006,Boston,MA,the2006SACNASConferenceOctober2006,Tampa,FL,theAMSFall2006EasternSectionalMeetingOctober2006,Storrs,CT,theConferenceonLogic,Computability,andRandomnessJanuary2007,BuenosAires,Argentina,andtheWorkshoponComputabilityandRandomnessDecember2007,Auckland,NewZealand.PortionsofthisworkwerealsopresentedbyD.CenzeratarandomnessworkshopattheAmericanInstituteofMathematicsAugust2006,Palo,Alto,CA.R.WeberalsopresentedportionsofthismaterialatarandomnessworkshopattheUniversityofChicagoSeptember2007,Chicago,IL.3.1OverviewTheliteratureaboundswithresultsinalgorithmicrandomessaspertainingtorealsoveranitealphabet,especiallywithinthelastfewyears.Littleisknown,orevendeveloped,however,withrespecttorandomnessforclosedsetsofbinaryreals.Thischapterisarstapproachinthisdirection.Inthischapter,weconsideranotionofeectivei.e.algorithmicrandomnessonthespaceCofnonemptyclosedsubsetsPof2N;toaccomplishthistask,wewillneedusethedenitionandmachineryofeectiverandomnessforreals,since,throughappropriate 54
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codingofclosedsets,wewilldeneaclosedsettoberandomiitscode,asareal,israndom.Infact,laterinChapter 4 ,wewillapproachadenitionofrandomnessforcontinousfunctionsinasimilarfashion.Consequentlywebeginthischapterwithanintroductiontoalgorithmicrandomness,includingabriefhistoricalbackground.Morespecically,thischapterisorganizedasfollows.InSection 3.2 ,weprovideanintroductiontoalgorithmicrandomnessforrealsoveranitealphabet.InSection 3.3 ,wegiveaprobabilitymeasureanddeneaversionoftheMartinLofTestforclosedsets,leadingtoadenitionofrandomnessforclosedsets.InSection 3.4 ,wetacklethequestionofwhichtypesofelementsnecessarilybelong,ordonotbelong,torandomclosedsets.Forinstance,everyrandomclosedsetcontainsrandomandnonrandomelements,butnonc.e.elements.InSection 3.5 ,weshowthatrandomclosedsetshavemeasurezeroandboxdimensionlog24 3.InSections 3.6 3.7 ,weexplorealternatenotionsforrandomness,suchastheproblemofcompressibilityoftrees.Finally,inSection 3.8 ,weconsidertheproblemofwhenarandomlychosenclosedsetmeetsaclosedQ;thisisthestudyofcapacities.3.2EectiveRandomnessofRealsInthissection,wepresentabasicintroduction,includingabriefhistoricalbackground,forrandomnessofrealsoveranitealphabet.3.2.1IntroductionThestudyofalgorithmicrandomnesshasbeenofgreatinterestinrecentyears.Thebasicproblemistoquantifytherandomnessofasinglerealnumber.Earlyinthelastcentury,vonMises[ 94 ]suggestedthatarandomrealshouldobeyreasonablestatisticaltests,suchashavingaroughlyequalnumberofzeroesandonesoftherstnbits,inthelimit.Thusarandomrealwouldbestochasticinmodernparlance.Ifoneconsidersonlycomputabletests,thentherearecountablymanyandonecanconstructarealsatisfyingalltests.Anearlyapproachtorandomnesswasthroughbetting.Eectivebettingonarandomsequenceshouldnotallowone'scapitaltogrowunboundedly.Thebettingstrategiesused 55
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areconstructivemartingales,introducedbyVille[ 93 ]andimplicitintheworkofLevy[ 65 ],whichrepresentfairdoubleornothinggambling.MartinLof[ 69 ]observedthatstochasticpropertiescouldbeviewedasspecialkindsofmeasurezerosetsanddenedarandomrealasonewhichavoidscertaineectivelypresentedmeasurezerosets;seeSection 3.2.4 .AtthesametimeKolmogorov[ 55 ]denedanotionofrandomnessfornitestringsbasedontheconceptofincompressibility.AstrongernotionofprexfreecomplexitywasdevelopedbyLevin[ 64 ],Gacs[ 48 ]andChaitin[ 27 ]andextendedtoinnitewords.Inthefollowingsections,weformalizethenotionsofconstructivemartingalerandomness,MartinLofrandomness,andprexfreerandomness.Aftertheirentryintotheliterature,Schnorrlaterproved[ 83 ]thatallofthesenotionsareequivlant;thisisafundamentalresultinthetheoryofalgorithmicrandomness.Whilethesedenitionsandresultsareusuallygivenforbinarystringsandsequences,theycarryovertokarystringsandsequencesaswell.See,forexample,Calude[ 17 18 ],orSection 3.2.4 below,wherewedothisfortheMartinLofdenitionofrandomness.3.2.2ConstructiveMartingaleRandomnessThebettingapproachtorandomnessisformalizedasfollows. Denition3.2.1Ville[ 93 ]. i Amartingaleisafunctionm:k
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Denition3.2.2. Amartingalemisconstructiveoreective,orc.e.ifitislowersemicomputable;thatis,ifthereisacomputablefunction^m:k
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domaindomM;thatis,ifvarestringsindomM,thenmustequal.Foranynitestring,theprexfreecomplexityofwithrespecttoMisKM=minfjj;1:M=g:ThereisauniversalprexfreefunctionUsuchthat,foranyprexfreeM,thereisaconstantcsuchthatforallKU6KM+c:WeletK=KUandcallittheprexfreecomplexityof. Denition3.2.5PrexfreeRandomness. x2f0;1g!iscalledprexfreerandomifthereisaconstantcsuchthatKxdn>n)]TJ/F24 11.955 Tf 11.955 0 Td[(cforalln.Thislatterinequalitymeansthattheinitialsegmentsofxarenotcompressible.3.2.4MartinLofnrandomnessAccordingtotheMartinLofdenitionofrandomness,arandomrealmustavoidcertaineectivelypresentedmeasurezerosets.Inherentinthedenition,therefore,isthechosenmeasurebeingused.Fixanalphabetk=f0;1;:::;k)]TJ/F15 11.955 Tf 12.58 0 Td[(1g.Wepresentthedenitionofageneralprobabilitymeasureonk!,aswellasdierentnamedmeasuretypes.Typically,however,wewillmakeuseofthealphabetsf0;1gorf0;1;2g. Denition3.2.6GeneralProbabilityMeasures. Letf:f0;1;2g![0;1]beafunctionsuchthatf;=1andf=Pi=0;1;2f_iforall.Thefprobabilitymeasurefisdenedsothatthefmeasureoftheinterval[]issuchthatf[]=f. Denition3.2.7DierentMeasuresTypes. LetfandfbeasinDenition 3.2.6 i fiscomputableiffiscomputable. ii fisnonatomic,orcontinuous,ifforallx23N,ffxg=0. iii fandfareboundedif9b;c2;188i[bf
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Denition3.2.8. Arealx2k!isMartinLofrandomifforeveryeectivesequenceS1;S2;:::ofc.e.opensetswithSn62)]TJ/F25 7.97 Tf 6.586 0 Td[(n,x=2TnSn.Thelatterconditionisequivalenttotheconditionwegetifwereplace2)]TJ/F25 7.97 Tf 6.587 0 Td[(nwithqnforacomputablesequenceqiofrationalssuchthatlimiqi=0.WecanalsoconsideranextendeddenitionofMartinLofrandomness,intermsof0nsets. Denition3.2.9. i A0ntestisacomputablecollectionfVn:n22Ngof0nclassessuchthatVk62)]TJ/F25 7.97 Tf 6.587 0 Td[(k; ii Arealis0nrandomornrandomifandonlyifitpassesall0ntestsi.e.,iffVn:n22Ngisacomputablecollectionof0nclassessuchthatVk62)]TJ/F25 7.97 Tf 6.587 0 Td[(k,then=2n>0Vn.Thus1randomrealsarejustMartinLofrandomreals.See[ 36 ]fordetailsonrandomandnrandomreals.Kurtz[ 59 ]andKautz[ 54 ]provedthefollowingresult.Let;ndenotethenthjumpof;. Theorem3.2.10. Letqbearationalnumber. i Foreach0nclassS,wecanuniformlycomputefromqanda0nindexforS,theindexofa;n)]TJ/F18 5.978 Tf 5.756 0 Td[(11classUSsuchthatUisanopen0nclassandU)]TJ/F24 11.955 Tf 11.955 0 Td[(S
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Theorem3.2.12vanLambalgen[ 92 ]. Thefollowingareequivalent. 1. ABisnrandom. 2. AisnrandomandBisnArandom. 3. BisnrandomandAisnBrandom. 4. AisnBrandomandBisnArandom.3.3MartinLofRandomnessofClosedSetsInthissectionwedeneameasureonthespaceCofnonemptyclosedsubsetsof2Nandusethistodenethenotionofrandomnessforclosedsets.Wethenobtainseveralpropertiesofrandomclosedsets.3.3.1TheHitorMissTopologyonCThestandardhitormisstopologyonChasasasubbasis,thefollowingtwotypesofsets,whereQisanyclosedset:VQ=fK:KQ6=;g;WQ=fK:KQg.Abasisforthehitormisstopology,then,isformedbytakingniteintersectionsofthese.WenowconsiderarenementofthesubbasissetsandobtainabasisfortheBorelsets.Wewillusethefollowingnotation.TndenotesthesetTf0;1gn,andT6ndenotesthesetTf0;1g6n Denition3.3.1. ForanytreeSandanyn,deneCnS=fQ2C:S6n=T6nQgThatis,CnSisthesetofclosedsetsQ2CthatagreewithSuptoleveln. Claim3.3.2. AbasisfortheBorelsetsisgivenbytheagreementsetA:A=fCnS:Sisatree;n2!g:Toseethis,rstnotethatanyclosedset[T]isthedecreasingintersectionofclopensets[Tn]:=[f[]:2Tng:ThereforewemayrewritesubbasiselementsV[T]andW[T]as V[T]=nV[Tn]bydenitionand 60
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W[T]=nW[Tn]bycompactnessButthen, V[Tn]=SfCnS:STn6=;gand W[Tn]=SfCnS:SnTgToseetherstequality,forexample,notethatQ[Tn]6=;ifandonlyifQ2CnSforsomeSwithSTn6=;.Thelatterequalityholdssimilarly.3.3.2TowardaMeasureTodeneVQandWQ,forsomexedmeasureandanyclosedsetQ,itsucestodeneVQnandWQnforclopensetsQnwhereQ=nQn.WewouldsimplydeneVQ=limnVQnandWQ=limnWQn.However,foranyclopenQ,WQisthecomplementVNnQ.HenceitfurthermoresucestodeneVQforclopensetstogetameasureonC.FromthejusticationofClaim 3.3.2 ,thelattertaskmaybeaccomplishedbydeningameasureonallBorelbasiselements,namelytheagreementsetsCnS.Toaccomplishthis,inthefollowingsectionwewillencodeallclosedsetsQ2CwithacanonicalcodexQ2C.ThenusingtheLebesguemeasureon3N,wewilldeneameasureonthesetsCnSwhich,infact,denesameasureonallclosedsets.3.3.3CanonicalCodingandMeasureTheCanonicalCoding.AneectiveonetoonecorrespondencebetweenthespaceCandthespace3Nisdenedasfollows.LetaclosedsetQbegivenandletT=TQbethetreewithoutdeadendssuchthatQ=[T].Denethecanonicalcodex=xQ2f0;1;2gNforQasfollows.Let=0;1;2;:::enumeratetheelementsofTinorder,rstbylengthandthenlexicographically.Wenowdenex=xQ=xTbyrecursionasfollows.Foreachn,xn=2if_n0and_n1arebothinT,xn=1if_n0=2Tand_n12Tandxn=0if_n02Tand_n1=2T.Forexample,ifQ=f0;1gN,thenxQ=;2;:::andifQ=fyg,thenxQ=y.LetQxdenotetheuniqueclosedsetQsuchthatxQ=x. Denition3.3.3TheMeasure. DenethemeasureonCbyX=fxQ:Q2Xg: 61
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whereistheLebesquemeasurei.e.theregularmeasuredwithb0=b1=b3=1 3seeDenition 3.2.7 on3N.Informallythismeansthatgiven2TQ,thereisprobability1 3thatboth_02TQand_12TQand,fori=0;1,thereisprobability1 3thatonly_i2TQ.Inparticular,thismeansthatQI6=;impliesthatfori=0;1,QI_i6=;withprobability2 3. Comment3.3.4. Atthisstage,wehavexedtheuniformmeasurei.e.allbi=1 3towardsdeningrandomnessofclosedsets.Thisallowsustomoreeasilydemonstratethevalidityofmanyresults.Later,inSection 3.7.1 ,wewillshowthattheresultsholdwithanyregularmeasure.Proposition 3.3.5 ,however,demonstratesthatthedenedmeasureonC,above,holdsforanygeneralizedprobabilitymeasuredseeDenition 3.2.6 .JusticationfortheCoding.Letusalsocommentbrieyonwhysomeothernaturalrepresentationswererejected.Supposerstthatwesimplyenumerateallstringsinf0;1gas0;1;:::andthenrepresentTbyitscharacteristicfunctionsothatxTn=1n2T.Theningeneralacodexmightnotrepresentatree.Thatis,oncewehave01=2Twecannotlaterdecidethat12T.Supposethenthatweallowtheemptyclosedsetbyusingcodesx2f0;1;2;3gandmodifyouroriginaldenitionasfollows.Letxn=ihavethesamedenitionasabovefori62butletxn=3meanthatneither_n0nor_1isinT.Informally,thiswouldmeanthatfori=0;1,2Timpliesthat_i2Twithprobability1 2.Theadvantagehereisthatwecannowrepresentalltrees.Butthisisalsoadisadvantage,sinceforagivenclosedsetP,therearemanydierenttreesTwithP=[T].Thesecondproblemwiththisapproachisthatwewouldhave[T]=;withpositiveprobability.WebrieyreturntothissubjectinSection 3.7.2 .Nowrecallthedenitionofageneralprobabilitymeasureon3NDenition 3.2.6 .Letd:f0;1;2g![0;1]beafunctionsuchthatd;=1andd=Pi=0;1;2d_iforall.Thend[]isdenedtobed.Wemaynowdene,foranysuchd,dexactlyasinDenition 3.3.3 .Furthermoredwebedeemedcomputableifdiscomputable. 62
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Proposition3.3.5. Foranyd,themeasuredisdenedonallBorelsetsinthehitormisstopologyonC.Furthermore,ifdiscomputable,thendiscomputableonthefamilyofclopensets.Proof.Asdiscussedinsection 3.3.2 ,itsucestoshowthatdCnSisdenedforallCnS2A.FixatreeSandsupposethatfv:2Sngisordered0<:::
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vi=PCUi,fori=0;1.ThenconsideringthethreecasesinwhichSincludesbothinitialbranchesorjustone,wecalculatethatPCU=1 3v0+v1+v0v1:ThusbyinductionwehavePCU61 3u0+u1+u0u1:Now2u0u16u20+u216u0+u1;andthereforePCU61 3u0+u1+u0u161 2u0+u1=U:ForaclosedsetQ,letQ=TnUn,whereUnisclopenandUn+1Unforalln.ThenPQifandonlyifPUnforalln.ThusPCQ=nPCUn;sothatPCQ=limn!1PCUn6limn!1Un=Q:Finally,foranopensetQ,letQ=SnUnbetheunionofanincreasingsequenceofclopensetsUn.Then,bycompactness,PCQ=[nPCUn;sothatPCQ=limn!1PCUn6limn!1Un=Q:Thiscompletestheproofofthelemma. 3.3.4GhostCodingWewishnowtointroduceasecondmethodofcoding,theghostcoding.AghostcodeofQisaninniteternarystringwhosetermscorrespondtoallnodesof2
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lexicographicalorder.ThetermscorrespondingtothenodesofQ'streethecanonicalnodes"agreewiththecorrespondingtermsinthecanonicalcode;theremainingghostnodes"mayholdanyvalues.Ghostcodesarenonunique,andeveryclosedsethasanonrandomghostcodeiftheclosedsetitselfisrandomtakethecodewithghostnodesallequaltozero,say.Thismethodofcodingismoreconvenientforsomepurposes;forexample,wewilluseittoshowthatifQ0;Q1areclosedsetsandQ=f0_x:x2Q0g[f1_x:x2Q1g,QisrandomifandonlyiftheQiarerandomrelativetoeachother.3.3.5CodingEquivalanceTheutilityoftheghostcodesrestsonthefollowingcorrespondence.RecallvanLambalgen'stheoremTheorem 3.2.12 Theorem3.3.9. ThecanonicalcodeofaclosedsetQ2NisrandomifandonlyifQhassomerandomghostcode.Furthermore,foranyy,thecanonicalcoderisyrandomifandonlyifQhasaghostcodewhichisyrandom.Proof.SupposethecanonicalcodeofQisnonrandom.Thenthereisac.e.martingalemthatsucceedsonit.FromanyinitialsegmentofaghostcodegforQ,thesubsequence^ofexactlythecanonicalnodesofiscomputable.Thereforeitiscomputablewhetherthebitofgafteriscanonicalorghost.Fromm,denethemartingalem0whichbetsasfollows:m0_i=8><>:m^_inextbitisacanonicalnodem0nextbitisaghostnode.Thatis,m0holdsitsmoneyonghostnodesandbetsidenticallytomoncanonicalnodes.Itisclearthatm0succeedsontheghostcodegandthusgisnonrandom.!NowsupposethecanonicalcoderforQisrandom,andletqbeaninniteternarystringthatisrandomrelativetorandsobyTheorem 3.2.12 rqisrandom.Weclaimtheghostcodegobtainedbyusingthebitsofrasthecanonicalnodesandthebitsofqintheiroriginalorderastheghostnodesisrandom.ItisclearthatgisaghostcodeforQ.Supposemisac.e.martingalethatbetsong.Frommitisstraightforwardtodeneanonmonotonicmartingalem0whichmimicsm'sbetsexactlybutperformsthemonrq, 65
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succeedingwhenevermsucceeds.Asrandqwerechosentoberelativelyrandom,thiswillshowgisrandom.Asdiscussedpreviously,fromgdnitiscomputablewhethergnwillbeaghostnodeoracanonicalnode,andwhichpositioningorritoccupiesineithercase.Therefore,assumingthebitsseensofarmaybeassembledintoaninitialsegmentofg,m0takesthevaluesm_i,i<3,asitsbetsonthecorrespondingbitofrorg,whicheverisappropriate.Havingseenthatbit,then,itcanassembleajj+1lengthinitialsegmentofgandrepeattheprocess.Asm0makesidenticalbetstomandhasidenticaloutcomes,sinceitcannotsucceedonrg,mcannotsucceedongandgisrandom.Torelativize!,supposethatrisyrandom,sothatryisrandombyVanLambalgen'sTheorem 3.2.12 .Thenintheproofsimplychooseqtoberandomrelativetory,andthengwillberandomrelativetoy.Theotherdirectionrelativizesinastraightforwardway. 3.3.6CodingandJoinsofClosedSetsTheprimarypurposeoftheghostcodesistoremovethedependenceontheparticularclosedsetunderdiscussionwheninterpretingbitsofthecodeasnodesofthetree.Thisisespeciallyusefulwhensubdividingthetree,asinthefollowingdenition. Denition3.3.10. ThetreejoinofclosedsetsP0andP1istheclosedsetQ=f0_x:x2P0g[f1_x:x2P1g:Givenghostcodesr0;r1forthePi,theirtreejoinr0r1isthecodeforQwiththecorrespondingghostnodevalues.Thestandardrecursiontheoreticjoinisdenedbyr0r1=r0;r1;r0;r1;::::Wewishtorelatetherecursiontheoreticjoinandthetreejoin. Lemma3.3.11. Giventwoghostcodesr0;r1,thetreejoinr0r1israndomifandonlyiftherecursiontheoreticjoinr0r1israndom. 66
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Proof.Itisclearthatthereisacomputablepermutationwhichuniformlymapsanytreejoinr0r1totherecursiontheoreticjoinr0r1.Thatis,inr0r1,theentriesofr0andr1alternate,whereasr0r1startswitha2,followedbyblocksfromr0andr1,asfollows.Firstr0,r1,thenr0,r0,r1,r1,andcontinuingwithpairsofblocksofsize4,8andsoon.TheresultnowfollowsfromtheVonMises{Church{WaldComputableSelectionTheorem[ 94 ];thetheoremstatesthat,foranyrandomsequencexandanycomputable11functiong,thesequencezn=xgnisrandom. WenowobtainthefollowingcorollaryofTheorems 3.2.12 and 3.3.9 andLemma 3.3.11 Corollary3.3.12. SupposePi,i=0;1,areclosedsetswithcanonicalcodesriandletPbethetreejoinofP0;P1.ThenPisrandomifandonlyifr0r1israndom.Proof.Supposethatr0r1israndom.ThenbyTheorem 3.2.12 ,r0andr1aremutuallyrelativelyrandom.ByTheorem 3.3.9 ,P0hasaghostcodeg0whichisrandomrelativetor1,andsoalsoviceversa,andthenP1hasaghostcodeg1whichisrandomrelativetog0.Againby 3.2.12 ,therecursiontheoreticjoing0g1israndom,sobyTheorem 3.3.11 thetreejoing0g1isalsorandom,andhencePpossessesarandomghostcodeandisrandom.!SupposenowthatPisrandom,andthereforepossessesarandomghostcodeg.Thecodegmaybethoughtofasatreejoing0g1,whichisthereforerandom,andsobyTheorem 3.3.11 ,g0g1israndom.ByTheorem 3.2.12 ,theindividualcodesg0;g1arethereforemutuallyrelativelyrandom.NowbytherelativedversionofTheorem 3.3.9 ,r0israndomrelativetog1.Butr1iscomputablefromg1andhencer0israndomrelativetor1aswell.Similarly,r1isr0randomandthus,againby 3.2.12 ,r0r1israndom. 3.4MembersofRandomClosedSetsInthissectionwetacklethequestionofwhichtypesofelementsnecessarilybelong,ordonotbelong,torandomclosedsets.TheformerisaddressedinSection 3.4.1 andthelatterinSection 3.4.2 67
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3.4.1PositiveResultsWeshallsee,asaconsequenceofTheorem 3.4.19 ,thateveryclosedsetisperfectandcontainscontinuummanyelements.Inthissection,wedemonstrateotherpositiveresults.Forexample,everyrandomclosedsetcontainsrandomandnonrandomelements.Otherexamplesabound.Webeginwiththerstexample. Theorem3.4.1. Everyrandomclosedsetcontainsarandomelement.Proof.SupposethataclosedsetQhasnorandomelementandconsiderthefollowingMartinLoftestonthespaceC:Ui=fPjP2CandPVigwhereViisauniversalMartinLoftestontheCantorspace.ByLemma 3.3.8 ,Ui6Vi62)]TJ/F25 7.97 Tf 6.587 0 Td[(isothatUiisaMartinLoftestonC.ButQ2iUi,soQisnotrandom. AsaconversetoTheorem 3.4.1 wehavethefollowing. Theorem3.4.2. Foranyrandomr22N,thereexistsarandomclosedsetcontainingrasapath.TheproofofthistheoremwasoriginallygivenbyJoeMillerandAntonioMontalbanandhasbeensubsequentlyimprovedthankstotheanonymousreferee.Proof.Letrbearandomrealandletxbethecanonicalcodeofanrrandomclosedset.Wealterxtothecodex0ofaclosedsetguaranteedtocontainrbutchangedaslittleaspossibletoachievethat.Todeterminex0n,assumex0nhasbeendened.Ifxn=2orxncorrespondstoanodenotalongr,setx0n=xn.Ifxn2f0;1gcorrespondstork,setx0n=rk.Theclosedsetdenedbyx0willclearlycontainr.Foracontradiction,assumex0isnonrandomandletm0beac.e.martingalethatsucceedsonit.Webuildanonmonotonicmartingalemtobetonxr.Onbitsofx,mwillbeatripleornothingmartingale;onr,itwillbedoubleornothing. 68
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Firstnotethatfrominitialsegmentsofxandrwemayreconstructaninitialsegmentofx0computably,andwealwaysknowfromaninitialsegmentofx0whetherthenextbitisalongrornot,andwhichbitofritis.Wewillconstructmsothataftereverystageofbettingwhichwillbeonebetbym0andoneortwobetsbym,thevalueofmisequaltothevalueofm0.Ateverystageitwillbeclearwehaverevealedenoughbitsofxandrtoreconstructx0totheneededlength.Supposeinductivelymandm0holdequalcapitalafterthestageofbettingonthelastnodeof@x0.Ifthebitx0nfollowingisnotonr,mbetsidenticallytom0;i.e.,mxn=i=m0_ifori<3.Inthatcasexn=x0nsoourinductivehypothesisholds.Ifx0nisonr,setmxn=2=m0_2andfori=0;1,setmxn=i=1 2[m0_0+m0_1].Ifx0n=2,thenthecapitalforbothmandm0ism0_2,sotheinductivehypothesisholdsandweproceedtothenextstage.Otherwisembetsonrkfortheappropriatek,settingmrk=i=m0_ifori=0;1.Onrk,thesumofm'scapitaloneachofthetwooutcomesmustaveragetothepreviouscapital;asthepreviouscapitalwas1 2[m0_0+m0_1]thisclearlyholds.Byconstructionrk=x0n=i,sobothmandm0nowhavecapitalm0_iandtheinductivehypothesisholds.Asm0isc.e.,mwillalsobe.Asthevaluesofm0alongx0areasubsequenceofthevaluesofmalongxr,ifm0succeedssodoesm,contradictingourassumptiononxr.Thereforex0isthecodeofarandomclosedsetcontainingthegivenrandompathr. Thepreviousresultsmightsuggestthateveryelementofarandomclosedsetisarandomreal.However,itturnsoutthateveryrandomclosedsetcontainsanonrandomreal.WeneedthefollowingclassicresultofCherno[ 28 ]aversionofBernoulli'sWeakLawofLargeNumbershereandalsoforanothertheoremtofollow.See[ 67 ]foranexposition. Lemma3.4.3Cherno. LetEbeaneventwhichwewillrefertoas`success'.IfEoccurswithprobabilityp,thenforanynaturalnumbersnandany"with06"61,the 69
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probabilitythatoutofnmutuallyindependenttrials,thenumberofsuccessesdiersfrompnby>"pnis62)]TJ/F25 7.97 Tf 6.586 0 Td[("2pn=3. Theorem3.4.4. Everyrandomclosedsetcontainsanonrandomreal;inparticular,theleftmostandrightmostpathsinarandomclosedsetarenotrandomreals.Proof.Wewillshowthat,forarandomclosedsetQ,theleftmostpathisnotstochasticallyrandom,thatis,theasymptoticfrequencyof0'sis2 3.Sinceaneectivelyrandomrealin2Nmusthaveasymptoticfrequenceof1 2for0'sand1's,thiswillsucetoprovethattheleftmostpathisnotrandom.WedeneaMartinLoftestasfollows.Fixarational"suchthat0<"<1.Foreachn,letSnbethefamilyofclosedsetsthatis,codesforclosedsetssuchthattherstnbitsoftheleftmostpathhaveeither<2 3)]TJ/F24 11.955 Tf 12.295 0 Td[("n,or>2 3+"noccurrencesof0.Bythedenitionofourprobabilitymeasure,wehaveSn=Xjm)]TJ/F18 5.978 Tf 7.782 3.259 Td[(2 3nj>2 3"n0B@nm1CA2 3m1 3n)]TJ/F25 7.97 Tf 6.587 0 Td[(m:ItnowfollowsfromCherno'sLemma 3.4.3 thatSn62e)]TJ/F25 7.97 Tf 6.586 0 Td[("22n=9:ThusthemeasuresofthetestsetsSnhaveeectivelimitzero.ItiseasytoseethatthesequencefSngiscomputablyenumerable.Foreachn,SnisaclopensetandinfacttheunionofthenitefamilyofintervalsIinCsuchthatcodesatreeuptolevelninwhichtheleftmostpathhaseither<2 3)]TJ/F24 11.955 Tf 11.955 0 Td[("n,or>2 3+"noccurrencesof0.Furthermore,S0n=Sp>nSpisalsoaMartinLoftest.ItfollowsthatforanyrandomclosedsetQ,andany">0,thereisannsuchthatforallm>n,thefrequencyof0'sintherstmbitsoftheleftmostpathisalwayswithin"of2 3.Thustheleftmostpathisnoteectivelyrandom. Recallthattheleftmostandrightmostelementsofanystrong02closedsetare02.GivenTheorems 3.4.1 and 3.4.4 ,weask:Doesa02randomclosedsetcontaina02randompath? Theorem3.4.5. Everyrandomstrong02closedsetcontainsarandom02real. 70
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Proof.LetQbearandomstrong02class.ByTheorem 3.4.1 ,Qcontainsarandomrealx.LetPbea01classintheCantorspacewhichcontainsonlyrandomsandcontainsxthisexistssincetheclassofrandomrealsisaneectiveunionof01classes.NotethatPQisanonemptystrong02classanditfollowsthattheleftmostpathofPQisa02realwhichmustberandomsinceitbelongstoP. Theabovetheoremdoesnotcombinewiththelowbasistheoremtoestablishtheexistenceofalowrandomrealinanyrandomstrong02class.Wecanusethelowbasistheorem,however,todemonstratetheexistenceofalowrandomrealinanyrandomclosedsetwithlowcanonicalcode. Theorem3.4.6. Everyrandomclosedsetwithlowcanonicalcodecontainsalowrandomelement.ProofKjosHanssen.LetQbearandomclosedsetwithlowcanonicalcode.ByTheorem 3.4.1 ,Qcontainsarandomelementx.Thereforex22!nUnforsomenandsomeopenUnfromtheuniversalMartinLoftest.So,inparticular,Q2!nUnisnonempty.NowQ2!nUnis01relativetoTQ.Bythelowbasistheorem,Q2!nUnis01hasamemberysuchthaty06TT0Q.Inparticular,sincey2Q2!nUn,itisrandom.Furthermore,sinceTQislow,yisalsolow. Itisopenwhethereveryrandomclosedsetwitha02canonicalcodehasalowrandomelement;weconjecturethattheanswerisno.Inthefollowingsection,wewillshowthatthereisarandomclosedsetnotcontainingany02path.Ournextresult,Theorem 3.4.8 ,usesamethodwhichwasusedin[ 56 ]toshowthateveryrandomrealiseectivelybiimmune.Werstdenethislatternotion. Denition3.4.7. i AsetAiseectivelyimmuneifitisinniteandthereisacomputablefunctiongxsuchthatifWxA,jWxj6gx. ii AiseectivelybiimmuneifAand Aarebotheectivelyimmune.Note,inparticular,thataneectivelyimmunesetcannotcontainaninnitec.e.set. Theorem3.4.8. IfPisarandomclosedsetthenallelementsofPareeectivelybiimmune. 71
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Proof.SupposethatPisarandomclosedsetandA2P.LetUii2NbeaMartinLoftestinthespace3NsuchthatthereisacomputablefunctionfwiththepropertythatifVeii2NistheethMartinLoftestundersomeeectiveenumerationofallMartinLofteststhenforalle;i,Vefe;iUiastandardconstructionofauniversaltestgivesonewiththisproperty.SincePisrandom,thereissomeksuchthatthecanonicalcodeofPisnotinUk;letU=Uk.Itsucestondacomputablefunctiongsuchthat[A2PandWxA]jWxj6gx3{1forallsetsAandallxtheproofthat Aiseectivelyimmuneisentirelysimilar.LetBx;nbe;ifjWxj6n,andotherwisetheclassofcanonicalcodesoftreeswhichcontainapathcontainingtherstn+1elementsinthestandardenumerationofWx.ThenBx;nisauniformdoublesequenceof01classesandbythedenitionoftheprobabilitymeasureon3N,Bx;n62 3n+1:Soforeachx,Bx;2nisaMartinLoftestinthespace3Nandfromxwecancalculatetheindexofit.Thenbyusingthecomputablefunctionfmentionedabovewegetacomputablefunctiongsuchthatforallx,Bx;gxU.Thismeansgsatises 3{1 Itiswellknownthateectivelyimmunesetscancomputeaxedpointfreefunction,sowehavethefollowing. Corollary3.4.9. Thepathsthrougharandomtreeareofxedpointfreedegree.Thatis,eachofthemcomputessomexedpointfreefunction.Itisknownthateveryrealcanbecomputedbysomerandomreal.Itisnotknown,however,whetheranyrealcanbecomputedbyallthepathsofsomerandomclosedset.Thenexttheorem,anobservationofTedSlamanatarandomnessworkshopinChicagoin2007,isastepinthatdirection.First,weneedthefollowingdenitions,whichare,infact,equivalentnotions. Denition3.4.10. i ArealxisKtrivialifKxdn6Kn+cforsomec. 72
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ii Arealxisabasefor1randomnessifthereissomey>Txsuchthatyis1xrandom. Theorem3.4.11. AnysetwhichiscomputablefromallpathsofarandomclosedsetisKtrivial.Proof.IfAiscomputablefromallpaths,thenAiscomputablefromtheleftmostandrightmostpaths.Notethateachoftheselatterpathsareeachcomputablefromthetwohalvesofthetree.Furthermorethesetwohalvesarerelativelyrandomtoeachother.Henceeachisrandomrelativetoanythingtheothercomputes.Sohalf1,forexample,computesAandhencehalf2israndomrelativetoA.Ontheotherhand,half2alsocomputesA.ThereforethisisanexampleofsomethingintheconeaboveAthatisArandom.SoAisabasefor1randomness. MuchinterestingworkhasbeendoneontheKtrivialreals.ChaitinshowedthatifAisKtrivial,thenA6T00.SolovayconstructedanoncomputableKtrivialreal.Downey,Hirschfeldt,NiesandStephan[ 39 ]showedthatnoKtrivialrealisc.e.complete.ThenotionofaKtrivialclosedsetwasintroducedin[ 9 ].ItwasshowninparticularthateveryKtrivialclasscontainsaKtrivialmember,butthereexistKtrivial01classeswithnocomputablemembers.3.4.2NegativeResultsRandomclosedsetscannevercontainnc.e.,isolated,or1genericpaths,orpathsofincompletec.e.degree.Webuildtothesefactsandproveothersalongtheway. Theorem3.4.12. Randomclosedsetscontainnocomputableelements.Proof.Foranynitestringoflengthn,theprobabilitythataclosedsetQmeetsIis2 3n.Foracomputablerealy,thesqeuencefQ:QIydn6=;gthusformsaMartinLoftestinthespaceCofclosedsets,whichshowsthatydoesnotbelongtoanyMartinLofrandomclosedset.Thatis,foreachn,fx:QxIydn6=;gisac.e.opensetandhasmeasure2 3ninf0;1;2gN,whereQxistheclosedsetwithcodex. WeproveanevenstrongerresultinTheorem 3.4.17 .First,however,recallthata01classPisdecidableifTPisdecidable.Itfollowsthatanonemptydecidable01class 73
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mustcontainacomputableelementforexample,theleftmostpath.ByTheorem 3.4.12 ,itfollowsthatnodecidable01classcanberandom.AseveryrandomclasscontainsarandomelementTheorem 3.4.1 andhas,asweshallshow,measurezeroTheorem 3.5.1 ,thefollowingpropositiondemonstratesthatthisextendstoarbitrary01classes. Proposition3.4.13. IfPisa01classofmeasure0,thenPhasnorandomelements.Proof.LetTbeacomputabletreesuchthatP=[T],andforeachn,letPn=SfI:2Tf0;1gng.ThenfPngn2NisaneectivesequenceofclopensetswithP=TnPnandlimnPn=P=0.Furthermore,Pn=2)]TJ/F25 7.97 Tf 6.586 0 Td[(njTf0;1gnj;thisisacomputablesequence.ThusfPngn2NisaMartinLoftestandPhasnorandomelements. Alternatively,wecanshowthatno01classisrandomthroughthefollowingstrongerresult,combinedwithanappealtoTheorem 3.5.1 ,fromthenextsection. Theorem3.4.14. LetQbea01classwithmeasure0.ThennosubsetofQisrandom.Proof.LetTbeacomputabletreepossiblywithdeadendsandletQ=[T].ThenQ=TnUn,whereUn=[Tn].SinceQ=0,itfollowsfromLemma 3.3.8 thatlimnPCUn=0.ButPCUnisacomputablesequenceofclopensetsinCandPCUnisacomputablesequenceofrationalswithlimit0.ThusPCUnisaMartinLoftest,sothatforanyrandomclosedset,thereexistsnsuchthatP=2PCUnandhencePisnotasubsetofUn. Corollary3.4.15. No01classcanberandom. WenowprovideanevenstrongerversionofTheorem 3.4.12 ;weneedthefollowingdenition. Denition3.4.16fc.e.reals. Foranycomputable,nondecreasingfunctionf,wesaythatareal2f0;1gNisfc.e.ifthereexistsacomputableapproximatingfunctionsuchthat,foralli2N, i i;0=0 ii limsi;s=i; iii fs:i;s+16=i;sghascardinality6fi. 74
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Therealswhicharefc.e.forsomecomputablefunctionfarepartofthewellknownErshovhierarchy[ 43 86 ]. Theorem3.4.17. Supposethatfiscomputableandboundedbyapolynomial.Thennorandomclosedsethasanyfc.e.paths.Proof.Letfbeasabove,anfc.e.realandPaclosedsetcontaining.Letbethefapproximatingfunctionfor.AlsoletMnf0;1gnbethesetofdierentapproximationstodnduringthestages.Apriori,jMnjisexponential.However,foraxedn,dncanchangeatmostPi
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Nopathofarandomtreeis1generic. Nopairof02pathsofarandomtreecanbeaminimalpair. Every02pathofarandomtreecomputesapromptlysimpleset. Nopathofarandomtreecanhaveincompletec.e.degree. Theorem3.4.19. IfQisarandomclosedset,thenQhasnoisolatedelements.Proof.LetQ=[T]andsupposebywayofcontradictionthatQcontainsanisolatedpathx.Thenthereissomenode2TsuchthatQI=fxg.Foreachn,letSn=fP2C:jf2f0;1gn:PI_6=;gj=1g:Thatis,P2SnifandonlyifthetreeTPhasexactlyoneextensionofoflengthn+jj.ItfollowsthatjPIj=18nP2SnNowforeachn,SnisaclopensetinCandagainbyinduction,Snhasmeasure2 3n.ThusthesequenceS0;S1;:::isaMartinLoftest.Itfollowsthatforsomen,Q=2Sn.ThusthereareatleasttwoextensionsinTQofoflengthn+jj,contradictingtheassumptionthatxwastheuniqueelementofQI. Asmentionedpreviously,itfollowsthateveryrandomclosedsetisperfectandhencecontainscontinuummanyelements.Nextwewanttondarandomclosedsetwhichdoesnotcontaina02path.Nowitiseasy[ 20 24 ]toconstructastrong02classPofpositivemeasurewhichcontainsno02elements;ofcoursePmustcontainarandomrealsinceithasmeasure1.Thedicultproblemistoconstructarandomstrong02classwithno02elements.Wehavethefollowingresultinthisdirection,whichyieldsarandomstrong03closedsetwithno02elements. Theorem3.4.20. ForanysetAthereisanArandomclosedsetQsuchthatTQ6TA00butQhasnoelements6TA0.Proof.ItisenoughifweprovetheclaimforA=;becausetheargumentrelativisestoanyoracleAinastraightforwardway.ForA=;weuseaniteinjuryconstructionover 76
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;0toconstructQwiththeaboveproperties.Intheconstructionwewill;0approximatethecanonicalcodeofatreeTwhichhasno02paths.TomakesurethatthetreeTisrandomwexa01classPofpositivemeasureinthespace3NwherethecodeforTlieswhichcontainsonlyrandoms,andwemakesurethatateverystageourapproximationasaniteternarystringtoT'scanonicalcodecanbeextendedtoapathinP.ThenbycompactnessthecanonicalcodeofourtreewillbeinPandsothetreewillberandom.Thechangesintheapproximationsaremotivatedbytherequirements:Re:if;0eistotalthentherealitdenesisnotin[T]:Letsbeanitestringapproximationofthecanonicalcodewearebuilding.Wewillhavejsj=s.StrategyRewillcomeintopowerafterstageeandwillrestrainuptosomere>ethedefaultvalueisre[0]=e.Alsoitmightrequestsomechangesinaftertheethbit.Westartwith0=;andatstages+1,assuminginductivelythats#and[s]P6=;weaskfortheleasti0andontheotherhand,if=;0ewehaveseenthatfj23Nandisthecanonicalcodeofatreewhichhasasapathg=0: 77
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ThismeansthatifatstagesetherequirementReisnotyetsatised,itwillreceiveattentionatalaterstageandgetsatisedpermanently. 3.5MeasureandDimensionInthissectionweshowthatrandomclosedsethavemeasurezeroTheorem 3.5.1 andboxdimensionlog24 3Theorem 3.5.2 .3.5.1Measure Theorem3.5.1. IfQisarandomclosedset,thenQ=0.Proof.WewillshowthatinthespaceCofclosedsets,theprobabilitythataclosedsetPhasLebesguemeasure0,is1.Thisisprovedbyshowingthatforeachm,P>2)]TJ/F25 7.97 Tf 6.587 0 Td[(mwithprobability0.Foreachm,letSm=fP:P>2)]TJ/F25 7.97 Tf 6.586 0 Td[(mg:Weclaimthatforeachm,Sm=0.Theproofisbyinductiononm.Form=0,wehaveP>1ifandonlyifP=2N,whichisifandonlyifxP=;2;:::,sothatS0isasingletonandthushasmeasure0.NowassumebyinductionthatSmhasmeasure0.ThentheprobabilitythataclosedsetP=[T]hasmeasure>2)]TJ/F25 7.97 Tf 6.587 0 Td[(m)]TJ/F23 7.97 Tf 6.587 0 Td[(1canbecalculatedintwoparts.iIfTdoesnotbranchattherstlevel,sayT0=fgwithoutlossofgenerality.NowconsidertheclosedsetP0=fy:0_y2Pg.ThenP>2)]TJ/F25 7.97 Tf 6.587 0 Td[(m)]TJ/F23 7.97 Tf 6.587 0 Td[(1ifandonlyifP0>2)]TJ/F25 7.97 Tf 6.587 0 Td[(m,whichhasprobability0byinduction,sowecandiscountthiscase.iiIfTdoesbranchattherstlevel,letPi=fy:i_y2Pgfori=0;1.ThenP=1 2P0+P1,sothatP>2)]TJ/F25 7.97 Tf 6.587 0 Td[(m)]TJ/F23 7.97 Tf 6.587 0 Td[(1impliesthatatleastoneofPi>2)]TJ/F25 7.97 Tf 6.587 0 Td[(m)]TJ/F23 7.97 Tf 6.587 0 Td[(1.Notethatthereverseimplicationisnotalwaystrue.Letp=Sm+1.Theobservationsaboveimplythatp61 3)]TJ/F15 11.955 Tf 11.955 0 Td[()]TJ/F24 11.955 Tf 11.956 0 Td[(p2=2 3p)]TJ/F15 11.955 Tf 13.15 8.088 Td[(1 3p2;andthereforep=0. 78
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ToseethatarandomclosedsetQmusthavemeasure0,xmandletS=Sm.ThenSistheintersectionofaneectivesequenceofclopensetsV`,whereforP=[T],P2V`[T`]>2)]TJ/F25 7.97 Tf 6.586 0 Td[(m:Sincethesesetsareuniformlyclopen,thesequencem`=V`iscomputable.Sincelim`m`=0,itfollowsthatthisisaMartinLoftestandthereforenorandomsetQbelongstoT`V`.Theningeneral,norandomsetcanhavemeasure>2)]TJ/F25 7.97 Tf 6.587 0 Td[(mforanym. 3.5.2DimensionSurprisingly,wecancomputetheKolmogorovboxdimensionofarandomclosedset,andinfactitturnsoutthatallrandomclosedsetshavethesamedimension.Theintuitionforthiscomesfromthefollowinglemma.ForanyfunctionFmappingthespaceCofclosedsetsinto<,theexpectedvalueofFonCistheintegralRFPwithrespecttotheprobabilitymeasure. Lemma3.5.2. InthespaceCofclosedsets,theexpectedcardinalityoff2f0;1gn:QI6=;gisexactly4 3nforeveryn,whereQischosenuniformlyatrandomaccordingto.Proof.LetSn=f2f0;1gn:QI6=;g,forarandomlychosenQfromC.Theproofisbyinductiononn.Forn=1,wehavetwocases.Withprobability2 3,cardS1=1andwithprobability1 3,cardS1=2.Thustheexpectedvalueisexactly4 3.Forn+1,thereareagaintwocases.Withprobability2 3,cardS1=1,sothattheexpectedcardSn+1equalstheexpectedcardSn,whichis4 3nbyinduction.Withprobability1 3,cardS1=2,inwhichcasetheexpectedcardSn+1istwicetheexpectedcardSn,thatis,24 3n.ThuswehavetheexpectedvaluecardSn+1=2 34 3n+1 324 3n=4 3n+1: 79
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TheboxdimensionofaclosedsetintheCantorspace,ifitexists,isgivenbythefollowinglimit:dimBFQ=limn!1log2cardTQf0;1gn n:See[ 6 ]forthisformulationoftheboxdimensioninf0;1gN.NowbyLemma 3.5.2 ,theexpectedvalueofcardTQf0;1gnforarandomclosedsetQis4 3n,whichsuggeststhattheboxdimensionofQshouldbelog24 3. Lemma3.5.3. LetQbearandomclosedset.Thenforany">0,thereexistsam2Nsuchthat,foralln>m,4 3n)]TJ/F24 11.955 Tf 11.955 0 Td[("nm,c6n>n.SincethetreeTQf0;1g66n)]TJ/F23 7.97 Tf 6.586 0 Td[(1hasatleast6nnodes,itfollowsfromCherno'sLemmathatthenumberofbranchingnodesislessthannwithprobability62)]TJ/F25 7.97 Tf 6.587 0 Td[(n=6.Thusc6nmislessthan1Xn=m2)]TJ/F25 7.97 Tf 6.586 0 Td[(n=6=2)]TJ/F25 7.97 Tf 6.587 0 Td[(m=6 1)]TJ/F15 11.955 Tf 11.956 0 Td[(2)]TJ/F23 7.97 Tf 6.586 0 Td[(1=6:ThisprovidesacomputablesequenceofclopensetswithmeasuresboundedbyacomputablesequencewithlimitzeroandhenceaMartinLoftest.ItfollowsthatforanyrandomclosedsetQ,thereexistsm0suchthatc6n>nforalln>m0.Nowforn>m0,thereareatleast6n2nodesinTQf0;1g612n)]TJ/F23 7.97 Tf 6.586 0 Td[(1)229(f0;1g66n)]TJ/F23 7.97 Tf 6.587 0 Td[(1,sothatagainbyCherno'sLemma,theprobabilitythat3suchthatc12n>n2foralln>m1.Nowsupposethatm>12m1andthat12n6m<12n+1<16n.Thenn>m1,sothatcm>c12n>n2>m=162:AgainbyCherno'sLemma,theprobabilitythatthenumberofbranchingnodesfromTQf0;1gndiersfrom1 3cnby>1 3c)]TJ/F18 5.978 Tf 7.782 3.259 Td[(1 4ncnis<2)]TJ 6.586 5.248 Td[(p cn=9.Butthisisexactlytheprobabilitythatcn+1diersfrom4 3cnby>1 3c)]TJ/F18 5.978 Tf 7.782 3.259 Td[(1 4ncn.Forn>m1,weknowthatcn>)]TJ/F25 7.97 Tf 8.34 4.977 Td[(n 162,sothatp cn>n 16andc)]TJ/F18 5.978 Tf 7.782 3.258 Td[(1 4n64 p nandhence2)]TJ 6.587 5.248 Td[(p cn=962)]TJ/F25 7.97 Tf 6.586 0 Td[(n=144.Thustheprobabilitypnthatcn+1 80
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diersfrom4 3cnbymorethancn 9p nis<2)]TJ/F25 7.97 Tf 6.587 0 Td[(n=144.Thentheprobabilitythatforanyn>m1,cn+1diersfrom4 3cnbymorethan4 3p ncnisboundedby1Xn=mpn=1Xn=m2)]TJ/F25 7.97 Tf 6.587 0 Td[(n=144=2)]TJ/F25 7.97 Tf 6.586 0 Td[(m=144 1)]TJ/F15 11.955 Tf 11.956 0 Td[(2)]TJ/F23 7.97 Tf 6.586 0 Td[(144:ThisagainprovidesaMartinLoftestwhichshowsthatforanyrandomclosedsetQ,thereexistsm2sothatforn>m2,4 31)]TJ/F15 11.955 Tf 18.699 8.088 Td[(1 p ncn6cn+164 31+1 p ncn:Nowgiven",choosem>m2sothat+1 p m2<1+"and1)]TJ/F24 11.955 Tf 11.955 0 Td[("<)]TJ/F23 7.97 Tf 18.307 4.707 Td[(1 p m2.Thenforanyk,cm4 32k)]TJ/F24 11.955 Tf 11.955 0 Td[(km+2k.Foroddn,thisinequalitywillholdbytheinequalityabove. Theorem3.5.4. ForanyrandomclosedsetQ,theboxdimensionofQislog24 3.Proof.Given">0,letmbegivenbyLemma 3.5.3 .Thenforn>m,wehavenlog24 3+nlog2)]TJ/F24 11.955 Tf 11.955 0 Td[("6log2cardTQf0;1gn6nlog24 3+nlog2+";sothatlog24 3+log1)]TJ/F24 11.955 Tf 11.955 0 Td[("6log2cardTQf0;1gn n6log24 3+log2+";andthereforedimBQ=limnlog2cardTQf0;1gn n=log24 3. 81
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3.6PrexFreeComplexityofClosedSetsInthissection,weconsiderrandomnessforclosedsetsintermsofincompressibilityoftrees.Ofcourse,Schnorr'stheoremtellsusthatPisrandomifandonlyifthecodexP2f0;1;2gNforPisprexfreerandom,thatis,K3xPdn>n)]TJ/F24 11.955 Tf 12.658 0 Td[(O.Schnorr'stheoremforarbitrarynitealphabetsisshownin[ 18 ].HerewewriteK3toindicatethatwewouldbeusingauniversalprexfreefunctionU:f0;1;2g!f0;1;2g.However,manypropertiesoftreesandclosedsetsdependonthelevelsTn=Tf0;1gnofthetree.Forexample,if[Tn]=[fI:2Tng,then[T]=Tn[Tn]and[T]=limn!1[Tn].SowewanttoconsiderthecompressibilityofatreeintermsofKTn.NowthereisanaturalrepresentationofTnasastringoflength2n.Thatis,listf0;1gninlexicographicorderas1;:::;2nandrepresentTnbythestringe1;:::;e2nwhereei=1ifi2Tandei=0otherwise.HenceforthweidentifyTnwiththisnaturalrepresentation.ItisinterestingtonotethatthecodeforTnwillhaveashorterlengththanthenaturalrepresentation.Forexample,if[T]=fygisasingleton,thenx=yandforeachn,thecodeforTnisxdn.Ifxisthecodeforthefulltreef0;1g,thenx=;2;:::andthecodeforTnisastringofn)]TJ/F15 11.955 Tf 11.957 0 Td[(12's,thoselabelsattachedtonodesoflength2n)]TJ/F24 11.955 Tf 12.273 0 Td[(cforalln.However,wewillseethatthisisnotpossibleforanytree.Ontheonehand,inSection 3.6.1 weachievealowerboundforincompressibility.Thatis,weshowthatifP=[T]israndomthenthereisaconstantcsuchthatKTn>)]TJ/F23 7.97 Tf 6.675 4.977 Td[(7 6n)]TJ/F24 11.955 Tf 12.446 0 Td[(cforalln.Ontheotherhand,inSection 3.6.2 ,weseethatthe2nistoohighofanincompressibilityboundsincethereissomecandsomerandomclosedsetsuchthatKTn62n)]TJ/F24 11.955 Tf 11.955 0 Td[(cforalln. 82
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Inalargersense,weseekaformulationofrandomness,intermsoftheincompressibilityofTn,forotherobjectssuchas01classesor02classes.Inthefollowingtwosectionsweconsiderthesequestionsandachievesomelowerandupperboundsfortheseclassesofobjects.3.6.1LowerComplexityBoundsFirstwegivealowerboundfortheprexfreecomplexityofarandomtree. Theorem3.6.1. IfPisarandomclosedsetandT=TP,thenthereisaconstantcsuchthatKTn>)]TJ/F23 7.97 Tf 6.675 4.977 Td[(7 6n)]TJ/F24 11.955 Tf 11.955 0 Td[(cforalln.Proof.LetP=[T]bearandomclosedset.LetmbegivenbyLemma 3.5.3 ,for"=7 6,sothatforn>m,cardTn>7 6n:ItfollowsthatthecodexnforTnhaslength>)]TJ/F23 7.97 Tf 6.675 4.977 Td[(7 6n.Sincexisrandom,weknowthat,forn>m,K3xn>7 6n)]TJ/F24 11.955 Tf 11.955 0 Td[(a;forsomeconstanta.NowwecancomputexnfromTn,sothatKTn>K3xn)]TJ/F24 11.955 Tf 11.956 0 Td[(b;forsomeconstantb.Theresultnowfollows.Thatis,letUmappingf0;1gtof0;1gbeauniversalprexfreeTuringmachineandletKTn=minfjj:U=Tng.LetMbeaprexfreemachineMmappingf0;1gtof0;1;2gsuchthatMTn=xn.ThendeneVbyV=MU:ThenKVxdn6KTn,sothatforsomeconstante,K3xn6KTn+eandhenceKTn>K3xn)]TJ/F24 11.955 Tf 11.955 0 Td[(e>7 6n)]TJ/F24 11.955 Tf 11.955 0 Td[(b)]TJ/F24 11.955 Tf 11.955 0 Td[(e: 83
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Thestandardexampleofarandomreal,Chaitin's[ 27 ],isac.e.realandtherefore02.Thusthereexistsa02randomtreeTandbyTheorem 3.6.1 ,KT`>)]TJ/F23 7.97 Tf 6.675 4.976 Td[(7 6n)]TJ/F24 11.955 Tf 12.409 0 Td[(cforsomec.Wehaveamoremodestresultfor01classes.Thatis,thereisaneectivelyclosedsetwithnottoomuchcompressibility,inthefollowingsense. Theorem3.6.2. Thereisa01classP=[T]suchthatKTn>nforalln.Proof.RecalltheuniversalprexfreemachineUandletS=f2DomU:jUj>2jjg.ThenSisac.e.setandcanbeenumeratedas1;2;:::.ThetreeT=TsTswhereTsisdenedatstages.InitiallywehaveT0=f0;1g.Wesaythattrequiresattentionatstages>twhen=Ut=Tsnforsomensothatjj=2nandn>jtj.Actionistakenbyselectingsomepatht2TsoflengthnanddeningTs+1tocontainallnodesofTswhichdonotextendt.Then6=Ts+1nandfurthermore6=Trnforanyr>s+1sincefutureactionwillonlyremovemorenodesfromTn.Atstages+1,lookfortheleastt6s+1suchthattrequiresactionandtaketheactiondescribedifthereissuchat.Otherwise,letTs+1=Ts.LetAbethesetoftsuchthatactionisevertakenont.RecallfromtheKraftInequalitythatPt2jtj<1.Sincejtj>jtj,itfollowsthatPt2A2jtj<1aswell.Now[T]=1)]TJ/F29 11.955 Tf 11.955 8.967 Td[(Pt2jtj>0andtherefore[T]isnonempty.Itfollowsfromtheconstructionthatforeacht,actionistakenfortatmostonce.NowsupposebywayofcontradictionthatU=Tnforsometwithjj6n.Theremustbesomestager>tsuchthatforalls>r,Tsn=Tnandsuchthatactionisnevertakenonanyt02p `forall`. 84
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Proof.WewillconstructatreeTsuchthatTn2cannotbecomputedfromfewerthan2nbits.WewillassumethatU;"totakecareofthecasen=0.Atstages,wewilldenethenonemptylevelTs2ofT,usinganoraclefor00.WebeginwithT0=f;g.Atstages>0,weconsiderDs=f2DomU:jj<2sg:SinceUisprexfree,cardDs<22s.Nowthereareatleast222s)]TJ/F18 5.978 Tf 5.756 0 Td[(1treesofheights2whichextendTs)]TJ/F23 7.97 Tf 6.586 0 Td[(12andwecanusetheoracletochoosesomeniteextensionT0=Ts2ofTs)]TJ/F23 7.97 Tf 6.587 0 Td[(12suchthat,forany2Ds,U6=T0andfurthermore,U6=TrforanypossibleextensionTrwiths26r.Thatis,sincethereare<22snitetreeswhichequalUforsome2Ds,thereissomeextensionT0ofTs)]TJ/F23 7.97 Tf 6.587 0 Td[(12whichdiersfromalloftheseatlevels2.Weobservethattheoraclefor00isusedtodeterminethesetDs.Atstages,wehaveensuredthatforanyextensionTf0;1gofTs2,anywithjj62s2andanyn>s2,U6=Tn.ItisimmediatethatKTn>2p n. 3.6.2UpperComplexityBoundsInTheorem 3.6.1 weachievedalowerboundof)]TJ/F23 7.97 Tf 6.675 4.977 Td[(7 6nfortheprexfreecomlexityofatreeTPofarandomclosedsetP.Itseemsplausiblethatwemightbeabletoachieveahigherboundof2n.Iftruethiswouldactuallyprovideandimmediatecharacterizationofrandomnessofclosedsetsintermsofprexfreecomplexityoftrees.Thatis,aclosedwouldberandomiKTn>2n)]TJ/F24 11.955 Tf 12.613 0 Td[(cforsomeconstantc.Toseethis,notethatfromxd2nwecancomputeTnuniformlysothatK3xd2n>KTn)]TJ/F24 11.955 Tf 12.307 0 Td[(bforsomeb.Howeverthefollowingtheoremprovidesanuppercomplexityboundlessthan2n,refutingsuchapossibility. Theorem3.6.4. ForanytreeTf0;1g,thereareconstantsk>0andcsuchthatKT`62`)]TJ/F15 11.955 Tf 11.955 0 Td[(2`)]TJ/F25 7.97 Tf 6.587 0 Td[(k+cforall`.Proof.Forthefulltreef0;1g,thisisclearsosupposethat=2Tforsome2f0;1gm.Thenforanylevel`>m,thereare2`)]TJ/F25 7.97 Tf 6.587 0 Td[(mpossiblenodesforTwhichextendandT`may 85
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beuniformlycomputedfromandfromthecharacteristicfunctionofT`restrictedtotheremainingsetofnodes.Thatis,xoflengthmanddeneaprexfreecomputerMasfollows.ThedomainofMisstringsoftheform0`1wherejj=2`)]TJ/F15 11.955 Tf 12.378 0 Td[(2`)]TJ/F25 7.97 Tf 6.586 0 Td[(m.MoutputsthestandardrepresentationofatreeT`suchthatnoextensionofisinT`andsuchthattellsuswhetherstringsnotextendingareinT`.ItisclearthatMisprexfreeandwehaveKMT`=`+1+2`)]TJ/F15 11.955 Tf 11.539 0 Td[(2`)]TJ/F25 7.97 Tf 6.587 0 Td[(m.ThusKT`6`+1+2`)]TJ/F15 11.955 Tf 11.539 0 Td[(2`)]TJ/F25 7.97 Tf 6.587 0 Td[(m+cforsomeconstantc.Now`+1<2`)]TJ/F25 7.97 Tf 6.586 0 Td[(m)]TJ/F23 7.97 Tf 6.586 0 Td[(1forsucientlylarge`andthusbyadjustingtheconstantc,wecanobtainc0sothatKT`62`)]TJ/F15 11.955 Tf 11.955 0 Td[(2`)]TJ/F25 7.97 Tf 6.586 0 Td[(m)]TJ/F23 7.97 Tf 6.586 0 Td[(1+c0: ThefollowingtheoremalsorefutesthepossibilitythatKT`>2`)]TJ/F25 7.97 Tf 6.587 0 Td[(cisacharacterizationofrandomclosedsetsintermsofprexfreerandomness.Itshowsthatclosedsetswithsmallmeasure,suchasrandomclosedsetswhichhavemeasurezeroseeTheorem 3.5.1 ,aremorecompressible. Theorem3.6.5. If[T]<2)]TJ/F25 7.97 Tf 6.587 0 Td[(k,thenthereexistscsuchthat,forall`,KT`62`)]TJ/F25 7.97 Tf 6.587 0 Td[(k+1+c:Proof.Supposethat[T]<2)]TJ/F25 7.97 Tf 6.587 0 Td[(k.Thenforsomeleveln,Tnhas<2n)]TJ/F25 7.97 Tf 6.586 0 Td[(knodes1;:::;t.Nowforany`>n,T`canbecomputedfromthexedlist1;:::;tandthelistofnodesofT`takenfromtheatmost2`)]TJ/F25 7.97 Tf 6.586 0 Td[(kextensionsof1;:::;t.ItfollowsasintheproofofTheorem 3.6.4 abovethatforsomeconstantcandall`,KT`62`)]TJ/F25 7.97 Tf 6.587 0 Td[(k+`+1+c:Thusforlargeenoughsothat`+162`)]TJ/F25 7.97 Tf 6.587 0 Td[(k,wehaveKT`62`)]TJ/F25 7.97 Tf 6.587 0 Td[(k+1+c;asdesired. 86
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Weconjecturethataboundof)]TJ/F23 7.97 Tf 6.675 4.977 Td[(4 3nwouldcharacterizerandomclosedsetsintermsofprefexfreecomplexity.Itwouldsuce,then,toshowthat)]TJ/F23 7.97 Tf 6.675 4.976 Td[(4 3nisalowerboundandthatthisboundimpliesrandomness.Wealsostillseekupperboundsfor01orclosed02classestowardsestablishingprexfreecomplexitycharacterizationsoftheseclasses.Itseemsplausiblethat01classesaremorecompressiblei.e.necessarilyhavesmallerlowerboundsthanrandomclosedsetsandwewouldliketoexplorethisnotionfurther.3.7OtherNotionsofRandomnessforClosedSetsOthernotionsofrandomnessthatdependondierentprobabilitymeasures,ortheinclusionoftreeswithdeadendsintheencoding,mightalsobeconsidered.3.7.1RandomnesswithRegularProbabilityMeasuresForanyregularmeasure,wecandenethenotionofaMartinLoftestandtheresultingnotionofaMartinLofrandomorjustrandomreal.Itiseasytoseethatrandomrealsexistforanyandhencerandomclosedsetsexist.Theresultsonghostcodesandjoinswillholdforanyregularmeasure.ThecorrespondingversionofLemma 3.3.8 willholdifisregularwithb0andb161 2.TheproofsofTheorem 3.4.14 andCorollary 3.4.15 ,thatnosubsetofameasurezero01classisrandom,alsogothroughunderthisassumption.Someoftheresultsinthischaptermayalsobeobtainedforfwheref_i61 2fori=0;1.Forexample,withrespecttof,arandomclosedsetwillhavenoisolatedelementsanditwillalwayscontainarandomelement.Foranyregularmeasure,eithertheleftmostortherightmostpathwillbenonrandom,sinceeitherb0+b2>1 2orb1+b21 2.TheproofofTheorem 3.4.19 thateveryrandomclosedsethasmeasure0seemstorequire,forfrandomness,thatf_261 2forall.3.7.2RandomnesswiththeInclusionofTreeswithDeadsEndsReturningtothenotionofrandomnesswhichallowstreeswithdeadends,letb3nowbetheprobabilitythatagivennodehasnoextensionsandlettheprobabilityberegularasabove.Thenasimplerecursionshowstheprobabilitypofagivenclosedsetbeing 87
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emptysatisestheequationp=b3+b0+b1p+b2p2:Solvingforp,weobtainp)]TJ/F15 11.955 Tf 11.955 0 Td[(1b2p)]TJ/F24 11.955 Tf 11.955 0 Td[(b3=0:Thuseitherp=1orp=b3 b2.Itfollowsthatifb26b3,thenp=1,thatis,almosteveryclosedsetisempty.Supposenowthatb3
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ii Forn>2andanyQ1;:::;Qn2CTni=1Qi6Xf)]TJ/F15 11.955 Tf 9.298 0 Td[(1jIj+1T[i2IQi:;6=If1;2;:::;ngg:Thisisthealternatingofinniteorderproperty. iii IfQ=nQnandQn+1Qnforalln,thenTQ=limn!1TQn.Wewillassume,unlessotherwisespecied,thatTN=1foragivencapacityT. Denition3.8.2ComputableCapacities. AcapacityTiscomputableifitiscomputableonthefamilyofclopensets.Itfollowsthatthecapacityofany01classisuppersemicomputable.Finally,thefollowingnotationaldenitionwillbeusedthroughout. Denition3.8.3TdQ,forQ2C. SupposeQ2C.DeneTdQ:=dVQ,whereVQisasubbasissetforthehitormisstopologyonCasgiveninsection 3.3.1 anddisagivenprobabilitymeasure.Thatis,TdQistheprobabilitythatarandomlychosenclosedsetmeetsQ.Wenowshow,inthefollowingtwotheorems,thatacomputablecapacityisalwaysobtainablefrom,oraconsequenceof,acomputableprobabiltymeasuredforsomed.Thefollowingtheorem,inparticular,iswellknown.Fordetailsoncapacityandrandomsetvariables,see[ 73 ]. Theorem3.8.4. IfdisacomputableprobabilitymeasureonC,thenTdisacomputablecapacity.Proof.Thisiseasilyveried.CertainlyT;=0.Thealternatingpropertyfollowsbybasicprobability.Foriii,supposethatQ=nQnisadecreasingintersection.Thenbycompactness,QK6=;ifandonlyifQnK6=;foralln.Furthermore,VQn+1VQnforalln.ThusTQ=VQ=nVQn=limnVQn=limnTQn: 89
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ThecomputabilityofTiseasilyveried.Thatis,foranyclopensetI1[[Ikwhereeachi2f0;1gn,wecomputetheprobabilitydistributionforalltreesofheightnandaddtheprobabilitiesofthosetreeswhichcontainoneofthei. Thisresulthasaconverse,duetoChoquet.See[ 73 ]forthegeneralresult. Theorem3.8.5. IfTisacomputablecapacity,thenthereisacomputablemeasuredonthespaceofclosedsetssuchthatT=Td.Proof.GiventhevaluesTUforallclopensetsI1[[Ikwhereeachi2f0;1gn,thereisinfactauniqueprobabilitymeasuredontheseclopensetssuchthatT=Tdandthiscanbecomputedasfollows.SupposerstthatTIi=aifori<2andnotethateachai61anda0+a1>1bythealternatingproperty.IfT=Td,thenwemusthaved+d=a0andd+d=a1andalsod+d+d=1,sothatd=a0+a1)]TJ/F15 11.955 Tf 12.337 0 Td[(1,d=1)]TJ/F24 11.955 Tf 12.012 0 Td[(a1andd=1)]TJ/F24 11.955 Tf 12.011 0 Td[(a0.ThiswillimplythatT=Tdwhenjj=1.Nowsupposethatwehavedeneddandthatisthecodeforanitetreewithelements0;:::;n=andthusd_iisgivingtheprobabilitythatwillhaveoneorbothimmediatesuccessors.Weproceedasabove.LetTI_i=aiTIfori<2.Thenasaboved_2=da0+a1)]TJ/F15 11.955 Tf 11.955 0 Td[(1andd_i=d)]TJ/F24 11.955 Tf 11.955 0 Td[(aiforeachi. 3.8.2RegularMeasuresandCapacitiesofClosedSetsInthissectionallresultsarewithrespecttodxedastheuniformmeasurei.e.theregularmeasurewithb0=b1=b2=1 3;seeDenition 3.2.6 .Withthismeasure,wewillconsiderthecapacitiesofrandomclosedsetsandeectivelyclosedsets.WesaythatQ2CisdrandomifxQisMartinLofrandomwithrespecttothemeasured. Theorem3.8.6. Fortheregularmeasuredwithbi=1 3,ifRisadrandomclosedset,thenTdR=0.Proof.Fixdasdescribedabovesothatd_i=d1 3andlet=d.Wewillcomputetheprobability,giventwoclosedsetsQandK,thatQKisnonempty.LetQn=[fI:2f0;1gn&QI6=;g 90
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andsimilarlyforKn.ThenQK6=;ifandonlyifQnKn6=;foralln.LetpnbetheprobabilitythatQnKn6=;fortwoarbitraryclosedsetsKandQ,relativetoourmeasure.Itisimmediatethatp1=7 9,sinceQ1K1=;onlywhenQ1=IiandK1=I)]TJ/F24 11.955 Tf 12.31 0 Td[(i.Nextwewilldeterminethequadraticfunctionfsuchthatpn+1=fpn.Thereare9possiblecasesforQ1andK1,whichbreakdowninto4distinctcases.CaseI:TherearetwochancesthatQ1K1=;.CaseII:TherearetwochancesthatQ1=K1=Ii,sothatQn+1Kn+16=;withprobabilitypn.CaseIII:TherearefourchanceswhereQ1=2NandK1=Iiorviceversa,sothatonceagainQn+1Kn+16=;withprobabilitypn.CaseIV:ThereisonechancethatQ1=K1=2N,inwhichcaseQn+1Kn+16=;withprobability1)]TJ/F15 11.955 Tf 12.118 0 Td[()]TJ/F24 11.955 Tf 12.118 0 Td[(pn2=2pn)]TJ/F24 11.955 Tf 12.118 0 Td[(p2n.ThisisbecauseQn+1Kn+1=;ifandonlyifbothQn+1IiKn+1=;forbothi=0andi=1.Addingthesecasestogether,weseethatpn+1=6 9pn+1 9pn)]TJ/F24 11.955 Tf 11.955 0 Td[(p2n=8 9pn)]TJ/F15 11.955 Tf 13.151 8.087 Td[(1 9p2n:Itfollowsthatthesequencehpnin2!iscomputableandwewillseethatthelimitiszero.Letfp=8 9p)]TJ/F23 7.97 Tf 13.605 4.707 Td[(1 9p2.Elementarycalculusshowsthatfhasxedpointsatp=)]TJ/F15 11.955 Tf 9.298 0 Td[(1andp=0andthatfor02)]TJ/F25 7.97 Tf 6.586 0 Td[(ng:LetCmbethenumberoftreesofheightmwithoutdeadends. 91
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ForeachQ2Am;nthereare2)]TJ/F25 7.97 Tf 6.586 0 Td[(nCmpossiblechoicesforKmsuchthatKmQm6=;andthusatleastDm;n=1 2)]TJ/F25 7.97 Tf 6.587 0 Td[(nAm;nC2mchoicesforK;Q2CCsuchthatKmQm6=;withQ2Am;nsinceeachpairmightbecountedtwice.Nowdeneacomputablesequencehmnin2!,sothatpmn<2)]TJ/F23 7.97 Tf 6.587 0 Td[(2n)]TJ/F23 7.97 Tf 6.587 0 Td[(1.Thenpmn>Dmn;n C2mn=2)]TJ/F23 7.97 Tf 6.586 0 Td[(n+1Amn;n:ItfollowsthatAmn;n62n+1pmn<2n+12)]TJ/F23 7.97 Tf 6.587 0 Td[(2n)]TJ/F23 7.97 Tf 6.586 0 Td[(1=2)]TJ/F25 7.97 Tf 6.587 0 Td[(n:LettingSn=[r>nAmr;ritfollowsthatSn62)]TJ/F25 7.97 Tf 6.587 0 Td[(naswell.NowletRbearandomclosedset.ThesequencehSnin2!isacomputablesequenceofc.e.opensetswithmeasure62)]TJ/F25 7.97 Tf 6.586 0 Td[(n,sothatthereissomensuchthatR=2Sn.Thusforr>n,fK:KmrRmr6=;g<2)]TJ/F25 7.97 Tf 6.587 0 Td[(randitfollowsthatfK:KR6=;g=limnfK:KmnRmn6=;g=0:ThusTdR=0,asdesired. Thisresultseemstodependonthemeasure.Fordierentregularmeasures,thecapacityofarandomclosedsetcanhavedierentvalues. Theorem3.8.7. Fortheregularmeasuredwithbi=1 3,thereisameasurezero01classQsuchthatTdQ>0:Proof.FirstletuscomputethecapacityofXn=fx:xn=0g.Forn=0,wehaveTdQ0=2 3.NowtheprobabilityTdXn+1thatanarbitraryclosedsetKmeetsXn+1maybecalculatedintwodistinctcases.LetKnbeasintheproofofTheorem 3.8.6 .CaseIIfK0=2N,thenTdXn+1=1)]TJ/F15 11.955 Tf 11.955 0 Td[()222(TdXn2. 92
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CaseIIIfK0=Iiforsomei<2,thenTdXn+1=TdXn.ItfollowsthatTdXn+1=2 3TdXn+1 3TdXn)]TJ/F15 11.955 Tf 11.095 0 Td[(TdXn2=4 3TdXn)]TJ/F23 7.97 Tf 12.29 4.707 Td[(1 3TdXn2.Nowthefunctionfp=4 3p)]TJ/F23 7.97 Tf 13.812 4.707 Td[(1 3p2hasthepropertythatfp>pfor0 3 4=c0andforeachk,choosen=nk+1suchthatTdXn0;:::;nk;n>ck+1.Thiscanbedonesinceck+1limkck=1 2. Thisresultcaneasilybeextendedtoanyboundedmeasure. 93
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CHAPTER4RANDOMCONTINUOUSFUNCTIONSThefollowingchapterisjointworkwithGeorgeBarpalias,DouglasCenzer,JereyB.Remmel,andRebeccaWeberandwillappearintheArchiveforMathematicalLogicasanarticleentitledAlgorithmicRandomnessofContinuousFunctions[ 8 ].ApreliminaryversionofthisresearchwasoriginallypresentedattheThirdInternationalConferenceofComputabilityandComplexityinAnalysisinGainesville,Floridain2006byJ.B.Remmel.ThispreliminaryworkwaspublishedinthereferredconferenceproceedingsasRandomContinuousFunctionsP.Brodhead,D.Cenzer,andJ.B.RemmelinProceedingsofCCA2006D.Cenzer,R.Dillhage,T.GrubbandKlausWeihrauch,eds.,InformationBerichte,FernUniversitat06,pages76{89andinSpringerElectronicNotesinTheoreticalComputerScience,ElsevierScience16707[ 15 ].PortionsofthisworkwerealsopresentedbyP.BrodheadattheAMSFall2006EasternSectionalMeetingOctober2006,Storrs,CTandtheConferenceonLogic,Computability,andRandomnessJanuary2007,BuenosAires,Argentina.4.1OverviewInChapter 3 ,weconsideredanotionofrandomnessforclosedsets.Wedothesameforcontinuousfunctionshere.AnintroductiontorandomnessforrealsisprovidedinSection 3.2 .Thischapterisorganizedasfollows.InSection 4.2 ,weprovideadenitionofrandomnessforcontinuousfunctionsandshowthatitissound.InSection 4.3 ,weprovevariousresultsforimagesofrandomcontinuousfunctions{perfectness,noninjectivity,andinstancesofnonsurjectivity;wealsostudyimagesofcomputableelements.InSection 4.4 ,wetierandomclosedsetstorandomclosedfunctionsthroughimages:inverseimagesof0!arerandomclosedsets,butimages,ingeneral,arenot.Continuingonthethemeofinverseimagesof0!,inSection 4.5 weconsiderpseudodistancefunctions.InSection 4.6 ,webrieyconsiderhowtheresultsofChapters 3 and 4 canberelativizedfornrandomness.Finally,inSection 4.7 ,wedescribesomedirectionsforfutureresearch. 94
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4.2DeniningRandomnessforContinuousFunctionsAfunctionF:2N!2Niscontinuousiithasaclosedgraph.Itseemsreasonable,then,todenecontinuousfunctionftoberandom,itheitsgraphGrF=fxy:y=Fxgisrandom.Howeverif[T]isthegraphofafunctionand2Thasevenlength,thenwemusthave_02Tand_12T.Thismeansthatthefamilyofclosedsetswhicharethegraphsoffunctionshasmeasure0inthespaceofclosedsetsandhencearandomclosedsetwillnotbethegraphofafunction.Weneed,therefore,adierentmethodtodenerandomnessforcontinuousfunctions.Wedothisbelow.4.2.1RepresentingFunctionsGivenacontinuousfunctionF:2N!2N,weareinterestedinrepresentingitinsuchawaysoastobeabletoconsideranotionofalgorithmicrandomness.WeshowbelowthatanysuchfunctionFmayberepresentedbyinnitelymanyrepresentingfunctionsoftheformf:f0;1g!f0;1;2g.Thiswillallowus,inthefollowingsection,tobeabletorepresentcontinuousfunctionaselementsof3!,socalledrepresentingsequences,andtoconsiderasuchafunctionasrandomifitpossessesarandomrepresentingsequence.Informationoutput,thekeytorepresentation.ForanycontinuousfunctionFon2Nandany2f0;1g,thereisanaturalnumbernandbinarystringoflengthnsuchthatforallu2I,Fudn=.Inparticular,Fun=nforeverysuchu.Ingeneral,thelengthofmaybemuchlargerthann,sowemayhavetoextendbyseveralbitstogetuniformityofFudn+1withintheintervalaround'sextension.Representingfunctions.Takingtheaboveintoconsideration,wemayrecursivelyrepresentanycontinuousfunctionF:2N!2Nbysomefunctionf:f0;1g!f0;1;2gasfollows.SupposeFisgiven.Letf;=;.Forjj=m+1,havingdenedfdi=eiforalli6m,let=n1;:::;nkbetheresultofdeletingall2sfrome1;:::;em.Ifforallu2I,Fudk=_j,j2f0;1g,wemayletem+1=j.Ifnotwemusthaveem+1=2;evenifsoweallowem+1=2. 95
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Thecanonicalrepresention.Notice,fromtheabove,thatanycontinuousFhasinnitelymanyrepresentingfunctionsf:f0;1g!f0;1;2g.Therepresentationwhichusesasfew2saspossibleweshallcallthecanonicalrepresentation.4.2.2RepresentingSequencesWewanttocodetherepresentingfunctionasanelementof3Ntodiscussitsalgorithmicrandomness.Todoso,rstenumeratef0;1g=f;gas0;1;:::,orderedrstbylengthandthenlexicographically.Thus0=,1=,2=0,etc.Wedenerepresentingsequencesbelow. Denition4.2.1. i INF,Rem2LetINFequalthesetofy2f0;1;2gnsuchthatfn:yn6=2gisinniteand,fory2INF,letRem2ybetheresultofremovingfromxalloccurrencesof2. ii RepresentingfunctionsAfunctionf:f0;1g!f0;1;2grepresentsafunctionF:2N!2Nifforallx22N,thesequencey,denedbyyn=fxdnbelongstoINFandRem2y=Fx. iii RepresentingsequencesAsequencer2f0;1;2gNrepresentsthecontinuousfunctionFwrittenF=Frifthefunctionfr:f0;1g!f0;1;2g,denedbyfrn=rn,representsF. iv Labelled2!treesGivenarepresentingsequencer,thefunctionfrgivesrisetoalabelled2!tree.Weattach,orassociate,thevalueoffrwitheachnode. Example4.2.1The2!treefortheIdentity,AGeometricIntrepretation. Theidentityfunctioncanberepresentedbyplacinganeonanynodewhichendsine.ThiscanalsobepicturedgeometricallyasrepresentingthegraphofFastheintersectionofadecreasingsequenceofclopensubsetsoftheunitsquare.Initiallythechoiceoffandfselectsfromthe4quadrants.Thatis,forexample,f==fimpliesthatthegraphofFisincludedinthebottomhalfofthesquareandf=;andf=impliesthatthegraphexcludesthelowerrighthandquadrant.SuccessivevaluesoffcontinuetorestrictthegraphofFinasimilarfashion. 96
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4.2.3ASoundDenitionInthissectionwedeneameasureonthespaceoffunctionsF:2N!2Nthatallowsustodenethenotionofrandomnessforfunctionson2N.Inshort,afunctionisrandomifitpossessesarandomrepresentingfunction,orequivalently,arandomrepresentingsequence.Wewillshowthatnocanonicalrepresentingfunctioncanberandom,sothatnecessarily,thedenitionofrandomnessforfunctionsisinthisexistentialformat.Itmaybe,however,thatnocontinuousfunctionhasarandomrepresentingfunction.Weshowthatthisisnotso.Infact,weshowthateveryrandomrepresentingfunctioniscontinuous.Clearlythen,randomcontinuousfunctionsexistand,infact,02randomcontinuousfunctionsexist.Thereforethedenitionissoundandthestructurebeginstomanifestitselfasrich.TheMeasureforRandomness.ThemeasurewhichisusedtodenerandomnessforcontinuousfunctionsistheLebesguemeasureonthespace3Nofrepresentingsequences.Thusforeachnewbitofinput,thereisequalprobability1 3thatfrgivesanewoutputof0forFr,givesanewoutputof1forFr,orgivesnonewoutputforFr.Thismeasurenowinducesameasure,say,onthespaceFofcontinuousfunctions. Denition4.2.2. AfunctionF:2N!2Nisrandomifthereisarepresentingsequencer23NforFthatisrandomwithrespecttothemeasure.Werstshowthateveryrandomrepresentingfunctioniscontinuous.Thefollowinglemmaisneeded. Lemma4.2.3. LetbeanitesetandletQNbea01classofmeasure0.ThennoelementofQisMartinLofrandom.Proof.Let=f0;1;2gwithoutlossofgenerality.LetQ=[T]whereTf0;1;2gisacomputabletreepossiblywithdeadends.Foreachn,letTn=Tf0;1;2gnandletQn=[fI:2Tng:Letgn=Qn=jTnj 3n.Thengnisacomputablesequenceandlimn!1gn=Q=0: 97
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ThisMartinLoftestshowsthatQhasnorandomelements.AsobservedbySolovay,itissucienttohaveacomputablesequenceapproachingzeroratherthanthestrictertestwithasequenceofmeasuresgn62)]TJ/F25 7.97 Tf 6.586 0 Td[(n. Theorem4.2.4. i Thesetofrepresentingfunctionsfortotalfunctionshasmeasureone. ii Everyrandomfunctioniscontinuous.Proof.iLetr23Nandsupposethatfrdoesnotrepresentatotalfunction.Thenthereissomex22Nandsome2f0;1gsuchthatfrxdn=foralmostalln.Withoutlossofgeneralitywemayassumethat=;.LetAbethesetoffunctionsf:f0;1g!f0;1gsuchthatf=;forarbitrarilylongstringsandletp=A.Thencertainlyp65 9,sinceifrandrarebothinf0;1g,thenfr=2A.Consideringthe9casesfortheinitialchoicesoffandf,weseethatp=4 9p+1 9[1)]TJ/F15 11.955 Tf 11.955 0 Td[()]TJ/F24 11.955 Tf 11.956 0 Td[(p2];sothat1 9p2+1 3p=0,whichimpliesthatp=0.Thatis,thereare4casesinwhichjfij=1fori=0;1sothatimmediatelyf=2A,thereare4casesinwhichonlyoneoffi=;,inwhichcasetheremainingfunctiong,denedbyg=fi_mustbeinA,andthereisonecaseinwhichfi=;fori=0;1,inwhichcaseatleastoneoftheremainingfunctionsmustbeinA.Consequently,thesetofrepresentingfunctionsfortotalfunctionshasmeasureone.iiObservethatAisa01class,sincefr2Aifandonlyif8n92f0;1gnfr=;.ItfollowsfromLemma 4.2.3 thatnorepresentingfunctionon2forarandomfunctionon2NcanbeinA.Asallfunctionsrepresentingpartialfunctionson2NoccurinA,itfollowsthateveryrandomfunctionistotal.Sincethegraphofatotalfunctionisaclosedset,itfollowsthatrandomfunctionsarecontinuous. NowthesetofMartinLofrandomelementsoff0;1;2gNhasmeasureoneandthereexistsa02MartinLofreal.Hencewehavethefollowing. Theorem4.2.5. Thereexistsarandomcontinuousfunctionwhichis02computable. 98
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Wealsorstobservethatanycontinuousfunctionwillhavearepresentationwhichisnotrandom.Infact,thecanonicalrepresentationitselfcanneverberandom. Proposition4.2.6. ForanycontinuousfunctionF,thecanonicalrepresentationisnotrandom.Proof.Theideaisthatwheneverthecanonicalrepresentationlabelsanodewith2,thenthetwolabelsonthesuccessornodes_0and_1cannotbeboth0,orboth1.ThuswehavethefollowingMartinLoftest.Assumebywayofcontradictionthatrisrandomandcanonical.LetSebethesetofr23Nsuchthatrhasatleasteoccurrencesof2andsuchthat,forthersteoccurrencesof2inr,thecorrespondingsuccessorvaluesarenotboth0orboth1.Sincerisrandom,itmusthaveinnitelymanyoccurrencesof2andsinceriscanonical,itmustbelongtoeverySe.ButeachSeisac.e.opensetandhasmeasure67 9e,sothatnorandomsequencecanbelongtoeverySe. Thetheorem,infact,demonstratestheneedfortheexistentialpartofdenitionofrandomfunctions.Inthefollowingsectionswewillobtainsomeadditionalpropertiesofrandomcontinuousfunctions.4.3RandomContinousFunctionsandImages4.3.1PerfectImages,ineveryinstanceInthissectionweshowthatallrandomcontinuousfunctionsalwayshaveperfectimages.Thisisaconsequenceofthefollowingtheorem. Theorem4.3.1. IfFisarandomcontinuousfunction,thentheimageF[2N]hasnoisolatedelements.Proof.LetfbetherandomrepresentingfunctionforFandletQ=F[2N].SupposebywayofcontradictionthatQcontainsanisolatedpathy.Thenthereissomenite@ysuchthatyistheuniqueelementofIQ.Fixsuchthatf=.Foreachn,letSnbethesetofallg2Fsuchthatforall1;22f0;1gn, 1. g_1iscompatiblewithg_2, 2. @g_1,and 3. @g_2 99
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Thenforanyeachm3 4.Proof.Theproofisbyinductiononjj.Withoutlossofgenerality,weassumethat=0n.Foreachn>0,letqnbetheprobabilitythatF[2N]meetsIn.LetfbetherepresentingfunctionforF.Forn=1,thereare9equallyprobablechoicesforthepairfandf,breakingdowninto4distinctcases.Case1.Iff==f,thenF[2N]doesnotmeetI.Thisoccursjustonce.Case2.Iff=0orf=0,thenF[2N]meetsI.Thisoccursin5ofthe9choices.Case3.Iffi=;andf)]TJ/F24 11.955 Tf 12.561 0 Td[(i=,thenF[2N]meetsIifandonlyifFi[2N]meetsI.Thisoccursin2ofthe9choices,withprobabilityq1.Case4.Iff=;=f,thenF[2N]meetsIifatleastoneofFi[2N]meetsI.Thisoccursin1ofthechoices,withprobability1)]TJ/F15 11.955 Tf 12.005 0 Td[()]TJ/F24 11.955 Tf 12.005 0 Td[(q12.Thatis,F[2N]failstomeetIifbothF[2N]andF[2N]failtomeetI. 100
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Puttingthesecasestogether,weseethatq1=5 9+2 9q1+1 9q1)]TJ/F24 11.955 Tf 11.955 0 Td[(q21;sothatq1satisesthequadraticequationx2+5x)]TJ/F15 11.955 Tf 11.956 0 Td[(5=0:Thusq1istheuniquesolutionin[0,1]ofthisequation,thatis,q1=p 45)]TJ/F15 11.955 Tf 11.955 0 Td[(5 2;whichisindeed>:75.Nowletqn=qandletqn+1=p.Onceagainweconsiderthe9initialchoices,nowbreakingdowninto6distinctcases.Case1.Iff==f,thenF[2N]doesnotmeetIn+1.Thisoccursjustonce.Case2.Iff=0=f,thenF[2N]meetsIn+1ifandonlyifatleastoneofFandFmeetsIn.Thisoccursjustonce,andwithprobability1)]TJ/F15 11.955 Tf 11.955 0 Td[()]TJ/F24 11.955 Tf 11.955 0 Td[(q2=2q)]TJ/F24 11.955 Tf 11.955 0 Td[(q2.Case3.Iffi=andf)]TJ/F24 11.955 Tf 12.023 0 Td[(i=,thenF[2N]meetsIn+1ifandonlyifFi[2N]meetsIn.Thisoccursin2ofthe9choices,withprobabilityq.Case4.Iffi=;andf)]TJ/F24 11.955 Tf 12.213 0 Td[(i=1,thenF[2N]meetsIn+1ifandonlyifFi[2N]meetsIn+1.Thisoccursin2ofthe9choices,withprobabilityp.Case5.Iff=;=f,thenF[2N]meetsIn+1ifatleastoneofFi[2N]meetsIn+1.Thisoccursjustonce,withprobability1)]TJ/F15 11.955 Tf 11.956 0 Td[()]TJ/F24 11.955 Tf 11.955 0 Td[(p2.Case6.Iffi=;andf)]TJ/F24 11.955 Tf 12.128 0 Td[(i=,thenF[2N]meetsIn+1ifatleastoneofthefollowingtwothingshappens.EitherFi[2N]meetsIn+1,orF)]TJ/F25 7.97 Tf 6.586 0 Td[(i[2N]meetsIn.Thisoccursin2ofthe9choices,withprobability1)]TJ/F15 11.955 Tf 11.955 0 Td[()]TJ/F24 11.955 Tf 11.955 0 Td[(p)]TJ/F24 11.955 Tf 11.955 0 Td[(q. 101
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Puttingthesecasestogether,weseethatp=2 3p)]TJ/F15 11.955 Tf 13.151 8.088 Td[(1 9p2)]TJ/F15 11.955 Tf 13.15 8.088 Td[(2 9pq+2 3q)]TJ/F15 11.955 Tf 13.15 8.088 Td[(1 9q2;sothatp=qn+1satisestheequationp2+3p+2pq)]TJ/F15 11.955 Tf 11.955 0 Td[(6q+q2=0:Wenotethatforp=q,thesolutionsarep=q=0andp=q=3 4.Thisexplainsthevalue3 4inthestatementoftheorem.Nowassumebyinductionthatq>3 4.Supposebywayofcontradictionthatp63 4.Itfollowsthat9 16+9 4+3 2q)]TJ/F15 11.955 Tf 11.955 0 Td[(6q+q2>0:Simplifying,thisimpliesthat16q2)]TJ/F15 11.955 Tf 12.138 0 Td[(72q+45>0.Butthisfactorsintoq)]TJ/F15 11.955 Tf 12.137 0 Td[(3q)]TJ/F15 11.955 Tf 12.138 0 Td[(15andisonly>0wheneitherq63 4orq>15 4.Sincethelatterisimpossible,weobtainthedesiredcontradictionthatq63 4. Theorem4.3.3. Norandomcontinuousfunctionisinjective.Proof.LetpbetheprobabilitythatanarbitarycontinuousfunctionFisinjective.ItfollowsfromTheorem 4.3.2 thatthereisa9 16chancethatFhasazeroinIandalsoinI,sothatp67 16.Nowingeneral,ifFisinjective,thenitmustbeinjectivewhenrestrictedtoIandwhenrestrictedtoI.Itfollowsthatp6p2,whichhappensonlyforp=0andp=1,giventhat06p61.Sincep67 16,itfollowsthatp=0,asdesired.ThiscanbereformulatedasaMartinLoftestasfollows.FirstweobservethatFisinjectiveifandonlyif,theimagesofeachpairofdisjointintervalsIandIaredisjoint.LetD;=fF:F[I]F[I]=;:ThenD;isuniformlyc.e.sinceF[I]F[I]=;ifandonlyifthereexistsnsuchthatforallextensions0ofand0ofoflengthn,f0andf0areincompatible.NowletSm=fF:8;2f0;1gm6=!F2D;g: 102
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ItfollowsfromtheobservationabovethatFisinjectiveifandonlyifF2TmSm.TheargumentaboveshowsthatS167 16andthatSm+16Sm2andhenceSm67 16m:ItfollowsthatfSm:m2!gisaMartinLoftestandthereforenorandomcontinuousfunctionmaybelongtoeverySmandhencenorandomcontinuousfunctioncanbeinjective. 4.3.3NonsurjectiveImages,ininstancesInthissectionweshowthatrandomcontinuousfunctionsarenotnecessarilyonto. Denition4.3.4F,therestrictionofFtoI. ForanyfunctionFon2Nandany2f0;1g,denetherestrictionFofFtoIbyFx=F_x:Clearlyanysuchrestrictionofarandomcontinuousfunctionwillberandom,butmorecanbesaid.RecallvanLambalgen'stheorem,Theorem 3.2.12 Proposition4.3.5. FisarandomcontinuousfunctionifandonlyifthefunctionsFandFarerelativelyrandom.Proof.LetrrepresentF.SupposerstthatFisrandom.Itfollows,asinCorollary 3.3.12 ,thatFFisrandomandhenceFandFarerelativelyrandombyvanLambalgen'stheorem.NextsupposethatFandFarerelativelyrandomandletrirepresentFifori=0;1.Letdbeanymartingale,whichwethinkofasbettingonr.Thenfori=0;1,wecandeneamartingalediwithoracler1)]TJ/F25 7.97 Tf 6.587 0 Td[(iasfollows.Wewillgivethedenitionford0andleaved1forthereader.Given=r0;:::;r0p+q)]TJ/F15 11.955 Tf 12.254 0 Td[(2where06q<2p,user1tocompute=r;:::;rp+1+q)]TJ/F15 11.955 Tf 12.133 0 Td[(2andthendeneditobetinthesameproportionasd.Thatis,di_j=di=d_j=dforj<3.ThusforanynodeontheleftsideofthelabelledtreeforF,d0ismakingthesamebetonthenextlabelthatdwouldhavemade,andsimilarlyford1andtherightside. 103
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SincetheFiarerelativelyrandomfori=0;1,itfollowsthatdidoesnotsucceedandhencethereexistupperboundsBiforfdiridngn2N.Butitfollowsfromtheabovedenitionsofdithatforanyp,drd2p+1)]TJ/F15 11.955 Tf 11.955 0 Td[(2=d0r0d2p)]TJ/F15 11.955 Tf 11.955 0 Td[(1d1r1d2p)]TJ/F15 11.955 Tf 11.955 0 Td[(1:Thisisbecausethemartingaledalternatesusingd0andd1andtheresultcanbeviewedineachalternationasmultiplyingthecapitalbysomefactor.Theningeneral,for0
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Dene^d=d,andfor2Band~thecorrespondingstringofA,^d_x=8><>:d~_x d~^dx2A^dx2B)]TJ/F24 11.955 Tf 11.955 0 Td[(AThefunction^disclearlyconstructive,sincedis.Toshow^disamartingale,considerthesumXx2Bd_x=Xx2Ad~_x d~^d+Xx2B)]TJ/F25 7.97 Tf 6.587 0 Td[(A^d=^dXx2Ad~_x d~+^djB)]TJ/F24 11.955 Tf 11.955 0 Td[(Aj=^d[jAj+jB)]TJ/F24 11.955 Tf 11.955 0 Td[(Aj]:Itremainstoshowthat^dsucceedsonX.However,thatisclear,asonbitswhichareinXbutnoteX,^dkeepsitscapitalconstant,andonbitsfromeX,itactsexactlyasdwould.ThereforesincedsucceedsoneX,^dsucceedsonXandXisnonrandom. Itiseasytoseethat,foranyrandomcontinuousfunctionFandanycomputablerealx,Fxisnotcomputable.Thisfollowsfromournextresult. Theorem4.3.8. IfFisarandomcontinuousfunction,then,foranycomputablerealx,Fxisnotcomputable.Proof.SupposethatFisrandomandletxandybecomputablereals.Foreachn,letSn=fG:Gxdnydng:ThenS0;S1;:::isaneectivesequenceofc.e.opensetsinF,andaneasyinductionshowsthatSn=2=3n.ThisisaMartinLoftestanditfollowsthatF=2Snforsomen,sothatFx6=y. Wenowstrengthenthisresulttoshowthattheimageofacomputableelementisrandom. Theorem4.3.9. IfFisarandomcontinuousfunction,then,foranycomputablerealx,Fxisarandomreal. 105
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Proof.SupposethatFisrandomwithrepresentingfunctionfr,letxbeacomputablerealandlety=Fx.Denethecomputablefunctiongsothat,foreachn,gn=xdn:BytheVonMises{Church{WaldComputableSelectionTheorem,thesubsequencezn=rgnisrandominf0;1;2gN.Nowy=Fxmaybecomputedfromzbyremovingthe2's.ThusFxisrandombyProposition 4.3.7 WenotethatFouche[ 45 ]hasusedadierentapproachtorandomnessforcontinuousfunctionsconnectedwithBrownianmotion,rstpresentedbyAsarinandProkovskiy[ 5 ],andhasshownthat,underthisapproach,itisalsotruethatforanyrandomcontinuousfunctionF,Fxisnotcomputableforanycomputableinputx.ItfollowsthatarandomfunctionFcanneverbecomputablycontinuousandhencethegraphofFisnota01class.4.4RandomClosedSetsarisingfromrandomcontinuousfunctions4.4.1APositiveResult:InverseImagesof0!InthissectionweprovethatforanyrandomcontinuousfunctionF,thesetZF=fx:Fx=0gisarandomclosedset.ForanysubsetSofC,letZS=fF2F:ZF2Sg. Lemma4.4.1. ForanyopensetS,ZS6S.Proof.ItsucestoprovetheresultforintervalsS=I.WewillshowbyinductiononjjthatZI=1 4jj,whereasofcourseI=1 3jj.RecallfromCorollary 4.4.3 that02F[2N]withprobabilityexactly3 4.Forjj=1,therearetwodistinctcases.CaseISupposerstthat=i,wherei2f0;1g.ThenF2ZSifandonlyifFhasazeroinIiandhasnozeroinI)]TJ/F24 11.955 Tf 12.137 0 Td[(i.NowFhasazeroinIiiffi2f0;2gandiftherestrictedfunctionhasazero,whichgivesprobability2 33 4=1 2.ThusthecombinedprobabilitythatF2ZSis1 4.CaseIISupposenextthat=.ThenF2ZSifandonlyifFhaszeroesinbothIandI.ItfollowsfromtheargumentinCaseIthatZS=1 4. 106
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NoticethatZf;g=fF:Fhasnozeroesghaspositivemeasure1 4butf;g=0.Nowsupposejj=nandlet=_i;supposebyinductionthatZI6I.Interpretasthecodeforanitebinarytreeandlet2f0;1gbetheterminalnodeofthattreesuchthatiindicatesthebranchingof.Againtherearetwocases.CaseISupposerstthati2f0;1g.ThenF2ZIifandonlyifF2ZIandfurthermoreFhasazeroinI_iandhasnozeroinI_1)]TJ/F24 11.955 Tf 12.189 0 Td[(i.ItfollowsasabovethatZI=1 4ZI=1 4n+1.CaseIISupposenextthati=2.ThenF2ZIifandonlyifFhaszeroesinbothI_0andI_1.ItfollowsasabovethatZI=1 4ZI=1 4n+1.Anarbitraryopensetisadisjointunionofintervalsandthusthedesiredinequalitycanbeextendedtoopensets. Theorem4.4.2. ForanyrandomcontinuousfunctionG:2N!2N,thesetofzeroesofGiseitheremptyorisarandomclosedset.Proof.SupposethatGisarandomcontinuousfunctionwhichhasatleastonezero,andletS0;S1;:::beaMartinLoftestinC.ThenthereisacomputablefunctionsuchthatSi=[nIi;n.WemayassumewithoutlossofgeneralitySi62)]TJ/F25 7.97 Tf 6.587 0 Td[(i)]TJ/F23 7.97 Tf 6.587 0 Td[(2andthateachSiisnotclopenandthat,foreachi,theintervalsIi;narepairwisedisjoint.WewilldeneaMartinLoftestS00;S01;:::inthespaceFandusethefactthatGmustsatisfyfS0igi2!toshowthatZGsatisesfSigi2!.FixanintervalIinCandletC=ZI.ObservethatthereisaclopensetB2Nandacorrespondingniteset0;:::;k)]TJ/F23 7.97 Tf 6.587 0 Td[(1ofstringssuchthatB=[j
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iiFhasnozeroesoutsideofB.Let2N)]TJ/F24 11.955 Tf 12.06 0 Td[(B=[2AI.Bycompactness,FhasnozeroesoutsideofBifandonlyif9`82A80[j0j=`9mf0dm=1]:{1NotethatthemeasureofCmaybecomputeduniformlyfromgiventhecalculationfromCorollary 4.4.3 thatwheneverf2f0;2gjj,thentheprobabilitythatFhasazeroinIisexactly3 4.Foreach,wewilluniformlycomputeac.e.opensetSFsuchthatCBandsuchthatB62C.TherearetwostagesintheconstructionofB.StageI:LetUbethesetofcodes0forpartialfunctionsf0suchthat 4{1 holdswithf0inplaceoff,andsuchthatfurthermoreforeveryjand`suchthatf0isdenedonalllength`extensionsofj,thereissuchawithf02f0;2g8.ItisclearthatforanyF2C,thereexists02UwithF2I0andhenceC[fI0:02Ug:Asusual,wemaythenuniformlycomputefromUasetU0suchthattheintervalsI0for02U0arepairwisedisjointinFand[fI0:02Ug=[fI0:02U0g:Foreach02U0,letQ0Ibethe01classinFconsistingofthoseextensionsof0whichactuallyhavezeroesineachIj.TheninfactwehaveC=[fQ0:02U0g:Asnotedabove,wecanactuallycomputethemeasureQ0uniformlyfrom0byexpressingQ0asaneectivedecreasingintersectionofclopensets.Thusforeach0,wecancomputeaclopensetB0suchthatQ0B0I0and 108
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B062Q0.LetB=[fB0:02U0g:ThenwehaveCBandB6C.Finally,foreachi,letS0i=[nB0i;n:ThenbyProposition 4.3.7 ,S0i62Si62)]TJ/F25 7.97 Tf 6.587 0 Td[(i)]TJ/F23 7.97 Tf 6.586 0 Td[(1andthereforethereexistssomeisuchthatG=2S0i,sinceFisrandom.ButthismeansthatZG=2SiandhenceZFmeetstheMartinLoftest.ThusZFisrandom,asdesired. 4.4.2ANegativeResult:Images,ingeneralIngeneral,theimageofarandomcontinuousfunctionneednotbearandomclosedset.Toseethis,recallthestatementofTheorem 4.3.2 .Thatis,given2f0;1g,theprobabilitythattheimageofacontinuousfunctionFmeetsIisalways>3 4.Weobtainthefollowingcorollary. Corollary4.4.3. Foranyy22N, a fF:y2F[2N]g=3 4; b thereexistsarandomcontinuousfunctionFwithy2F[2N].Proof.aLetpbetheprobabilitythaty2F[2N].Itfollowsthatforeach2f0;1gn,theprobabilitythaty2F[I],giventhatfisconsistentwithy,alsoequalsp.ItfollowsfromtheproofofTheorem 4.3.2 thatp=3 4.bSincetherandomcontinuousfunctionshavemeasure1inCN,itfollowsthatsomerandomcontinuousfunctionhasyintheimage. Thisallowsustodemonstrateourresult. Theorem4.4.4. Theimageofarandomcontinuousfunctionneednotbearandomclosedset.Proof.ItwasshowninTheorem 3.4.12 thatarandomclosedsethasnocomputablemembers.LetFbearandomcontinuousfunctionwith0!intheimage,asgivenbyCorollary 4.4.3 .ThenF[2N]isnotarandomclosedset. 109
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4.5PseudoDistanceFunctionsInsection 4.4.1 weshowedthatifisarandomcontinousfunction,then)]TJ/F23 7.97 Tf 6.586 0 Td[(1!isarandomclosedset,ifitisnonempty.Thismotivatesthestudyofpseudodistancefunctions. Denition4.5.1. :2N!2NisapseudodistancefunctionforQ2Nifiscontinuousand)]TJ/F23 7.97 Tf 6.586 0 Td[(1!=Q. Comment4.5.2Background. ThenamecomesfromamodicationofthedistancefunctiondistQ:2N![0;1]foraclosedsetQ.Forx22N,distQxisdenedtobeminfdx;y:y2Qgwheredisametricon2Ngivenbydx;y=8>><>>:0ifx=y;2)]TJ/F25 7.97 Tf 6.587 0 Td[(nifnistheleastsuchthatxn6=yn:Thismaybeviewedasacomputablemappingfrom2N2Ninto2Nbyrepresenting0as0!and2)]TJ/F25 7.97 Tf 6.587 0 Td[(nas0n10!.Fromthisviewpoint,wemayviewdistQsimilarly:distQx=8>><>>:0!ifx2Q;0n10!otherwise,wherenistheleastsuchthatxdn=2TQ:Everyclosedsethasacharacterizationintermsofpseudodistancefunctions,asfollows. Theorem4.5.3. Q2NisclosedithereispseudodistancefunctionforQ.Proof.FirstsupposethatQisclosedandQ=[T].Deneamapb:2><>>:b_0if_i2T;b_1otherwise:Letting:2!!2!bedenedsothatXistheuniqueY2n[bXn],weobtaintherequiredpseudodistancefunction. 110
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Nowsupposethat:2!!2!isapseudodistancefunctionforQ.Thismeansthatthereissomeb:2
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4.7FutureWorkWeclosethischapternotingthatrandomBrownianmotionsasstudiedbyFouche[ 45 ]areaspecialcaseofrandomcontinuousfunctionsontherealline.Thisisanotherareaofinterestforfurtherresearch.Thatis,wewouldliketoextendthenotionofarandomcontinuousfunctiontofunctionsontherealunitinterval[0;1]andtherealline
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CHAPTER5CONTINUITYOFCAPPINGINCBTThefollowingchapterisjointworkwithAngshengLiandWeilinLiandwillappearintheAnnalsofPureandAppliedLogicasanarticleentitledContinuityofCappinginCbT[ 16 ].Forthisproject,P.BrodheadacknowledgessupportfromtheNationalScienceFoundationundergrantnumber0714151astheprincipalinvestigatortoconductthisjointworkinBeijingduringthesummerof2007aspartoftheEastAsiaandPacicSummerInstitutesEAPSI.ThisworkwaspresentedbyP.BrodheadattheFirstJointAMSNZMSMeetingDecember2007,Wellington,NewZealand.5.1IntroductionGivensetsA;B!,wesaythatAisTuringreducibletoB,ifthereisanoracleTuringmachinesay,suchthatA=BdenotedbyA6TB.Furthermore,ifthebitsoforaclequeriesareboundedbyacomputablefunction,thenusingrecentnomenclaturefromSoare[ 88 ]wesaythatAisboundedTuringreducibletoB,writtenA6bTB.Theliteratureoftenreferstothisastheweaktruthtablereducibility,written6wtt.ATuringandaboundedTuringorbT,forshortdegreeistheequivalenceclassofasetundertheTuringreductionsandtheboundedTuringreductionsrespectively.Adegreeiscalledcomputablyenumerablec.e.,ifitcontainsac.e.set.LetCandCbTbethestructuresofthec.e.degreesundertheTuringreductionsandtheboundedTuringreductionsrespectively.Duringthepastdecades,thestudiesofthestructuresC;CbTfocusedonthatofthealgebraicproperties,leadingtomajorachievementssuchasthedecidabilityresultsofthe1theoryofC,andthe2theoryofCbTAmbosSpies,P.Fejer,S.LemppandM.Lerman[ 3 ],andtheundecidabilityresultsofthe3theoryofCLempp,Nies,andSlaman[ 63 ],andofthe4theoryofCbTLemppandNies[ 62 ].Thisprogressbringsthedecidabilityproblemsofthe2theoryofC,andthe3theoryofCbTintosharperfocus,forwhichnewingredientsarewelcome. 113
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Intherecentyears,thestudyofthecomputablyenumerabledegreeshasfocusedonTuringdenabilityinthestructureC.Forinstance,Slamanaskedin1985ifthereareanyc.e.degreesthatareincompleteandnonzerowhicharedenableinthec.e.degreesC.ThisquestionofSlamanisstillopen.AnaturalapproachtothisproblemistondsomedenablesubstructuresofCthathavenontrivialminimal/maximaland/orleast/greatestmembers.Asaresult,topicssuchatthecontinuityofthec.e.degrees,startedbyLachlanin1967,haverenewedinterest.Inthischapter,wedemonstratethecontinuityofcappinginCbT.Thisrefutestheexistenceofamaximalnonboundingdegree.Italsobringsthequestionofthe3theoryofCbTintosharperfocus,asthestatementisoneof3complexity.Tomotivatetheseideasfurther,webeginwithabriefhistoryofrelevantcontinuityresultsinSection 5.2 .Thismotivatesourmainresultandmethodofproof,describedinSection 5.2.2 .ThemainsubstanceoftheproofinvolvesdemonstratingthatTheorem 5.2.3 holds,thatlocalnoncappabilityholdsinCbT.Sections 5.3 { 5.6 aredevotedtoprovingthistheorem.5.2ContinuityResultsTheTuringdegreesformanuppersemilattice;thatis,eachpairofelementsa;bhasaleastupperboundorjoina_b.Agreatestlowerbounda^bmayormaynotexist.GivenaboundedTuringdegreea,wesaythataiscappableifaboundedTuringdegreeb6=0existssuchthata^b=0.Wesaythataiscuppableifthereisadegreeb6=00suchthata_b=00.ThestudyofcontinuitypropertiestheboundedTuringdegreesiswithrespecttomeetsandjoins,andrelatednotionssuchascappingandcupping.5.2.1ContinuityResultsinCIn1979,Lachlan[ 60 ]provedtheexistenceofanonboundingc.e.degree{namely,anoncomputablec.e.degreewithnominimalpairbelowit;Cooperdemonstratedin1974,thatnohighc.e.degreecanbenonbounding[ 30 ].ReturningtotheSlamanquestion,Downey,Lempp,andShoredemonstratedin1993thatCooper'snonboundingdegree 114
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couldbemadehigh2[ 41 ],leadingtothepossibilityofamaximalnonboundingc.e.degree.Suchadegreecouldbeusedtoshowtheexistenceofadiscontinuity,whichcouldbeusedtoprovethedenibilityofac.e.singleton.Toseetheformer,notethatifb>0isnonbounding,thenforanyc.e.a>b,thereissomeminimalpairr^sbelowa.Theneitherb^r=0orb^s=0,butneithera^rnora^sequals0.However,Seetapunrefutedthispossibilityalbeitearlierin1991,demonstratingthenonexistenceofamaximalnonboundingdegree[ 84 ].Welchprovedacomplimentaryresultin1981:thereisnomaximalboundingdegree,inthesensethatforalla6=00,thereisareb;csuchthatb^c=0andb;c66a[ 95 ].Continuingwithcappingresults,HarringtonandSoare[ 51 ]provedin1989,thenonexistenceofmaximalminimalpairs{thatis,foranynontrivialminimalpaira;bofc.e.Turingdegreesa;b,thereexistsac.e.Turingdegreec>asuchthatc;bisstillaminimalpair.Seetapun[ 84 ]showedanevenstrongerresult,thecontinuityofcapping:foranyc.e.Turingdegreeb6=0;00,thereexistsac.e.Turingdegreea>bsuchthatforanyc.e.Turingdegreex,ifx6a,thena^x=0ifandonlyifb^x=0.AmbosSpies,Lachlan,andSoare[ 4 ]provedthedualcaseoftheHarringtonandSoare'sresult:foranynontrivialsplittingx;yof00,thereexistsac.e.degreea0,witha0beingtheuniquecomplementofa1intheinterval[0;a0_a1],suchthatifb
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anyxbsuchthatforanyc.e.bTdegreex,b^x=0$a^x=0.Forthis,itsucestoproveTheorem 5.2.3 below,ananalogoftheSeetapunlocalnoncappabilitytheoremforthec.e.Turingdegrees[ 84 ]. Denition5.2.2LocalNoncappability. Adegreeb6=0islocallynoncappableifthereissomea>bsuchthatforallxbsuchthatifx6aisnoncomputable,thenx^b6=0.ProofofTheorem 5.2.1 .AssumingTheorem 5.2.3 ,wecanseeTheorem 5.2.1 .Givenb,letabethedegreeinTheorem 5.2.3 .Foraxedx2CbT,byTheorem 5.2.3 ,weconsideronlythecasewherex66a.Clearlyifx^a=0,thenx^b=0.Assumea^x6=0.Wecanchooseac.e.bTdegreeysuchthaty6=0andy6a;x.Therefore0bbethedegreeinTheorem 5.2.1 .WeclaimthattherearenobTminimalpairsbelowa.Supposetothecontrarythatx;yisaminimalpairbelowa.Sincea^x=x6=0anda^y=y6=0,wecanchoosenonzerox1tobebelowbothxandb,andnonzeroy1belowbothyandb.Thenx1;y1isaminimalpairbelowb,acontradiction.Analternativeapproachistousethefactthatamaximalnonboundingc.e.degreeisequivalenttoanonboundingdegreewhichisnotlocallynoncappable[ 49 84 ].Consequently,byTheorem 5.2.3 ,nomaximalnonboundingdegreecanexist.Ourapproachtotheproofoftheorem 5.2.3 issimilartoSeetapun'sapproachforthec.e.Turingdegrees,butitisnonobviousduetothecomputableboundsoforaclequery 116
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bitsinboththeconditionsandconclusionsofrequirements.Thatis,bTreductionsarestrongerthanTuringreductions.SowhenwerequirethereductionsbeingbuilttobebTreductions,wemustsatisfystrongerconditionsand,inthissense,theproblembecomeshardertosolve.Forexample,A.Li,W.Li,Y.Pan,andL.Tang[ 66 ]haveshownthatthesolutiontothemajorsubdegreeprobleminCbTi.e.thedualtothecontinuityproblemiscompletelydierentfromtheresultforCseeCooperandLi[ 31 ].TheyshowthatthestatementofthesolutioninCfailsbadlyinCbT:thereexistc.e.bTdegreesa;bsuchthat0a.Ourapproachmightnotbetheonlyone.KlausAmbosSpiesprovedthatforanyc.e.set,itsTuringdegreeiscappableintheTuringdegreesiitsbTdegreeiscappableinthebTdegrees[ 2 ];wethankananonymousrefereeforpointingthisout.Therefore,anotherpossibleapproachmightbetoprove,ifpossible,thatforanytwoc.e.sets,theirTuringdegreesformaminimalpairintheTuringdegreesitheirbTdegreesformaminimalpairintheboundedTuringdegrees.Asconsequence,ourcontinuityresultwouldimmediatelyfollowfromSeetapun'scontinuityresult.Wecommentthatalthoughourcontinuityproofmightbenonobvious,fromtheaboveperspective,oftentimesbTdegreescanbehandledmuchmoreeasilythanTuringdegrees[ 1 ].AmbosSpiesprovidesvariousexamples[ 1 ].Forexample,densityofthec.e.bTdegreescanbeprovedbyaniteinjurypriorityargument,whereasthesameresultrequiresaninniteinjuryargumentforthec.e.Turingdegrees.TherestofthischapterisdevotedtoprovingTheorem 5.2.3 ,themainresult.Insection 5.3 ,weformulatetheconditionsofthetheorembyrequirements;insection 5.4 ,wearrangeallstrategiestosatisfytherequirementsonthenodesofatree,ormoreprecisely,theprioritytreeT.Insection 5.5 ,weusetheprioritytreetodescribeastagebystageconstructionoftheobjectsweneed.Finally,insection 5.6 weverifythattheconstructioninsection 5.5 satisesalloftherequirements,nishingtheproofofthetheorem.OurnotationandterminologyarestandardandgenerallyfollowSoare[ 86 ].Duringthecourseofaconstruction,notationssuchasA;areusedtodenotethecurrentapproximationstotheseobjects,andifwewanttospecifythevaluesimmediatelyat 117
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theendofstages,thenwedenotethembyAs,[s]etc.Forapartialcomputablep.c.,orforsimplicity,alsoaTuringfunctional,say,theusefunctionisdenotedbythecorrespondinglowercaseletter.Thevalueoftheusefunctionofaconvergingcomputationisthegreatestnumberwhichisactuallyusedinthecomputation.ForaTuringfunctional,ifacomputationisnotdened,thenwedeneitsusefunction=)]TJ/F15 11.955 Tf 9.298 0 Td[(1.Duringthecourseofaconstruction,wheneverwedeneaparameter,psay,asfresh,wemeanthatpisdenedtobetheleastnaturalnumberwhichisgreaterthananynumbermentionedsofar.Inparticular,ifpisdenedafreshatstages,thenp>s.5.3RequirementsandStrategiesInthissectionweprovidetherequirementsandstrategiesforprovingTheorem 5.2.3 .Werestateithereforconvenience.Theorem 5.2.3 LocalNoncappabilityinCbT.Foranyc.e.bTdegreeb,ifb6=0;00,thenthereisac.e.bTdegreea>bsuchthatifx6aisnoncomputable,thenx^b6=0.5.3.1TherequirementsGivenac.e.setB,wewillbuildac.e.setAtosatisfythefollowingrequirements: Pe: A6=eB_K6bTB Re: Xe=eA;B)167(!9c.e.De[De6bTXe;B&8iSe;i] Se;i: De6=i_Xe6T;_B6T;wheree;i2!,fe;e;Xe:e2!gisaneectiveenumerationofalltriples;;XofallboundedTuringbT,forshortreductions;,andofallc.e.setsX;fiji2!gisaneectiveenumerationofallpartialcomputationfunctions;andKisaxedcreativeset.Deforalle,arec.e.setsbuiltbyus.Leta;b;x;dbethebTdegreesofAB,B,X,D,respectively.BythePrequirements,a>bunlessbwasalreadythedegreeof00,andbytheRrequirements,ifx6athereisadbelowbothxandbsuchthatd6=0unlesseitherx=0orb=0.Thereforetherequirementsaresucienttoprovethetheorem. 118
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Beforedescribingthestrategies,weintroducesomeconventionsoftheboundedTuringreductions.WewillassumethatforanygivenboundedTuringreductionor,theusefunctionsandwillbeincreasinginarguments.5.3.2APstrategyAPstrategywilltrytosatisfyaPrequirement,Psaywedroptheindexinthefollowingdiscussion.Weuseanodeonatree,say,todenoteaPstrategy.ItaimstoensurethatifA=B,thenthereisaboundedTuringreductionsuchthatB=K.ThereforethePstrategywilltrytobuildaboundedTuringreduction.willbebuiltbyan!sequenceofcyclesk.EachcyclekofwillberesponsiblefordeningB;kasfollows:rstchoosesafreshwitnessakandwaitsforastage,vsay,atwhichwehaveB;ak#=0=Aak.Whenthisoccurs,wedeneB;ktobeKkwithusefunctionk=ak.Sinceispartialcomputable,soistheusefunctionof.WewillalwaysassumethatwheneverBchangesbelowtheuse,ksay,thecorrespondingcomputationB;kbecomesundenedsimultaneously.Supposethatatalaterstages>v,kisenumeratedintoK,andBhasnotchangedsinceB;kwaslastcreated,thenB;k6=Kk.Inthiscase,weenumerateakintoAsothataninequalityB;ak6=Aakiscreated.Thekeypointisthat,ifB;k6=Kkisapermanentinequality,soisB;ak=06=1=Aak.ThePstrategywillstartcycleskinincreasingorderofk.Cyclekactsonlyifthefollowingconditionsoccur: 1. Forallk0
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5.3.3AnRstrategyBeforedescribingtheRstrategy,weintroduceaconventionoftheboundedTuringreduction.Weassumethatforanyxandanys,ifxentersXatstages,thenA;B;x[s]#=1.GivenanRrequirement,Rsay,wedenethelengthfunctionofagreementasusual.Thatistosay:Atstages,thelengthfunctionofagreement`betweenA;BandXisdenedasthelargestxsuchthatA;BandXagreeonallvaluesy`[v].AtRexpansionarystages,anRstrategybuildsboundedTuringreductionsX;Bwithusefunctions,,respectivelysothatfortheleastundenedx<`:X;x#=Dxwithx=xandB;x#=Dxwithx=x{1whereistheuseofA;B,andDisac.e.set,whoseelementsareenumeratedintoitbylowerprioritySstrategiesassociatedwithR,tosatisfytheRrequirement.SupposethatisanRstrategy.Asabove,theusefunctionsandoftheboundedTuringreductionsandbuiltbyaretheidentityfunctionandrespectively,sothatbothandareboundedTuringreductions.NoticethatA;BisaboundedTuringreduction,sothattheuseisapartialcomputablefunction.TosatisfyX=DandB=D,theRstrategywillimposethefollowingconstraintsonallSstrategieswiththesameglobalindexastheRstrategy:Foranys,andanyd,disallowedtobeenumeratedintoDatstagesonlyifbothX;dandB;dareundenedduringstages.WeassumethatfortheboundedTuringreductionsand,anycomputationwillautomaticallybecomeundened,whenevertheoraclechangesbelowthecorrespondinguse. 120
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Bythebuildingofand,andbytheconstraintsof,wehavethatifandarebuiltinnitelymanytimes,thenbothXandBaretotal,andbothequalD.HenceRissatised.ThereforethekeypointtowardsthesatisfactionofRisthatifthereareinnitelymanyRexpansionarystages,thenbothandarebuiltinnitelymanytimes.WethusdenethepossibleoutcomesoftheRstrategyby0
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setftobetotallyundenedthefproveswrong,soitiscancelled, droptheArestraintbydeningrA=)]TJ/F15 11.955 Tf 9.299 0 Td[(1,and createalink;.[Noticethatatstagev,B;disundenedduetotheBchangeinthedomainoff.WeregardthisasaBpermissionfortheenumerationofdintoD.ThisBpermissionwillbekeptuntilthecurrentlink;iseithertravelledorcancelledsothat,ineithercase,thelinkisremoved.]Supposethatcreatesalink;atstagev.Thenthelink;willbetravelledatthenextexpansionarystages>v.Nowweconsidertwocases:Case1.ThereisanerrorbetweengandX.Inthiscase,thereisanx6dwhichhasenteredXsincestagev.ThereforeX;discurrentlyundened.TogetherwiththeconditionthatB;d",foundatthestagewecreatedthecurrentlink;,isqualiedtoenumeratedintoD.Sissatisedbyd=06=1=Dd.Case2.Otherwise,weknowthatgiscorrectduringthegap.ThereforewepreservegonitsdomainuntilopensanotherAgap.Forthis,weimplement: foreveryy6d,iffyisundened,thendenefy=By,and denetheArestraintrAoftobed.[NoticethatalthoughwehaveArestraintatthisstage,XmaychangeduetoaBchangebelowd.ThedenitionoffatthisstageallowsustoimmediatelyopenanAgaponcesuchaBchangeoccurs,inwhichcase,wedonotincreasethedomainofgbutresumewiththeoldcandidated.]ThereforetheSstrategyisagap/cogapstrategy.Itwillbuildpartialcomputablefunctionsfandgandwillproceedasfollows: 1. Deneapossiblecandidatecasfresh. 2. BuildingfWaitforastagevatwhich c#=0=Dc, X;c#=0=Dc,and 122
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B;c#=0=Dc:Thenforc=c, foreveryy6c,iffy",thendenefy=By, dened=c, setctobeundened,andgobacktostep1. 3. Creatingalink;Letsbethecurrentstage.SupposethatthereisabinthedomainoffthatentersBatstages.Noticethatthedomainoffispreciselyeverything6d=d,sothatB;dbecomesundenedatstages.Then: foreveryx6d=d,ifgxisundened,deneittobeXx, denetheArestraintofbyrA=)]TJ/F15 11.955 Tf 9.299 0 Td[(1, setftobetotallyundened,and createalink;. 4. Travellingthelink;Wetravelthelink;atthenextexpansionarystaget>s.Therearetwocases:Case4a.SuccessfulclosureThereisanxsuchthatgx#=06=1=Xx.ThisxmustenterXsincethecurrentlink;wascreated.Then: enumeratedintoD,andstop.Case4bUnsuccessfulClosureOtherwise,then foreveryy6d,iffyisundened,thendenefy=By, deneanArestraintofbyrA=d.ThePossibleOutcomesWeconsiderthefollowingcases.Case1.Case4aoccursatsomestaget.Inthiscase,limsd[s]#=d
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Noticethatgisneversettobetotallyundened,andthatforaxednumberd,disaxednumber,sothatBchangesbelowdonlynitelymanytimes,andsothatStep3occurswiththesamedonlynitelymanytimes.SinceCase4boccursinnitelymanytimes,wehavethatd[s]willbeunboundedoverthecourseoftheconstruction,andthatwheneverStep3occurs,webuildgontheinitialsegmentofthecurrentd.Thereforegisbuiltasacomputablefunction.Foranarbitrarilygivenx,weprovegx#=Xx.Letsbethestageatwhichgxiscreated.Supposethatsiareallstagess0>satwhichStep3ofoccurs,andthatforeachsi,ti2si;si+1isthestageatwhichthelink;createdatstagesiistravelledthroughCase4b.Bythechoiceofsi,s0=s.SinceCase4boccursatstaget0,andt0isexpansionary,wehavethat i gx=X[s0]xwillneverbevisitedatstages0. ii Foranys2[s0;t0],gx=X[s]x. iii gx=A;B;x[t0]=X[t0]x.BytheArestraintrA[t0],andtheconventionof,wehavethatforanyt2[t0;s1, iv gx=A;B;x[t]=X[t]x.Supposebyinductionthatforn,wehavethat A Foranys2[sn;tn],gx=X[s]x. B gx=Xx[tn]=A;B;x[tn], C Foranyt2[tn;sn+1,gx=A;B;x[t]=X[t]x,and D gx=X[sn+1]x.ByC,Dforn,andbythechoiceoftn+1,Aholdsforn+1.ByAforn+1,andthechoiceoftn+1,wehaveBforn+1.ByBforn+1,bytheArestraintatstagetn+1,andbytheconventionof,Choldsforn+1.Dforn+1followsfromCforn+1andtheassumptionthatisnotvisitedatstagesn+1.[Remark.WewillarrangetheconstructionsothatanAgapcanbeopenedonlyatoddstages,andthatnoRstrategiescanbevisitedatthesestages.Thismeansthatno 124
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XforanyRstrategycanreceiveelementsatoddstages,sinceweassumethatXisenumeratedonlyatstagesatwhichisvisited.]Thereforeforanys>s0,eithergx=Xx[s]orgx=x[s].SinceA;B;xequalsXx,wehavethatgx=Xx.ThisisaglobalwinfortherequirementR,sothatwedon'tconsideranyotherSrequirementswiththesameglobalindexwiththerequirementR.Case3.Otherwise,andStep2occursinnitelymanytimes.Sincestep3occursonlynitelymanytimes,fissettobetotallyundenedonlynitelymanytimes.Letfbethenalversionoff.Bytheassumptionthatstep2occursinnitelymanytimes,fisbuiltasacomputablefunction.Bythechoiceoff,foranyx,oncefxiscreated,wehavethatfx=Bx.Biscomputable,contradictingtheassumptionofthetheorem.Soweassumethatthiscasewillneveroccur.Case4.Otherwise,thenbythestrategy,wehavethatlimsc[s]#=c
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Se;iwithi>e.LetP
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Proposition5.4.5. LetfbeaninnitepaththroughT.ThenforanyrequirementX,thereisanode0fsuchthateitherioriibelowholds, i Xissatisedatforanywith0f. ii Xisactiveatforanywith0f.Proof.LetfXi:i2!gbethepriorityrankingoftherequirementssothatXi
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Denition5.4.6. IfisanSe;istrategyforsomee;i,thendenetoptobethelongestRestrategysuchthat^h0i.5.5TheConstructionOurconstructionwillperformdierentactionsatevenandoddstages.Atevenstages,strategiesonthetreewillacttosatisfytherequirements.SupposethatBisenumeratedatoddstagesonly,andthatateveryoddstage,thereisexactlyoneelementthatentersB.GivenaBpermissionintheconstruction,wewanttoopenAgapsforasmanySstrategiesaspossible.ThisallowsustospecifyaPstrategysothatweenumerateitswitnessintoA.SowewillensurethatArestraintsdropatoddstages,andalsothatAisonlyenumeratedatoddstages.Duringthecourseoftheconstruction,wemayinitializeanode,say,whichmeansthatalltheactionstakenbypreviously,arecancelled,orsettobetotallyundened.Precisely,ifanRstrategyisinitialized,thenbothandaresettobetotallyundened,Dissettobetheemptyset;,andalllinksassociatedwitharecancelled.IfanSstrategyisinitialized,thenbothgandfaresettobetotallyundened,parametersdandcarebothsettobeundened,andanylinkassociatedwithiscancelled.IfaPstrategyisinitialized,thenissettobetotallyundened,andallwitnessesofarecancelled.NoticethatanSstrategyopensitsAgap,exactlyatstagesatwhichanerrorbetweenfandBoccurs,whichgivesaBpermissionforitscurrentcandidated.OurproblemistomakesurethatthereareinnitelymanystagesatwhichalltheSstrategiesonthetruepathorthecurrentapproximationofthetruepathdroptheirArestraintssimultaneously.Givenanode,supposethat12nareallSstrategieswith^hgi.ToguaranteethatifisaPstrategy,thenthereareinnitelymanystagesatwhichforalli=1;2;;n,theArestraintsrAiofidropto)]TJ/F15 11.955 Tf 9.299 0 Td[(1innitelyoften,proceedasfollows.Letfibefiforalli=1;2;;n.Wewillarrangethebuildingofvariousfisuchthat:foranys, 128
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1. foranyi,iffi[s]isempty,thenthecurrentArestraintrAiis)]TJ/F15 11.955 Tf 9.299 0 Td[(1,and 2. foranyi
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letkbetheleastxsuchthatB;x#6=Kxandax62A, enumerateakintoA,and initializeallnodeswith>,andgotostages+1. 5. Otherwise,thengotostages+1.Stages=2n+2.Werstspecifytherootnodetobeeligibletoactatsubstaget=0.Ateachsubstaget,weallowthestrategywhichiseligibletoactatthissubstagetotakeaction,andtheneitherclosethecurrentstageorspecifyanewnodetobeeligibletoactatthenextsubstageofstages.Substaget.Letbethenodewhichiseligibletoactatsubstagetofstages.Ifhaslengths,theninitializeallnodes066,andclosethecurrentstage.Otherwise,therearethreecasescorrespondingtodierenttypesofstrategy.Case1.=isaPstrategy.Thenrunthefollowing:Program:willbuildaboundedTuringreduction,anddenewitnessesak.Forsimplicity,wedropthesubscriptioninthedescriptionoftheprogram. 1. Ifthereisannsuchthatanisdened,andlB;A6>an,thenlet^h1ibeeligibletoactnexti.e.atsubstaget+1ofstages. 2. Otherwise,letkbetheleastxsuchthatB;xisundened.Then: ifak#,thendeneB;k=Kkwithk=ak, otherwise,thendeneaktobefresh,and initializeallnodes>,andgotostages+1.Case2.=isanRestrategyforsomee.RunthefollowingProgram: 1. Ifsisnotexpansionary,let^h1ibeeligibletoactnext. 2. Otherwise,andthereisalink;whichwascreatedandhasneverbeencancelledortravelled.Let0bethe<leastsuch,andlet0beeligibletoactatthenextsubstage. 3. Otherwise,then, letxbetheleastysuchthateitherX;yorB;yisundened, 130
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ifX;x",deneX;x=Dxwithx=x, ifB;x",thendeneB;x[s]:=Dx[s]withusex:=x,whereistheuseofA;B,and let^h0ibeeligibletoactnext.Case3.=isanSe;istrategyforsomee;i.Let=top.Weperformthefollowing,Program: 1. Ifhasalreadybeensatised,asdenedin2abelow,then^hdiiseligibletoact. 2. Travelalink;Ifalink;wascreatedandithasneverbeencancelledortravelledsinceitwascreated,thentravelthelink;bycases.Case2a.Successfulclosure9xgx#6=Xx.Noticethatgwascorrectatthestagewecreatedthecurrentlink;,sothiserrormustoccurduringtheAgapoftheSstrategy.Then: enumerated,thelargestconrmedcandidateof,intoD, wesaythatissatisedatstages, initializeallnodes>,andgotostages+1.Case2b.UnsuccessfulclosureOtherwise.Then: deneu=maxfy:gy#g=d, setfd+1=Bd+1, setrA=u+1,and initializeallnodestotherightof^hgi,andgotostages+1.Ineithercase,thelink;isremoved.[Remark.Wehavethatifstep2ofprogramoccursatstages,thenisvisitedatstages.] 3. BuildingfIfc#=c,c#=Dc=0,X;c#,andB;c#then:Case3a.900^hgianddomf0[0;c].Then: initializeallnodes>^hwi,andgotostages+1.Case3b.Otherwise,then: 131
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foranyx6c,iffx",thendenefx#=Bx, setd=c;dissaidtobeconrmed, cancelc,sothatc",and let^hgibeeligibletoactnext. 4. Ifc",thendenecasfresh,initializenodes>^hwi,andgotostages+1. 5. Otherwise,let^hwibeeligibletoactatthenextsubstage.Thiscompletesthedescriptionoftheconstruction.5.6TheVericationInthissection,weverifythesatisfactionoftherequirements.Firstweinvestigatesomeglobalpropertiesthatholdattheendofanarbitrarystage.Thesepropertiesensurethattheconstructionisimplementedproperly. Proposition5.6.1. Letsbeastage. i Thereisatmostonelinkthatistravelledduringstages. ii Ifalink;istravelledatstages,thenbeforewetravelthelink,isvisitedandstep2ofprogramoccursatstages. iii Thereareno1;2;1;and2suchthat1212andbothlinks1;1and2;2existattheendofstages.Proof.Itiseasytoseethatbothiandiiholdbyobservingtheconstruction.Foriii,supposetothecontrarythatsistheleaststagesuchthatthereare1;2;1;and2with1212,andsuchthatbothlinks1;1and2;2existattheendofstages.Bytheminimalityofs,exactlyoneofthetwolinks1;1,and2;2iscreatedduringstages.Weconsidertwocases.Case1.Thelink1;1iscreatedatstages.Bytheconstruction,s=2n+1forsomen,andf1issettobetotallyundenedatstages.Weanalyzethelocationof2.If1^hwi2,thenbytheconstructionatstages,2isinitializedduringstages,sothelink2;2isremovedduringstages,contradictingthechoiceof2.If1^hgi2,thenbythedenitionoftheprioritytreeT,top26=2,sothatthereisnolink2;2whichcanbecreatedintheconstruction.If1^hdi2, 132
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thenthecurrentd2mustbedenedafter1createditsinequalityatargumentd1,afterwhichnolink1;1canbecreated,since1hassatiseditsrequirementthrough1d1=06=1=Dd1.Socase1doesnothappen.Case2.Thelink2;2iscreatedatstages.Lets11^hwiwereinitializedatstages1.Therefore2cannotbevisitedatanystage>s1unlessthelink1;1[s1]hasbeenremoved.Thiscontradictsthechoiceofs.iiiholds.ThePropositionfollows. Proposition5.6.2. i LetbeanSstrategy.Thenforanys;t,iffistotallyundenedatsubstagetofstages,thenrA=)]TJ/F15 11.955 Tf 9.298 0 Td[(1holdsattheendofsubstagetofstages. ii LetbeanSstrategy,andsbeastage.Lets)]TJ/F15 11.955 Tf 10.986 4.338 Td[(bethegreateststaget
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Letb=domf[s],andlets0betherststageatwhichfwasdenedon[0;b].Byprogram,case3bofprogramoccurredatstages0.Bytheconstruction,therewasnolinkthatwastravelledbythesubstageatwhichwasvisitedduringstages0.Therefore,0wasvisitedatstages0,andcase3bofprogram0occurredatstages0.Bytheassumptionincase3bofprogram,wehavethatbothf0andfarenonemptyattheendofstages0,anddomf0[s0]domf[s0].Bythechoiceofs,0hasnotbeeninitializedduringstages[s0;s],byiiiff0[s]isnotempty,thendomf0[s]domf0[s0],iiifollowsincase1.Case2.Case3bofprogramoccursatstages.Asthesameasthatincase1,byobservingtheconstruction,wehavethatforany,isvisitedatstages,sothat0isvisitedatstages,andfurthermore,case3bofprogram0occursatstages.Bytheassumptionofcase3bofprogram,thedomainoff0islargerthanthatoffattheendofstages.iiifollowsincase2.Thepropositionfollows. Denition5.6.3. SupposethatK66bTBandB66T;. i Letsbethelastnodewhichiseligibletoactatstages. ii DenethetruepathTP2[T]oftheconstructionbyTP=liminfss.Hereafterwheneverweconsideranodeonthetruepath,wewillusethenotation2TPratherthanTP. Proposition5.6.4. ExistenceofthetruepathSuppose2TP.Thenthereissomeasuchthat^haiisvisitedinnitelyoftenandinitializedonlynitelymanytimes.Hence^hai2TP.Proof.Weprovebyinductiononthelengthof.Supposebyinductionthatthepropositionholdsforall0and2TP.Lets0beminimalafterwhichwillneverbeinitialized.Bytheinductivehypothesis,willbevisitedinnitelyoften.Weprovethepropositionforbycases.Case1.=isaPestrategyforsomee.Byprogram,therearetwosubcasestoconsider. 134
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Subcase1a.Thereareinnitelymanyexpansionarystages.Bytheconstruction,step2ofprogramoccursinnitelymanytimes,sothatisbuiltinnitelymanytimes,andthatBisbuiltasatotalfunction.SinceK66bTB,thereissomemsuchthatB;m#6=Kmholdspermanently.Letnbetheleastsuchm,andletB;nbecreatedatstagev>s0.Bythechoiceofn,B;an[v]=0anditwillholdpermanently.Letu>vbethestageatwhichnentersK.Supposethat12lareallSstrategieswith^hgi.Bythechoiceofs0,fjwillneverbesettobetotallyundenedafterstages0byinitializationforanyj.Therefore,foreveryj2f1;2;;lg,fjissettobetotallyundenedafterstages0onlyifanerroroccursbetweenfjandB.Byinductivehypothesis,case3bofprogramjoccursinnitelymanytimes,sothatfjwillbebuiltinnitelymanytimesforallj2f1;2;;lg.Inparticular,flwillbebuiltinnitelymanytimes.BytheassumptionofB66T;,wecanchooseastages1>uatwhichthereisanumberbsuchthatflb#=06=1=Bboccurs.ByProposition 5.6.2 iii,foranyj2f1;2;;lg,iffjisnotemptyatthebeginningofstages1,thenthereisanerrorbetweenfjandBthatoccursexactlyatstages1.Wehavethatforeveryj2f1;2;;lg,iffj6=;atthebeginningofstages1,thenjopensitsAgapduringstages1.ByProposition 5.6.2 i,foranyj2f1;2;;lg,iffjisemptyatthebeginningofstages1,rAj=)]TJ/F15 11.955 Tf 9.298 0 Td[(1holdsatboththebeginningandtheendofstages1.Byprogram,wehavethatrequiresattentionatstages1,andweletreceiveattentionbyenumeratingitswitnessanintoA.Bythechoiceofn,foranys>s1,wehavethatB;an=06=1=Aanholdsduringstages,contrarytotherebeinginnitelymanySexpansionarystages.Thiscaseisimpossible.Subcase1b.Otherwise. 135
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Inthiscaseisbuiltonlynitelymanytimes.Lets1>s0beminimalafterwhichwillneverbebuilt.Bythechoiceofs1,^h1iwillneverbeinitializedafterstages1,andbyprogram,foranys>s1,ifisvisitedatstages,sois^h1i.Thepropositionfollowsincase1.Case2.=isanRstrategy.Observingprogram,weconsidertwosubcases.Subcase2a.Step3ofprogramoccursinnitelymanytimes.Then^h0i2TP.Bychoiceofs0,^h0iwillneverbeinitializedafterstages0.Furthermore^h0iisvisitedinnitelyoften.Therefore^h0i2TPandthepropositionholdsinthiscase.Subcase2b.Otherwise.SupposethatStep3ofprogramoccursatmostnitelymanytimessothat^h0iisvisitedatmostanitenumberoftimes.Wewillshowthat^h1i2TP.Toshowthat^h1iisinitializednitelyoften,rstnotethatbythechoiceofs0,onlynodes^h0icaninitialize^h1i.Lets1>s0beminimalafterwhichstep3ofprogramwillneveroccur.Thenforanys>s1,ifanode0^h0iisvisitedatstages,thenthereisanRstrategy0,andalink0;0whichistravelledatstages.Supposethat12n)]TJ/F23 7.97 Tf 6.587 0 Td[(1areallRstrategies0with0^h0i.Letn=.Weprovebyinductionthatforeachi6n,thereisastageviafterwhichtherewillbenolinksj;jthatcanbeeithercreatedortravelledforallj>iandforallj^h0i.Fori=n.Denebn=maxfnx[t]jt6s1;nx[t]#gForeverys>s1,dene 136
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pn[s]=maxfyjfy[s]#;top=ngBytheconstruction,wehavethatforeverys>s1, ifalinki;0istravelledforsomeis1,pn[s]6bn.Bytheconstructionatoddstages,alinkn;ncanbecreatedatastages>s1onlyifthereisanelementb6bnthatentersBatstages.Sincebnisaxednumber,Bchangesbelowbnonlynitelymanytimes.Thereforethereareonlynitelymanystagesatwhichwecreatelinksn;n.Sinceoncealinkistravelled,itisremovedimmediately,thereareonlynitelymanystagesatwhichalinkn;iseithercreatedortravelled.Letvn>s1beminimalafterwhichtherewillbenolinkn;nthatcanbeeithercreatedortravelled.Supposebyinductionthatvi+1isaminimalstageafterwhichtherewillbenolinkj;jwhichiseithercreatedortravelledforallj2fi+1;i+2;;ng,andallj^h0i.Denebi=maxfix[t]jix#;t6vi+1gForanys>vi+1,denepi[s]=maxfyjfy[s]#;^h0i;top=igBytheconstruction,itiseasytoseefromaninductiveargumentthatforalls>vi+1, ifalinkj;jistravelledforsomej
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ifalinki;iforsomei^h0iistravelledatstages,thenpi[s]6bi.Thisshowsthatforalls>vi+1,pi[s]6bi.Bytheconstruction,ifalinki;iforsomei^h0iiscreatedatastages>vi+1,thenthereisanumberb6biwhichentersBatstages.Sincebiisaxednumber,thecreationoflinksi;ifori^h0ioccursonlynitelymanytimes,sothatthereisastagevi>vi+1say,afterwhichtherewillbenolinkoftheformi;iforanyi^h0iwhichcanbeeithercreatedortravelled.Thereforethereisastagev1sayafterwhichnolinki;canbecreatedortravelledforanyi6nandany^h0i.Sothereareonlynitelymanystagesatwhichsomenode^h0iisvisited.Thus^h1iisinitializedonlynitelymanytimes.Bytheproofabove,thereareonlynitelymanystagesatwhicheitherStep2orStep3ofprogramoccurs.^h1iisvisitedatalmosteverystageatwhichisvisited.ThereforeinSubcase2b,wehavethat^h1iisinitializedonlynitelymanytimesandvisitedinnitelyoften.Case3.=isanSstrategy.Let=top.Subcase3a.alink;istravelledonceandsuccessfullyclosed.Then^hdiisvisitedinnitelyoftenandonlyinitializednitelymanytimes.So^hdi2TP.Subcase3b.;istravelledinnitelyoftenandunsuccessfullyclosed'sAgap.AsinSubcase3a,^hgi2TP.Subcase3c.Otherwise.Bytheassumptionofthiscase,fisbuiltonlynitelymanytimes,sinceifthisisnottrue,thenfisbuiltasacomputablefunction,andf=B,contradictingthehypothesisB66T;.Byprogram,limsc[s]#=cs1,ifisvisitedatstages,sois^hwi. 138
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Therefore^hwiisinitializedonlynitelymanytimes,andvisitedinnitelyoften,^hwi2TP.ThepropositionfollowsinCase3. SincethetruepathexistsonlyifbothK66bTB,andB66T;holdasprovedinProposition 5.6.4 ,wealwaysassumethetwoconditionsfromnowon. Proposition5.6.5PossibleoutcomesalongTP. Given2TP: i If=isaPestrategyforsomee,then^h1i2TPandA6=eB. ii If=isanRstrategy,then a if^h0i2TP,thenD=X=B; b if^h1i2TP,thenispartialorA;B6=X. iii If=isanSstrategy,thenfor=top,wehave: a if^hdi2TP,thenlimsd[s]#=ds0bethestageatwhichthecomputationB;nwascreated.NoticethateB;an[v]=0andBwillneverchangebeloweanafterstagev.Bytheproofincase1ofProposition 5.6.4 ,thereisastages1atwhichwecanenumerateanintoA.Bythechoiceofnands1,eB;an=06=1=Aanwillbepreservedforever.Acontradiction.ThereforewehavethatA6=eB.Forii.Let^h0i2TP. 139
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ByProposition 5.6.3 ,bothandarebuiltinnitelymanytimes.ByStep3ofprogram,itsucestoprovethatthefollowingconstraintsofaresatisedduringthecourseoftheconstruction:Foranyd,andanys>s0,ifdisenumeratedintoDatstages,thenbothX;dandB;dareundenedduringstages.ByProposition 5.6.1 i,thereisatmostonelinkwhichistravelledduringstages.LetdbeenumeratedintoDatstages.ThenthereisanSstrategysuchthattop=andcase2aofprogramoccursatstages,andd=d[s].ByProposition 5.6.1 ii,isvisitedatstages,andthereisalink;whichwascreatedatastages)]TJ/F15 11.955 Tf 7.085 4.338 Td[(>s0s0afterwhichstep3ofprogramwillneveroccur.Howeverthereareinnitelymanystagesatwhichwetravelalink;forsome.Bythechoiceofs1,isaniteset,lets2bethestage>s1afterwhichBwillneverchangebelowmaxfx[s1]jB;x[s1]#g. 140
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BytheSstrategies,foranywithtop=,iffiscreatedafterstages2,thentherewillbenolink;whichcanbecreatedbyusingthedierencebetweenBandf.Thereforewecanchooseastages3>s2afterwhichthereisnolink;whichcanbecreatedforany.Bytheconstruction,oncewetravelalink;,itisremoved.Thereisastages4>s3afterwhichthereisnolinkfromtoanySstrategywhichcanbeeithercreatedortravelled.Thiscontradictstheassumptionthatstep2ofprogramoccursinnitelymanytimes.WehavethatXe6=eA;B.Foriiia.If^hdi2TPthenthereissomestageswhereCase2aofprogramenumeratesd=limsd[s]intoD.Thenatallstagest>s,programenactsCase1.Furthermoresincedisonlydenedwhend=0;wehavethatd=06=1=Dd.Foriiib.Bythechoiceofs0,andbytheassumptionofthiscase,^hgiisneverinitializedafterstages0.Byprogram,case3bofprogramoccursinnitelymanytimes.Bythechoiceofs0,fbecomestotallyundenedatanystages>s0onlyifalink;iscreatedatstages,andthislinkwillcertainlybetravelledunsuccessfully,insteadofbeinginitialized.Foraxednumberd,disanednumber,sothatBchangesbelowdonlynitelymanytimes.Thereford[s]willbeunboundedintheconstruction.Bytheconstructionatoddstages,ifalink;iscreatedatstages,thend[s]isdened,andgisbuiltontheinitialsegmentd[s].Thereforgisbuiltasacomputablefunction.Noticethatgwillneverbesettotallyundenedafterstages0.Weprovethatforanyx,ifgxiscreatedatastagev>s0,thenforanys>v,gx=X[s]x.Givenanx,letv>s0bethestageatwhichgxiscreatedanddenedas0ifitis1,thengx=Xxtakesalreadythepermanentvalue.Supposethatv0vatwhichalink;iscreated,andlettibethestageatwhichthelink;[vi]istravelled.Thenv0=v. 141
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Bythechoiceoft0,thelink;[v0]isunsuccessfullytravelledatstaget0,thismeansthatforanyt2[v0;t0],gx=X[t]x.Sincet0isexpansionary,wehavegx=X[t0]x=A;B;x[t0].BytheArestraintrA[t0],thedenitionoff[t0],andbytheconventionof,wehavethatforanys2[t0;v1,gx=A;B;x[s]=X[s]xispreserved.Supposebyinductivehypothesiswehave: 1. foranys2[vn;tn],gx=X[s]x. 2. gx=X[tn]x=A;B;x[tn]. 3. foranys2[tn;vn+1,gx=A;B;x[s]=X[s]x. 4. gy=X[vn+1].Sincethelink;[vn+1]isunsuccessfullytravelledatstagetn+1,holdsforn+1,andsincetn+1isexpansionary,2forn+1holds,andfurthermore,bytheArestraintatstagetn+1,forn+1holds,holdssincewillneverbevisitedatoddstages,sotherearenoelementswhichenterXatoddstages.Thisprovesthatgx=Xx.Sincexisarbitrarilygiven,wehavethatforalmosteveryx,gx=Xx,Xiscomputable.iiibfollows.Foriiic.Bytheassumptioninthiscase,gisbuiltonlynitelymanytimes.Iffisbuiltinnitelymanytimes,thenthenalversionoff,denotedbyf,isbuiltasacomputablefunction,andf=B.Acontradiction.Thereforefisbuiltonlynitelymanytimes.Lets1>s0besuchthatfwillneverbebuiltatanystages>s1.Furthermore,wecanchooseastages2>s1afterwhichnoneofthesteps1,2,3,or4ofprogramwilloccur.Thereforelimsc[s]=cmustbechosenbeforestages2.Clearlyc62D.Sinceisvisitedinnitelymanytimes,theonlyreasonthatcase3bwillneveroccurafterstages2isthatc6=0.iiicfollows. Proposition5.6.6PsatisfactionProposition. Foreache,Peissatised. 142
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Proof.Givene,letbethePestrategy2TP.ByProposition 5.6.5 i,necessarilyA6=eBsothatPeissatised.Thepropositionfollows. Proposition5.6.7. Foreache2!,ifXe=eA;BandXeandBarenotcomputable,thentheydonotformaminimalpair.Proof.ByProposition 5.4.5 ,letbethelongestRestrategyonthetruepathTP.ByProposition 5.6.5 ii,^h0i2TPandD=X=B.ByProposition 5.6.5 iii,foranySstrategy,iftop=and2TPtheneither^hwi2TPor^hdi2TP.Ineithercase6=DsothatXandBdonotformaminimalpair.Thepropositionfollows. 143
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BIOGRAPHICALSKETCHPaulStevenBrodheadwasborninOakPark,ILin1980.HemovedtoRichlandCenter,WIin1989andcontinuedhisschoolingthereuntilhegraduatedfromRichlandCenterHighSchoolin1997.AsaChancellor'sscholar,PaulattendedtheUniversityofWisconsinMadisonfrom1997until2000,whenheearnedaB.S.inmathematics.Duringthesummerof2000,PaulparticipatedintheNSFfundedundergraduatemathematicsresearchexperienceSIMU,attheUniversityofPuertoRicoHumacao.In2003PaulstartedgraduateschoolattheUniversityofFloridaandearnedamaster'sdegreeinmathematicsin2005.Duringthesummerof2006,PaulwasagraduateassistantforanNSFSEAGEPfundedundergraduatemathematicsresearchexperienceattheUniversityoftheVirginIslands.HewenttotheChineseAcademyofSciences,InstituteofSoftwareduringthesummerof2007asanNSFfellow,participatingintheEastAsiaandPacicSummerInstitutes.AsanNSFSEAGEPfellowduringthefallof2007,PaulwenttoVictoriaUniversityofWellingtonWellington,NewZealand.Asaninvitedvisitorandlecturer,PaulwenttotheUniversityofHawaiiatManoainthespringof2008.PaulearnedhisPh.D.inmathematicsfromtheUniversityofFloridain2008. 150
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