Citation
Natural compactifications of lattices

Material Information

Title:
Natural compactifications of lattices
Creator:
Smith, Linda Joe Wahl, 1944- ( Dissertant )
Strecker, G. E. ( Thesis advisor )
Jensen, G. A. ( Reviewer )
de Groot, J. ( Reviewer )
Clark, W. E. ( Reviewer )
Nevill, Gale ( Reviewer )
Place of Publication:
Gainesville, Fla.
Publisher:
University of Florida
Publication Date:
Copyright Date:
1969
Language:
English
Physical Description:
vi, 119 leaves : illus. ; 28 cm.

Subjects

Subjects / Keywords:
Algebra ( jstor )
Compactification ( jstor )
Conceptual lattices ( jstor )
Embeddings ( jstor )
Homomorphisms ( jstor )
Ions ( jstor )
Isomorphism ( jstor )
Mathematics ( jstor )
Topological theorems ( jstor )
Topology ( jstor )
Dissertations, Academic -- Mathematics -- UF
Lattice theory ( lcsh )
Mathematics thesis Ph. D
Topology ( lcsh )
Genre:
bibliography ( marcgt )
non-fiction ( marcgt )

Notes

Abstract:
Abbreviated introduction: Armed only with the definition of compactification, and asked to compactify the open disc in the plane, a novice topologist would most likely simply add its boundary. However, among the "standard" methods which have been developed for obtaining compactifications of spaces (Stone-Cech, Alexandroff, Freudenthal , Wallman, etc.) not one yields this seemingly "natural" compactification. The Stone-Cech compactification (although it has nice mapping properties) adds so many points that the disc's "intrinsic character" is not preserved, and it is in some sense "lost" in the new space. At the other extreme, the Alexandroff compactification does not add enough. What we seek, then, is a method for producing compact extensions which yields "natural" compactifications in the sense that the closed interval [a, b] "naturally" compactifies the open interval (a, b), the extended real line "naturally" compactifies the rationals, the closed disc "naturally" compactifies the open disc, and the Hilbert cube "naturally" compactifies Hilbert space. In this dissertation we define and investigate such a method.
Thesis:
Thesis - University of Florida.
Bibliography:
Bibliography: leaves 117-118.
General Note:
Manuscript copy.
General Note:
Vita.

Record Information

Source Institution:
University of Florida
Holding Location:
University of Florida
Rights Management:
Copyright [name of dissertation author]. Permission granted to the University of Florida to digitize, archive and distribute this item for non-profit research and educational purposes. Any reuse of this item in excess of fair use or other copyright exemptions requires permission of the copyright holder.
Resource Identifier:
021886833 ( AlephBibNum )
13406186 ( OCLC )
ACX9312 ( NOTIS )

Downloads

This item has the following downloads:


Full Text










NATURAL COMPACTIFICATIONS OF LATTICES















By
LINDA JOE WAHL SMITH














A DISSERTATION PRESENTED TO THE GRADUATE COUNCIL OF
THE UNIVERSITY OF FLORIDA
IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE
DEGREE OF DOCTOR OF PHILOSOPHY















UNIVERSITY OF FLORIDA
1969






















































UNIVERSITY OF FLORIDA


3 1262 08552 2935


































To Nevins













ACKNOWLEDGMENTS


The author wishes to extend her sincere and grateful appreciation

to Dr. G. E. Strecker and Dr. G. A. Jensen for their helpful suggestions

in the preparation of this paper, and to acknowledge Dr. J. de Groot,

Dr. W. E. Clark, and Dr. Gale Nevill for serving on her Supervisory

Committee.

To her husband, Nevins, and her parents, Mr. and Mrs. J. K. Wahl

and Mrs. Nevins Smith, she extends many thanks for all the love and

moral support they have given over the years and especially during her

work on this paper. For typing assistance, she is very much indebted

to Mrs. Margaret Parramore.














TABLE OF CONTENTS


Page

. iii


ACKNOWLEDGMENTS . . . .


KEY TO SYMBOLS . . . . . . .


v


1. INTRODUCTION. . . . .... . . . . . . . 1

2. PRELIMINARIES . . . .. . . . . . . .. 5

3. PRODUCTS OF SUPEREXTENSIONS . . . ... . . . .. 15

4. LATTICE SUPEREXTENSIONS . . . ... . . . . 20

5. THE COMPLETION BY CUTS. . . . .... . . . .. 29

6. TOPOLOGICAL LATTICE COMPLETIONS . . . . . . .. 48

7. LARGEST AND SMALLEST LATTICE SUPEREXTENSIONS. . . .. .68

8. EXAMPLES. . . . . . .... . . . . . . 86


APPENDIX I KAUFMAN'S ORDERABLE COMPACTIFICATIONS. ... . .111

APPENDIX II LATTICE CLOSURES. . . . ..... . . 113


BIBLIOGRAPHY. . . . . .. . . . . . . . . 117


BIOGRAPHICAL SKETCH . . . . . . . . .


* 119













KEY OF SYMBOLS


X ............ a topological lattice (2.8 (iii))

XD(a) = {X'E X: x : a} and X (a) = {x E X: a s x}

D ........... (2.3 (ii)) the collection of all closed ideals of X

FI ............ (2.3 (ii)) the collection of all closed dual ideals of X

r = D U rI

TD ........... (2.3 (iii)) the collection of all principal ideals of X
together with X

TI ........(2.3 (iii)) the collection of all principal dual ideals
of X together with X

T = TD U T .. (4.1 (iii)) is the usual subbase for X

Z .......... an arbitrary admissible subbase for X (4.2 (i))

AX .......... (2.15) the superextension of X with respect to E

S (or sometimes S* to avoid confusion) ... (2.14) the set of all
maximal linked systems of E which contain S

+ (or sometimes E*) = {S+: S E Z}. (2.14)

e ........... (2.16 (v)) the natural embedding of X into X X

B X .......... (2.17) the topological closure of e [X] in X X
(called the de Groot compactification of X with respect
to Z)

n ........... the collection of arbitrary intersections of Y

2 ........... the power set of X





D(A) = n{X (a): A c X (a)} where A E 2X (5.3)

X............ (5.2) the completion of X by cuts

d: X X ..... (5.4) the map defined by a -+ D({a})

= {X (d(a)): a E X} U{X (d(a)): a E X} U{X} (5.7)

wX ........... the Alexandroff one-point compactification of X

z = Zla Zb (6.1)

Z2 = 2a r Z2b (6.1)

T = {YD(i(a)): a E X} V{Y (i(a)): a e X} U{Y} (where Y = i[X]I) (6.6)

E ) ......... (6.15)














1. INTRODUCTION


Armed only with the definition of compactification, and asked

to compactify the open disc in the plane, a novice topologist would

most likely simply add its boundary. However, among the "standard"

methods which have been developed for obtaining compactifications of

spaces (Stone-Cech, Alexandroff, Freudenthal, Wallman, etc.) not one
1 v
yields this seemingly "natural" compactification. The Stone-Cech

compactification (although it has nice mapping properties) adds so many

points that the disc's "intrinsic character" is not preserved, and it

is in some sense "lost" in the new space. At the other extreme,

the Alexandroff compactification does not add enough. What we seek,

then, is a method for producing compact extensions which yields

"natural" compactifications in the sense that the closed interval [a, b]

"naturally" compactifies the open interval (a, b), the extended real

line "naturally" compactifies the rationals, the closed disc "naturally"

compactifies the open disc,and the Hilbert cube "naturally" compactifies

Hilbert space. In this dissertation we define and investigate such a

method.


Frink [8] has developed a Wallman-type method for obtaining
many compactifications for a given space. It has been shown by
Alo and Shapiro [2] that the compactification of the open disc by the
closed disc is of Wallman-type.







The method which we use is due to de Groot. For a given space,

this method produces compact extensions (called superextensions) and

compactifications (called de Groot compactifications) which are de-

pendent upon both the space and a chosen subbase for its closed sets.

Hence,in general, for each space there may be many superextensions

and de Groot compactifications. (See Chapter 2 for details.) In

particular, for locally compact Hausdorff spaces, it is possible (by

astutely choosing the subbases) to obtain both the Stone-Cech and the

Alexandroff compactifications as de Groot compactifications. Another

aspect of the versatility of the de Groot compactifications is that

they (unlike any of the "standard" compactifications mentioned above)

are productive. Specifically, we will show in Chapter 3 that the

topological product of any collection of de Groot compactifications

:(respectively, superextensions) of a family of spaces is homeomorphic

with a natural de Groot compactification (respectively, superextension)

of the topological product of the family.

In order to obtain the desired "natural" compactifications, we

consider only topological products of topological lattices (i.e., lattices

with the interval topology) and restrict our attention, when dealing

with lattices, to those superextensions which arise with respect to

certain "admissible" subbases. This restriction is not overly severe

since, as can be seen in Chapter 8, for a given topological lattice

there are, in general, many "lattice superextensions" (i.e., superex-

tensions with respect to admissible subbases). In Chapter 4 we will








show how the order of the original lattice can be extended to a

partial ordering of any lattice superextension in such a way that

the superextension becomes a complete lattice (in the interval

topology). Thus each lattice superextension of a topological

lattice turns out to be not only a topological compactification, but

also a lattice completion--hence a very "natural" type of compacti-

fication.

In Chapter 5 we consider the algebraic structure of a

topological lattice X and show that its completion by cuts can be

realized as a particular lattice superextension. In Chapter 6 we

;attempt to algebraically characterize other lattice superextensions of

X. This is done using the notion of topological lattice completions

of X, these being certain complete lattices containing X as a sublattice.

We show that all lattice superextensions of X are in fact topological

lattice completions of X and that those topological lattice completions

of X satisfying a particular "density" condition are lattice super-

extensions of X.

In [12] Kaufman considers (totally) orderable compactifications

for totally ordered spaces X; i.e., compactifications Y of X for which

the given topology on Y is the same as the order (interval) topology

on Y. Among other things,he establishes the existence of a "largest"

and a "smallest" orderable compactification for each totally ordered

space. In Chapter 7 we consider the notion of "domination" for lattice

superextensions of topological lattices and generalize Kaufman's results

to these compactifications of lattices. Hence each topological lattice








has a lattice compactification which is maximal relative to mapping

properties (analogous to the Stone-tech compactification for more

general spaces) and a lattice compactification which is minimal rela-

tive to mapping properties (analogous to the Alexandroff compactifi-

cation for locally compact Hausdorff spaces).

Chapter 8 consists of examples of lattice superextensions of

various kinds of lattices. Also, examples are given which show the

necessity of certain restrictions which are made in the text.













2. PRELIMINARIES


The purpose of this chapter is to state most of the basic

definitions, propositions, and theorems which will be needed in the

following chapters. Algebraic terms which are not specifically defined

here can be found in Birkhoff [ 5] and topological terms in Kelley [13].



DEFINITION 2.1. (i) A set X is called partially ordered if

there is a binary relation 5 on X which is reflexive, antisymmetric,

and transitive.

(ii) If X is a partially ordered set with more than one element

in which each pair of elements a and b have a supremum (denoted by

a v b) and have an infimum (denoted by a A b), then X is called a

lattice.

(iii) If X is a lattice and each subset C of X has a supremum

in X and an infimum in X, then X is said to be complete.2

(iv) If X is a lattice and a, b E X, then a and b are said to

be comparable if either a < b or b < a.

(v) If X is a lattice and a, b E X, then a < b will mean that

a 5 b and a / b.



1We consider only non-trivial lattices in order to avoid the
possibility that vX = AX.

Throughout we will use the standard conventions that for a
lattice X, A0 = vX and v0 = AX, and for a topological space X, f0 = X.

5







(vi) X is said to be totally ordered if for every pair of

elements x and y of X, x and y are comparable; i.e., x y or y x.



To avoid confusion when more than one lattice is considered,

the symbols "v, A and 5" may be subscripted such as "vX, AX, and :X"

to denote supremum in X, infimum in X, and order in X, respectively.



DEFINITION 2.2. Suppose that X and Y are lattices and

i: X Y is a function.

(i) The map i is called a lattice homomorphism whenever i

satisfies the following: For each non-void subset C of X, if v C

exists in X, then v i[C] exists in Y and i(vXC) = v i[C]. Dually, if

A C exists in X, then A i[C] exists in Y and i(A C) = A i[C].

(ii) The map i is called a lattice inclusion and X is called

a sublattice of Y if i is an injective lattice homomorphism.

(iii) The map i is called a lattice isomorphism and X is said

to be lattice isomorphic with Y if i is a surjective lattice inclusion.


DEFINITION 2.3. Let X be a lattice and I a non-void subset of

X.

(i) I is-increasing if a E I whenever a E X and i a for some

i E I. I is decreasing if a E I whenever a E X and a S i for some i I.

(ii) I is an ideal of X if I is decreasing and a v b E I for

each pair a, b E I. I is a dual ideal if I is increasing and a A b E I
3
for each pair a, b E I.



We do not consider 0 to be an ideal or dual ideal in order to
simplify the statements of proofs.







(iii) I is a principal ideal of X if there is some fixed a E X

such that I = {x E X: x 5 al. I is a principal dual ideal of X if

there is some fixed b E X such that I = {x E X: b 5 x}.



PROPOSITION 2.4. Let X be a lattice.

(i) Arbitrary intersections of ideals of X are either void or

ideals of X, and arbitrary intersections of dual ideals of X are either

void or dual ideals of X.

(ii) Finite intersections of ideals of X are ideals of X, and

finite intersections of dual ideals of X are dual ideals of X.

(iii) If 1 is a finite collection of ideals of X such that

X i f, then Uyl X, and dually if is a finite collection of dual

ideals of X such that X then U 9 X.


PROOF. (i) Suppose C is a collection of ideals of X such that

n X 0. Let I = n If b E I and a e X with a < b, then since b E E

for each E E ; and each E is decreasing, a E E for each E E Hence

a-E I; i.e., I is decreasing. Finally, if a, b e I, then since a, b e E

for each E E C and each E is an ideal, it follows that a v b E E for

each E, so that a v b E I. By Definition 2.3 (ii), I is an ideal of X.

A dual argument shows that the intersection of an arbitrary collection

of dual ideals of X is either void or a dual ideal of X.


(ii) Suppose II, 12, 1 -, I are ideals of X. By definition,

I. I 0 for each j = 1,2,-.-,n so there exists x. E I. for each j. It

follows that A{x.: j = 1,2,-' ,n} e I = {I.: j = 1,2,---,n} since
] J








each I. is decreasing, and hence I / 0. By (i) above, I is an ideal

of X. A dual argument shows that the intersection of a finite collection

of dual ideals of X is a dual ideal of X.


(iii) Since X g for

x = v{xF: FE % }. If x E

F is decreasing and xF < x,

x E X U and so U X.


each F E there exists xF E X F. Let

U-', then x e F for some F E,, and since

we see that xF E F--a contradiction. Thus

A dual argument proves the other statement.


PROPOSITION 2.5. If X and Y are lattices and i: X Y is a lattice

homomorphism, then i preserves the order of X.


PROOF. Let i: X -+ Y be a lattice homomorphism and suppose

x, y E X such that x
lattice homomorphism i(y) = i(x vX y) = i(x) v i(y); therefore

i(x) < i(y) and so i preserves the order of X.


DEFINITION 2.6.

topological closure of A

confusion seems likely.


Let X be a topological space and A c X. The
in X will be denoted by A or when no
in X will be denoted by A or A when no


DEFINITION 2.7. Suppose X and Y are topological spaces and


f: X -+ Y is a function.

(i) f is called

continuous, and a closed

(ii) f is called

embedding and f- X = Y.


an embedding of X into Y if f is injective,

map onto its image.

a dense embedding of X into Y if f is an







(iii) f is called a homeomorphism and X is said to be

homeomorphic with Y whenever f is a surjective embedding.

(iv) Y is called a compactification of X if Y is compact and f

is a dense embedding.



DEFINITION 2.8. Suppose that X is a lattice.

(i) A closed interval of X is a subset A of X such that either

A is a principal ideal of X, A is a principal dual ideal of X, or A is

the non-void intersection of a principal ideal of X and a principal

dual ideal of X. (In the latter case, there exist a, b E X such that

A = {x E X: a x 5 b}.)

(ii) The interval topology on X is the topology on X which has

all closed intervals of X as a subbase for the closed subsets of X.

(iii) A lattice X will be called a topological lattice when it

is also considered as a topological space having the interval topology.

(iv) E c 2 will be called a subbase for X when E is a subbase

for the closed sets of the topological lattice X and X e E.



The following proposition is due to Frink [9].


PROPOSITION 2.9. If X is a topological lattice, then X is compact

if and only if X is complete.



DEFINITION 2.10. Suppose that X and Y are topological lattices

and f: X Y is a function. f is called an iseomorphism and X is said

to be iseomorphic with Y if f is both a lattice isomorphism and a

homeomorphism.







The remainder of this chapter contains definitions and some

fundamental results concerning superextensions of T1-spaces. These

results are due to de Groot, Jensen, and Verbeek [ 7 ]. It is within

the framework of superextensions that we will study lattices in the

following chapters.


DEFINITION 2.11. Suppose that X is a T1-space

subbase for the closed subsets of X.

(i) If A and B are subsets of X, then A and B

screened by a subcollection Y of Z and is called a sc

B if UV = X and no member of Y meets both A and B. Ti

and y are screened by T if {x} and {y} are screened by

x E X and a subset A c X are screened by V if A and {xE

by '.


and Z is a



are said to be

:reen for A and

Jo elements x

', and an element

}are screened


(ii) E is called a T -subbase for X if for each x e X,

r{S e Z: x E S} = {x}, and whenever x E X and S E Z with x J S, there

is some T E with x E T and T n S = 0.

(iii) E is said to be weakly-normal if each pair of disjoint mem-

bers of E can be screened by a finite collection from E.

(iv) A non-void subcollection X of E is said to be linked if

each pair of elements of / has a non-void intersection; X is called a

maximal linked system of E if X is maximal in E with respect to the

property of being linked.


The next two propositions are found in [7 ]. Their proofs are

included only to acquaint the reader with the terminology.








PROPOSITION 2.12. A snace X is Tl if and only if there is a

T -subbase for the closed sets of X.


PROOF. Suppose that X is a T -space and let E be the collection

of all closed sets of X. Clearly E is a subbase for the closed subsets

of X. Let x e X. Since X is T1, {x} E E, and it follows that

{x} c n{S E Z: x e S} c {x}; i.e., n{S E E: x E S} = {x}. If x E X

and S E E with x S, then clearly {x} E E, x E {x}, and {x} S = 0.

Hence E is a T -subbase for X (2.11 (ii)).

Conversely, suppose that E is a T -subbase for X. Let x E X.

To see that {x} is closed, let y E X such that y d {x}. Hence

y (1{S E E: x E S} since E is a T -subbase for X. Thus there is some

closed set S E E with {x} c S and y V S. It follows that {x} is a closed

subset of X.



PROPOSITION 2.13. If E is a subbase for the closed subsets of

a T1-space X, then E is a T -subbase for X if and only if for each x E X,

{S E E: x E S} is a maximal linked system of E.


PROOF. Suppose that E is a T -subbase for X and x E X. Let

S= {S E: x E S}. Clearly /f is a linked system of E. If T c E and

T i {, then x / T. Thus since E is a T -subbase for X there is some

S E such that x E S and T ( S = 0. It follows that S E : and so

U {T} is not linked; therefore fH is a maximal linked system of E.

Conversely, suppose now that E is a subbase for the closed

subsets of a T1-space X such that {S E: x E S} is a maximal linked

system of E for each x E X. Let x be a fixed but arbitrary point of X.







Clearly {x} c n{S E Z: x e S}. If y E X and y / x, then it follows

from the fact that X is a T1-space that there is some T E E such that

x E T and y V T. Therefore y V f{S E E: x e S} and so

({S E Z: x e S} = {x}. Finally, let x E X and T E Z with x V T. Thus

T i {S E E: x e S} and since {S E : x e S} is a maximal.linked

system of E, there is some S E E such that x E S and S n T = 0. Hence

by Definition 2.11 (ii), E is a T -subbase for X.



NOTATION 2.14. Let E be a T -subbase for a space X.

X X will denote { c : K is a maximal linked system of E}.

For each S E S+ will denote {Y E X X: S eZ}, and E+ will

denote {S : S E Z1.



DEFINITION 2.15. Suppose that E is a T -subbase for a space X.

The set X X equipped with the topology which has ZE as a subbase for

the closed sets is called the superextension of X with respect to E.



The following proposition states some results found in [7 1

which follow readily from the definitions. The proof is thus omitted.


PROPOSITION 2.16. Suppose that Z is a T -subbase for the space X.

(i) If S, T E E with S f T = 0, then S+ nf T = 0.

(ii) If E E X and S E E such that / U {S} is linked, then S .

(iii) If c is a linked system of E, then o is contained in some

maximal linked system of E.

(iv) If X E X X and CZ / 0, then f x = {x} for some x e X.








(v) If e : X X X is defined by e (x) = {S E : x e S},

then e is an embedding of X into X X, and e [S] = S+ r e [X] for

each S 5 E.

(vi) The space X X is TI.

(vii) The space X X is compact.

(viii) If S and T are in E and S U T = X, then S UT = X X.



DEFINITION 2.17. Let E be a T -subbase for a space X.

(i) Let 8 X denote the topological closure of e EX] in X X;

this will be called the de Groot compactification of X with respect to

E. Note that {B X n1 S : S E Z} is a subbase for the closed subsets

of B X.

(ii) If;f c E, then o is said to he prime if and only if

whenever S1, S2, **-, Sn E such that U{S.: i = 1,2,---,n} = X, there

exists some j E {l,2,***,n} such that S. e .

(iii) A subset e of E is said to be centered if each non-void

finite subcollection from Z has a non-void intersection; is called

a maximal centered system of E in case 9 is maximal in E with respect

to the property of being centered.



It should be clear from Definition 2.7 (iv) that 3 X is indeed

a compactification of X. However, & X is not necessarily Hausdorff

even if X is Hausdorff.


PROPOSITION 2.18. Suppose that E is a T -subbase for a space X.








(i) If q is a finite subcollection of Z, then Uf = X if

and only if U{B1XT +: T E } X.

(ii) If Z is weakly-normal, then B X is Hausdorff.

(iii) If e E X X, then d~ 8 X if and only if Z is prime.

(iv) If ; is any maximal centered system of E, then is prime.

*(v) If E is weakly-normal and 'e A XX, then / e B X if and

only if Z contains a maximal centered system of E.



The proof of Proposition 2.18 can be found in [ 7] Chapter 1,

Proposition 1.5 through Proposition 1.9.

Propositions 2.16 and 2.18 together with the fact that a T1-space

is completely regular if and only if it can be embedded in a compact

Hausdorff space outline a proof for the following statement from [ 6]:

A T1-space X is completely regular if and only if there exists

a weakly-normal T -subbase for the closed subsets of X.



EXAMPLE 2.19. Suppose that X is a completely regular T1-space

and E is the collection of all non-void zero sets of X. Then E is a
V
weakly-normal T -subbase for X, and E X is the Stone-Cech compactification

of X. (This follows from Proposition 2.16 (i) and Theorem 6.5 of [10].)














3. PRODUCTS OF SUPEREXTENSIONS


In this chapter we investigate the productivity of superexten-

sions of T1-spaces and show that a natural superextension of the product

of an arbitrary family of T1-spaces is the product of the superextensions

of the family. We also show that the corresponding de Groot compacti-

fication of the product is the product of the de Groot compactifications.

Throughout this chapter we let {Ya }aA be a family of T1-spaces, Ea

be a T -subbase for the closed sets of Ya, Y be the topological product

of the family {Y } ,E and for each a E A, a : Y Y be the projection
a aEA a a
function. Also without loss of generality we may assume that for each

a E A, the set Y is itself a member of the subbase .
a a


-1
a a 1
for the closed sets of Y.


PROOF. It is clear that E is a subbase for the closed sets

of the product space Y. Suppose that y, z E Y and z E nr{7i [T]: w [T] E E
a a
-1
and y E na [T]}. Thus for each a E A, a (z) E rl{T: T E E and a (y) E T}.
a a a a
Since Ea is a T -subbase for Y a, ({T E Za: 7a (y) E T} = { a(y)};

therefore T (z) = T (y) for each a E A and hence z = y. Now let y E Y
a a
an -i
and iT [S] E with y i -1 [S]. If follows that a (y) 1 S, and
a a a







since E is a T -subbase for Y there is some T E E
a 1 a a
-1
S(y) 'E T and S ( T 0. Therefore, T [T] e E, y E
a a
a [S] n Ta [T] = 0. Hence E is a T -subbase for Y.
a a 1
2.11 (ii)).


such that

n I[T] and
a
(cf. Definition


In view of the above proposition we may consider the super-

extension of Y with respect to E.


PROPOSITION 3.2. Let P be the topological product of

family {X Ya For each (Va ) E P, the set A = {Q E E:
a aEA a aEA
a
a E A, na [Q] E } is a maximal linked system of E.
a a --_ _ _ _ _


the

for each


PROOF. Suppose that Q, R E V. For each a E A, a [Q], T [R] E ~a

so that IT [Q] n iT [R] 0; therefore, Q n R $ 0 so that I is linked.
a a
-1 -1
Suppose now that Eb [S] E and 0 U{7b [S]} is linked. If a E A and
-1 -1
a / b, then T [ S]] = Y If a = b, then T [ib [S]] = S. Thus for
ab a a b
-1 -1
each a E A it follows that na [i [S]] E ja and so 7i [S] C f; i.e., ff
a b a b
is a maximal linked system of E.



THEOREM 3.3. There is a homeomorphism f from the product P of

the superextensions {XA Y } aA to the superextension X Y of the product
a
Y. Furthermore, if g is the unique continuous map which makes each of

the diagrams

g
Y --> P


A
a a
)f I


a E a
e a


commute,








then the diagram


P




e




commutes.


PROOF. By Proposition 3.2 there is a function f: P X\ Y

defined by f((a ) a) = {Q E: for each a e A, E [Q] E X }.
a aEA a a
We will show that f o g = e For each y E Y, f(g(y)) = {Q E E: for

each a E A, a [Q] E T (g(y))} = {Q e E: for each a E A, Ta [Q] e (Ta ())}
a
= {Q e E: for each a E A, a (y) E a [Q]} = {Q E : y E Q} = e (y).

To see that f is injective, let 8 and X be members of P such
A A
that f / Thus there must be some b E A such that rb() ^ hb(A/), and
A A
since Tb(Y) and nb( () are maximal linked systems of Eb, it follows that
A A
there exist T T b(0) and S E Tb () such that S n T = 0. Hence

Tb [S] n T] = 0 and so = [S] and [T] cannot be contained in the
b b b b
same maximal linked system of E. Recall that if a E A and a / b, then

Tn [ [T]] = Y and if a = b, then na [nb [T]] = T. It follows that for
a b a a b
-1 A
each a E A, a[ b [T]] E Aa(). Hence by the definition of f,
a b a
-1 -1
[T [T] E f(9). By a similar argument we see that wb [S] E f(A). Thus

f(T) / f(A).

To see that f is surjective, let E e X Y. For each a E A, let

Xa = {r [Q]: Q E /}. For each a E A, a is clearly linked, and if
-1
S E E such that / U{S) is linked, then it. follows that U{Tr [S]}1
a a







-1 -1
is linked and hence a1 [S] E C. Therefore S = a [Wa [S]] E i so that
a a a
ta is a maximal linked system of Ea. Thus ( a)aEA E P and by the defi-

nition of f, f((a ) ) = {Q E E: for each a E A, w [Q] E X } "
a aEA a a
= {Q e E: Q E } = 4.

The continuity of f is an immediate consequence of the fact that
-1 -1 +
for each b E A and each TE E f- [(T [T]) ] = {(a ) a P:
b b a aeA
f(( ) ) E ( [T])+}= {( P: [T] f(( ) )}
a )aEA b E } afaEA [P: Tb a aEA
-1
= { )aA E P: for each a E A, a E[v [T]] E a} = {()aeA E P:
-1 A-1 A-1i +
T(= b b [T]) E b = A b {E x Y : T E }] = b[T ] which is a

closed subset of P.

Finally we show that f is a closed map. Using the fact that f

is injective and applying f to the above equality we know that for each

a E A and T E E f[ 1[T+]] = f[f- [( T]) ]] = (T- [T]) It follows
a a a a
A-1 + + +
from the fact that {- E[T ]: a E A, T+ E Z+} is a subbase for the closed
a a
sets of P and that f is bijective,that f is a closed map.

Consequently, XEY is homeomorphic with P.




Let Q be the topological product of the family {7 Y : a E A}.
L a
a
Thus Q is a subspace of the product space P.



COROLLARY 3.4. There is a homeomorphism h from the product Q

of the compactifications {8 Y } a to the de Groot compactification
------ LtE a aeA ---
a
B Y of the product Y.








PROOF. Let h be the restriction of-f in Theorem 3.3 to Q.

Since f is a homeomorphism we need only show that h[Q] = B Y.

To see that B Y c h[Q] let e B6 Y. Recall that in the proof

-of Theorem 3.3 we showed that for each a E A, a = {n [S]: S E-} is
a a
a maximal linked system of E and that f((.a)) )= To see that
a a aEA

(a)aEA E Q we will show that for each a E A, a is prime and Hence

a E Ya (cf., Proposition 2.18 (iii)). Let a E A and suppose that
a
{T T **', T } c E such that U{T.: i = 1, 2, ***, n} = Y It
1 n a 1 a
-1 -1 -1
follows that {T [T ], Tr [T ], .., [ T ]} c E and
a 1 a 2 a n -
-1
U{T- [T.1: i = 1, 2, ***, n} = Y. Since e is in B Y and hence is
a 1
-1
prime, there is some j E {1, 2, **', n} such that nT [T.] E Thus
a ]
-1
7T [n [T.]] = T. E e and so Z is prime.
a a ] a a

To see that h[Q] c Y, let (X ) E Q. Thus for each a E A,
E a aEA

SE 8 Y and so a is prime. Suppose now that Y is a finite subset
a E a a
a
of E such that UV = Y. Recall that each member of Y is of the form
-1
aT [T] for some a E A and T E Let B = {a E A: there is some T E E
a a a
-1
with n- [T] e '}. Suppose that for each b E B there is some
a

Yb E Yb U{T e b: nb [T] e T}. Then we can pick y E Y such that
-1
b(y) = yb for each b E B. It follows that for each 7a [T] E T,
-1
y i Ta [T] and so y E Y UT--a contradiction. Hence there must be
a
-1
some b e B such that U{T Eb : T b 1[T] e T = Y Since is prime,
-l
there is some S E E such that ib [S] E and S E b. It follows from
b b b
-1
the definition of f that l[I S] f((a ) ). Thus f((a ) ) is prime
b a aEA a aEA
and so h((a ) ) = f((a ) ) e B X. Consequently, h is a homeomorphism
a aEA a aEA C
from Q onto B Y.














4. LATTICE SUPEREXTENSIONS


For the remainder of the paper we restrict our attention to

superextensions of topological lattices with respect to T1-subbases

which consist of ideals and dual ideals. For these "admissible"

subbases we show that the superextensions are actually de Groot

compactifications (Theorem 4.6). Moreover, (with respect to a

"naturally induced" ordering), they are shown to be complete lattice

extensions of the original lattice (Theorem 4.11).



NOTATION 4.1. (i) For each a E X, XD(a) = {x e X: x : a) and

X (a) = {x E X: a x}. (D denotes decreasing and I denotes increasing.)

(ii) r = {C c X: C is a closed ideal of X}, r = {C C X: C is

a closed dual ideal of X}, and r = FD U I.

(iii) T = {X} U {XD(a): a E X} U {X (a): a E X}. (We refer

to T as the usual subbase for X.)

(iv) If E is a subbase for the closed sets of X and E c r, then

E = E F and ED D'



DEFINITION 4.2. (i) If E is a T -subbase for the closed subsets

of X and E c r,then E is called an admissible subbase for X.

(ii) A lattice superextension of X is any superextension of

the form X X where E is an admissible subbase for X.

20








PROPOSITION 4.3. If T c E c F, then E is an admissible subbase

for X.


PROOF. It is clear from the definition of the interval topology

(2.8 (ii)) that T is a subbase for the closed subsets of X. It is also

clear that any collection of closed subsets of X containing T is a

subbase for the closed sets of X. Hence E is a subbase for X. To

see that Z is a T1-subbase for X, let a E X. Thus since T c Z,

{a} c f{SE E : a E S} c {U E T: a E U} X (a) X (a) = {a};

i.e., n{S E E: a E S} = {a}. Finally, suppose a E X and S E Z such

that a / S. If S E ZD, then X (a) E ZI, a E X (a) and X (a) r S = 0.

Dually, if S E Z_, then XD(a) E ZD, a E X (a) and XD(a) n S = 0. Thus

E is a T -subbase for X (cf., Definition 2.11 (ii)) and hence is an

admissible subbase for X (4.2 (i)).



COROLLARY 4.4. T and F are admissible subbases for X.'




By the above corollary X X and X X are lattice superextensions

of X. Examples of these can be found in Chapter 8. Next we establish

that all lattice superextensions of X are in fact compactifications of

X which can be considered as complete lattices having X as a sublattice.



PROPOSITION 4.5. Let Z be an admissible subbase for X.

(i) ED and E are centered systems of E.

(ii) Every linked system of E is centered.







PROOF. (i) is immediate from Proposition 2.4 (ii).


(ii) Suppose that 9 is a linked system of E and

FI, F2, ***, Fn e If all of the F. belong to ED, then by (i) above

n{F.: i = 1, 2, ***, n} / 0. Similarly, if all of the F. belong to
1 1
EI, then n{F.: i = 1, 2, .**, n} / 0. In the remaining case we may

assume, by at most rearranging the subscripts, that there is some integer

m such that 1 5 m < n 1, F. E D for i = 1, .**, m, and F. EI for

i = m + 1, *.*, n. For each i e {1, ..., m} and j e {m + ***, n} we

can pick f. F. F F. because Z is linked. Since the members of D

are decreasing and those of I are dual ideals, it follows that for each

j e {m + 1, ., n}, gj = A{f. .: i = 1, *.., m} e F. r
J,] ]
(n{F.: i = 1, ***, m}). In the same way, since the members of E

are increasing and n{F.: i = 1, ***, m} is an ideal (2.4 (ii)), it
1
follows that v{g.: j = m + 1, .**, n} e (rn{F.: i = 1, .**, m}) n

(f{F.: j = m + 1, ..*, n}); i.e., is centered.




THEOREM 4.6. Each superextension of X with respect to an admis-

sible subbase Z is a de Groot compactification of X; i.e., X X = E X.


PROOF. Suppose that E is an admissible subbase for X. It is

clear from the definitions in Chapter 2 that EX XA X; therefore, we

need only show that X Xc_ C X. If X is a maximal linked system of E, then

by Proposition 4.5 (ii), is centered; i.e., e is a maximal centered

system of E. Thus Y is prime (2.18 (iv)), and so e E X (2.18 (iii)).








In view of Theorem 4.6 every lattice superextension of X is

actually a topological compactification of X. To see that it is also

a complete lattice with the interval topology and that it has X as a

sublattice we continue as follows.



DEFINITION 4.7. Suppose that E is an admissible subbase for X and

X, /m x X. We say that 5 X whenever 7i E c D E or, dually,

whenever n EI c n n E.



Since containment is a partial ordering, it follows readily that

4.7 defines a partial ordering on the elements of A X. Throughout the

remainder of this paper, X X will be understood to be partially ordered

according to Definition 4.7 whenever E is an admissible subbase for X.

The following theorem shows that X X is a complete lattice whenever

E is admissible.




THEOREM 4.8. Suppose that E is an admissible subbase for X.

Every non-void subset of X X has a supremum in X X and an infimum in

X X. Specifically, if 0 / C c X X and if


CD = D{ r D: C} and

CI = n{r n zx : E e C},







then


v C = CD U{S E ZI: CDU{S} is linked} and

A C = CI U{S e D: C U{S} is linked}.


PROOF. It is easily seen that ) = CD U{S E : CD U{S} is

linked} is a linked system of E. Suppose that T E ZD and U{T} is

linked. Since n ZI c n ZI for each eE C, it follows that

X U{T} is linked for each Z e C. Thus T E for each ;e C and so

T e CD; hence T E ). If T E I and B U{T} is linked, then CD U{T}

is linked and so by definition, T E Therefore ) is a maximal linked

system of Z. A dual argument shows that- = CI U{S e ZD: CI U{S} is

linked} is a maximal linked system of Z.

We continue by showing that P = v C. Since e n Zi c_ ~ Zi

for each E C, then by Definition 4.7, f for each f C. If

me A X and 7W< ', then there is some T in (T ZD) ( D r D) Since

T i n ZD = CD, there is some e C such that T k n D. It follows

from this that Le 4 M; therefore, = v C. A dual argument shows that

,=: A C.




PROPOSITION 4.9. Let E be an admissible subbase for X. Then

the natural embedding e : X -+ X X defined by x -- {S E : x e S} is a

lattice inclusion; thus X is a sublattice of X X.


PROOF. That e is injective follows from the fact that it is

a topological embedding (Proposition 2.16 (v)). To see that it







preserves supreme, suppose that C c X such that vX C exists in X. If

S E e (VX C) n ED, then vX C E S, and since S is decreasing it follows

that C c S. Therefore S E e (c) n ED for each c E C and by Proposition 4.8,

S E v e [C]. Hence by Definition 4.7, v e [C] 5 e (vX C). Conversely,

if SE v e[C] E ,D then S E e (c) f E-D and c E S for each c E C.

Thus C c S and,since S is a closed ideal of X, it follows from Frink

[ 9; Theorem 12] that vX C E S. Hence S E e(vX C), and by Definition 4.7,

e (vX C) < v e [C]; i.e., e (vX C) = v e [C]. A dual argument shows

that if C c X and AC exists in X, then e (X C) = A e [C]. By
X E X
Definition 2.2 (ii), eE is a lattice inclusion.



PROPOSITION 4.10. For any admissible subbase E of X, the

following hold: (Recall that for each S E, S = { E X X: S EZ}.)

(i) If S E E and T E E, then S+ = {aE X: v S } and
D I
T = {t E X X: A T+ < a}.

(ii) For each ~ZE X X, { E X X: t 5 } r = n{S+: S E m ) E }

and {E E AX: m < ;} = f){S+: S E m n Er }.
-1
(iii) For each flE A X, e- [{1 E X X: x r (L}] = (t rL E )
-1
and e.[{C E XX: m < }] = n(r) n E).

(iv) For each T EE D, v T = v e [T], and for each T E I,

A T = A e [T].

(v) If T E D and S E then S r T / 0 if and only if

A S+ v T.

(vi) If X E A X, T E E, and S E then T E if and only if

t 5 v T+, and S E t if and only if A S+ < .

(vii) For each E A X, v {A S+: S n E } =

= A {v S+: S E e n }.







PROOF. (i) Note first that by Proposition 4.8, v S+ X X

and S E v S. Clearly, if e E S, then s v S. Conversely, if

S v S, thenS SEcVS+ ED C C D and so ES The second

equality is proved dually.


(ii) Suppose that MV X X. It follows from Definition 4.7 that

{ x e X X: Z} c {S +: S E f n E }. Conversely, if E X X and

A /, then again by Definition 4.7 there is some T E (V r ED) (V \ D).

Thus / T+ and so {S +:S E 7M N The second equality is proved

dually.


(iii) Recall from Proposition 2.16 (v) that for each T E E,

T + e [X] = e [T]. Thus since eE is injective, it follows that for

each T E T = e- [e [T]] = e [T ] e X]] = el [T+ Let 1 X X.
-1
By the above remark and (ii) we see that e E[{ E X X: ~'}]
-1 + -1 +
= e [ n{S : S c: n n zD}] = n{e [S : S E n ED

= n{S: S e NI = n( in %). The other equality is proved similarly.


(iv) Let T E D. Again using Proposition 2.16 (v) we see that
+ +
e [T] cT+ and so v e [T] < v T. If S E v e [T] n Z then it follows

from Proposition 4.8 that T c S and so S E v T+; therefore by

Definition 4.7, v T < v e [T] and hence v T = v e [T]. A dual argument

shows that if T E Zi, then A T = A e [T].


(v) If there is some x E T r S, then by (iv) above

A S e e (x) v T Conversely, suppose that X = A S < v T By

(i) above Z E S and E T Thus S n T 0 and hence S n T i 0

(2.16 (i)).







(vi) Suppose that e X X and T E ED. By (i) above X < v T

if and only if E T and by definition, fc T if and only if T E .

Thus Te Zi if and only if < v T. Dually, if S E E then A S + <

if and only if S E .


(vii) Suppose that E X X. Let = v {A S+: S n },

and X = A {V T : T E X E D}. By (vi) above, for each S E X) I

and each T E E ZD, A S V
R n ZD. Thus by (vi), < v R+. Since for each S E ) (h ZI

A S 5 <, it follows from (v) that S f R f 0 for each S E Z C E;

i.e., (n~ E Z ) U{R} is linked. Therefore U {R} is linked, and so

R ce Hence by (v) and the definition of R R E J. By Definition 4.7,

Z <_ and hence P = X = 4.


THEOREM 4.11. Any lattice superextension of X is a complete

topological lattice having X as a sublattice.


PROOF. By Theorem 4.8 and Proposition 4.9 any lattice super-

extension of X is a complete lattice having X as a sublattice. By

Proposition 4.10 (i), the given subbase for a lattice superextension

X X is a collection of principal ideals and principal dual ideals of

X X, and by Proposition 4.10 (ii) each principal ideal and principal

dual ideal of X X is closed in the given topology on X X. Thus the

topology on X X generated by + is the interval topology on X X;

i.e., X X is a topological lattice.







We can conclude this chapter by showing that in the case where

X is totally ordered, all of the lattice superextensions of X are also

totally ordered.



PROPOSITION 4.12. If X is totally ordered and E is an admissible

subbase for X, then E is weakly-normal and X X is totally ordered.


PROOF. To see that E is weakly-normal, let S, T E E such that

S n T = 0. Either S u T = X, in which case {S,T} is a screen for S

and T, or there is some x E X -(S U T). In the latter case, since

E is a T -subbase for X we may pick S', T' E such that x e S' r T',

S r S' = 0, and T n T' = 0. It now follows from the facts that X is

totally ordered and that S and T are not both in ED or EI that

X = X (x) U X (x) c S' U T'. Thus {S',T'} is a screen for S and T.

Hence by Definition 2.11 (iii), E is weakly-normal. To show that X X

is totally ordered, suppose that ) and kare members of X X with X .

Then there, exist T E E n ED and S E X h E such that S T = 0. Since

X is totally ordered it must be the case that t s for each t E T and

s E S. If R E X ED, then it follows from R n S 0 that T c R,

R E E D, and thus .
















5. THE COMPLETION BY CUTS


In this chapter we characterize the completion by cuts of

a topological lattice as a particular lattice superextension

(Theorem 5.13). In doing so we prove a theorem (5.11) which is

used extensively in proving the main results in this and the

succeeding chapter.

Throughout this chapter we let X denote a topological lattice.

We first characterize the completion of X by cuts as a particular com-

plete sublaitice of the power set, 2X, of X which is ordered by

inclusion. Then using the theorem mentioned above we show that the

completion of X by cuts is iseomorphic with the superextension of X with

respect to the usual subbase T. We conclude the chapter by considering

the case where X is totally ordered and investigating the conditions

on X under which the completion by cuts is homeomorphic with the

Alexandroff one-point compactification.



DEFINITION 5.1. A completion-operator on a lattice L is a map
L L
2: 2 2 which satisfies the following conditions:

Cl. A c O(A), for each A E 2

C2. O(A) = O(Q(A)), for each A E 2.

C3. If A and B are in 2L and A c B, then O(A) c O(B).







REMARK 5.2. It follows from Ward [15] that if D is a completion-

operator on L, then A-.= {A E 2 : A = $(A)} is a complete lattice,ordered

by inclusion. Moreover, for each T c A,

(i) A Y = n and

(ii) v y = (U\ ),

and the mapping from L to A defined by x 4({x}) is an injective,

order-preserving map.



Next, we consider a specific completion-operator on X.



PROPOSITION 5.3. The map D: 2X 2 defined by

A n{X (a): A c X (a)} is a completion-operator on X.


PROOF. Cl: Clearly, for each A 2X, A c r{X (a): A c X (a)}.

C2: For each A E 2 by Cl, D(A) c D(D(A)). Conversely, if

A c X (a) for some a E X, then by definition, D(A) c X (a); therefore,

D(D(A)) c D(A).

C3: Suppose that A c B c X. If D(B) = X, then clearly D(A) c D(B).

If, however, B c X (a) for some a e X, then,since A c B c X (a), it follows

that D(A) c D(B).




It should be noted that for each a E X, D({a}) = XD(a) and that

D(A) is the set of all predecessors of all successors of all members of

A; i.e., D(A) = {x E X: x S a for all a E X with the property that b a

for all b E A}.








It follows from Proposition 5.3 and Remark 5.2 that

X = {A c X: D(A) = A) (when partially ordered by inclusion) is a

complete lattice and that the map d: X X defined by a D({a}) is

injective and order-preserving. In view of the equivalence for D(A)

noted in the preceding paragraph and Birkhoff [ 5; Chapter V, 9],

X is the completion of X by cuts.



PROPOSITION 5.4. The injection d: X + X defined by x ~ D({x})

is a lattice inclusion.


PROOF. Since d is injective, we need only show that d is a

lattice homomorphism (2.2 (i)). Suppose that C is a non-void subset

of X such that v C exists in X. It follows from Remark 5.2 that

v d[C] = D(U{d(x): x E C}) = D(C). Let c = v C. Thus C c X (c)
X X D
and hence D(C) D({c}); i.e., v d[C] 5 d(v C). Finally, if a E X
X x
such that C c XD(a), then a is an upper bound of C; hence c = v C : a

so that c e X (a). It follows that d(c) = d(v C) 5 D(C) = v d[C].
D X X
Thus d(v C) = v d[C]. A dual argument shows that if 0 / C c X such
XX
that AXC exists, then d(A C) = A d[C]. Thus d is a lattice
X
homomorphism (2.2 (i)).




Next, we demonstrate that X (when it is equipped with its

interval topology) is also a compactification of X.








PROPOSITION 5.5. For each A E X

(i) A {d(a).: A : d(a)} = A = v {d(a): a E A),

(ii) D(A) = n{ D(d(a)): A r d(a)},

(iii) X (A) = n{X (d(a)): a E A}.


PROOF. Let A E X.

(i) Since a E A if and only if d(a) 5 A, it is clear that

v {d(a): a E A} A A {d(a)? A d(a)}.

To see that v {d(a): a E A} = A, let B E X such that B < A.

Since X is ordered by inclusion there is some a E A such that a / B;

therefore X (a) B so that d(a) i B. Thus v {d(a): a E A} = A.

To see that A {d(a): A d(a)} = A,suppose that B E X such that

A < B. Since D(A) / D(B), it follows from the definition of D that

there is some a e X such that A c X (a) and B X X (a). Thus A 5 d(a)

and B $ d(a). It follows that A = A {d(a): A < d(a)}.


(ii) Since A d(a) implies that XD(A) c X (d(a)), it is clear

that

XD(A) n{X (d(a)): A d(a)}.

To see the other inclusion,suppose that B E X such that B E n{XD(d(a)):

A d(a)}. Thus B : d(a) for each a E X such that A : d(a). By (i),

A = A {d(a): A 5 d(a)} and it follows that B A; i.e., B E X (A).


(iii) Since a E A if and only if d(a) A, (iii) can be proved

by dualizing the argument in (ii).








PROPOSITION 5.6. For each a E X, d[X (a)] = X (d(a)) and

d[XI(a)] = X (d(a)).


PROOF. Let a E X. Since d preserves the order of X, it is

clear that d[X (a)] c X (d(a)), and since X (d(a)) is closed,

d[X (a)] c X (d(a)). To see the other inclusion, let A E X (d(a))

and let W be a basic open set containing A.

If W is of the form X U{X (Bi): i = 1, -**, n} for some

finite set {BI, ***, B} c X, then since A E W, it must be the case

that for each i, A X < B.. Since, by Proposition 5.5 (i),
1
A X = A {d(x): A N : d(x)}, it follows that for each i E {l, -**, n}

there is some b. X such that AX < d(b.) and B. S d(b.). Let
1 1 1
c = (A {b.: i = 1, "*', n}) A a. Then c E X, c < a, and since d is

a lattice inclusion, d(c) : d(a). Thus d(c) e d[XD(a)]. Finally,

note that d(c) e W, for if not, then d(c) e X (B.) for some

i E {1, **', n}, and so B. d(c) : d(b.) -a contradiction. Hence
1 1
W n d[XD(a)] i 0.

Consider now the case that W has the form X U{D(C): j = *,

for some finite set {C, ***, C } c X. Since for each j = 1, **', m,

A e C., and since A < d(a), it follows that for each j, d(a) % C.. Thus

d(-a) e W d[X (a)]. Hence W n d[XD(a)] / 0.

Finally, we consider the case that W = V n U where

V = X U{X (Cj): j = 1, ***, m} and U = X U{X(.): i = 1, n}
Dj 3
for some finite set {B, .**, B, C *, C }c :. Since A c V, it must
1 n 1 m -
be the case that A / A X and so A i 0. Since A i C. and A = v {d(x): x e A}

(5.5 (i)), it follows that for each j {1, **., m} there is some







x. E A such that d(x.) % C.. Let x = v {x.: j = 1, **., m}. Since

d is a lattice inclusion, it follows that for each j, d(x.) d(x).

Also d(x) A, and since A < d(a), d(x) E d[X (a)]. Note that

d(x) e V, for if not, then there is some j E {1, **., m} with

d(x) E X (Cj) and thus d(x.) d(x) 5 C. -- a contradiction. Note

also that d(x) E U, for if not, then there is some i E {1, **', n} with

d(x) E X (B.); thus B. 1 d(x) A so that A i U -- again a contradic-

tion. Consequently, d(x) E W n d[X (a)]. Hence, in each of the

possible cases, W 0 d[XD(a)] X 0, so that, since W was arbitrarily chosen,

A E d[XD(a)]. A dual argument shows that d[X (a)] = X (d(a)).



LEMMA 5.7. The collection T = {(D(d(a)): a E X}
-1
U {X (d(a)): a X} U {X} is a subbase for X, {d [A]: A E T = T ,
-l
and {d-[A]: A T T = TI.


PROOF. Clearly T is a collection of closed subsets of X and

T is contained in the usual subbase for X. Thus to show that T generates

the interval topology on X we need only verify that all principal ideals

of X and principal dual ideals of X,which are not in T, are closed in

the topology generated by T. This follows immediately from Proposition 5.5

parts (ii) and (iii).

To see that {d-l[A]: A T} = TD and {d-l[A]: A T } = T ,
-1 -
note that d-[X] = X and since d is a lattice inclusion-,

d- [X(d(a))] = XD(a) and d-l[X (d(a))] = X (a) for each a E X.







PROPOSITION 5.8. The completion of X by cuts is a compacti-

fication of X; specifically, the mapping d: X X is a topologically

dense embedding.


PROOF. Since X is complete, we know that it is compact (2.9).

That d is continuous follows immediately from the facts that T is a sub-

base for the closed subsets of X and {d-1[A]: A E T} = T (5.7) is a

collection of closed subsets of X.

That d is injective follows from the fact that d is a lattice

inclusion (2.2 (ii)).

To see that d is an embedding, we need only show that d is

closed onto its image. Recall that d is a lattice inclusion and is

(consequently) injective. Thus for each a E X, d[XD(a)]

= D(d(a)) n d[X] and d[X (a)] = X (d(a)) n d[X]; therefore, d[XD(a)]

and d[X (a)] are closed in d[X]. Hence d is an embedding since T is

a subbase for the closed sets of X.

Finally, to see that d[X] is topologically dense in X, recall

that we require that X 1 0. Let A E X. If A X < A, then A / 0 and so

there exists some a E A. Since d(a) 5 A, it follows from Proposition 5.6

that A E d[X (a)] c dX]. If A = A, then let a E X. Since A = A X < d(a),

it follows again from Proposition 5.6 that A e d[X (a c d[X]. Hence

X c d[X]; i.e., d[X] is topologically dense in X.


We return now to the discussion of lattice superextensions.







PROPOSITION 5.9. T is an admissible subbase for X.


PROOF. By Lemma 5.7, T is a subbase for X. To see that

T is a T -subbase for X let A E X. By the definition of T,

n{U e T: A E U} = [n{XD(d(a)): A d(a)}] n [n{X (d(a)): d(a) A}] n X.

Since nr{XD(d(a)): A d(a)} = XD(A) (5.5 (ii)) and r {X (d(a)): d(a) s A}

= {{X (d(a)): a E A} = X (A) (5.5 (iii)), it follows that


l{U e T: A E U} = XD(A) n X (A) = {A}.


Finally, suppose.that A E X and U E T such that A / U. We

consider first the case that U = XD(d(a)) for some a E X. Since A i U,

it follows that A % d(a). Since A = v {d(x): x e A} (5.5 (i)) there is

some x E A such that d(x) % d(a). Thus X (d(x)) e T and it follows

that A E X (d(a)) and X (d(a)) r X (d(a)) = 0. Since x E A if and

only if d(x) A, we may dualize the above argument in the case that

U e T I

Hehce T is a T -subbase for X (2.11 (ii)). Since T consists

of ideals and dual ideals of X, it is an admissible subbase for X (4.2 (i)).



PROPOSITION 5.10. If a, b E X, then X (d(a)) n X (d(b)) = 0

if and only if X (a) n X (b) = 0.


PROOF. Since d[X (a)] c X (d(a)) and d[X (b)] c X (d(b)), it

is easily seen that X (d(a)) n X (d(b)) = 0 implies that

XD(a) n X (b) = 0. Conversely, if XD(a) n X (b) = 0, then since

b % a and d is a lattice inclusion, it follows that d(b) % d(a). Hence

XD(d(a)) n X (d(b))= 0.







The following theorem will have several applications in this

cn-ater and Ch .r7ers 6 and 8.



T 5.11. Suppose that L and M are topological lattices

wi.ch satisfy the following conditions:

(i) f: L M is a continuous lattice inclusion.

(ii) 0 and Y are admissible subbases for L and M, respectively,

such that D = {f-lV]: V E YD} and 0 = {f-[V]: V E }.
D I
(iii) If V1, V2 E such that V1 r V2 i 0, then

f[v ] n f-[v] i 0
f- 1 1 2 '
Then X L is iseomorphic with A M via an iseomorphism f" which

is an extension of f; i.e., such that the following diagram commutes.




e
L > OL


I
f f*


M > AM
e





PROOF. Definition of f*:: Suppose that e X L. Clearly,
.-l
l' = {V E Y: f-l[V] E t} is a linked system of Y. If V a V and

{V} U f' is linked, then by conditions (ii) and (iii), Z U{f-[ V]} is

linked. Since X is a maximal linked system of 0, f- [] e and so
-l
V E '. 7 u-. {V E : f-[EV] c } is a maximal linked system of Y.

We define f*': X L A M by f"*() = {V E Y: f-l[V] e } for each c XA L.







f* is an extension of f; i.e., e o f = f' o e : In view

of condition (ii) we see that for each x E X, f* o e (x)

= f'({W E 0: x E W}) = f*({f-l[V]: x E f-1[V], V E })

= {V e T: f(x) E V} = e (f(x)). Thus the diagram commutes.


f is surjective: For each Mte X M it follows from (ii) and

(iii) that {f- [V]: V E %} is a maximal linked system of 0. By the

definition of f*, f*({f-l[V]: V E 17}) = 1; therefore, f* is surjective.


f* is a lattice inclusion: To see that f* is injective suppose

j,J E X L such that O Thus it follows from (ii) that there are

V, V' E with f-1] E, f-1[V'] E Xf, and f-EV] n f-l[V] = 0. By

condition (iii), V n V' = 0 and since, by the definition of f*,

V E f*(;) and V' E f*(.), it follows that f*(P) / f*(~).

To see that f* is a lattice homomorphism, suppose that C is

a non-void subset of X L, and let ZC = v C. Let V E YD. By the

definition of f*, V E f*(C ) if and only if f-[V] E C. Since

~C = v C, f- EV E C if and only if f-[EV] et for each te C (4.8).

For each t E C, it follows from the definition of f* that f-[EV] c Z

if and only if V E f*(C). Finally, by Proposition 4.8, V E f*(Z) for

each Ce C if and only if V E v f*[C]. It follows from the above

equivalences that V E f*(oC) if and only if V E v f*[C]. Thus

f*() ) N 'D = v f*[C] n gD and consequently, f*({) = v f*[C] (4.7).

A dual argument shows that f* preserves infima of X L. Hence

f* is an injective lattice homomorphism; i.e., f* is a lattice inclusion

(2.2 (ii)).







f" is continuous: It follows from the definition of f": that

for euch f-[V] c 0 and each AeX L, f-1[V] c if and only if

V e f (2). Thus for each V c Y,

f'-l[V+] = { E L: f*(e) V +

E {e X L: V E f"'(g)} (2.14)

= {E X L: f-EV] EX} = (f-lEV)T

which is a member of the subbase for the closed sets of X L. Since

{V : V E ?} is a subbase for the closed subsets of X M, it follows

that fr is continuous.


f" is a closed map: For each V e y, it follows from the above

equality and the injectivity of f* that

f*-[(f-1E[])] = fC[f '*-l[V]] = V+

By condition (ii), {f-l[V]: V E T} = 0 and so {(f-l[V1)+: V E Y} = 0+

is a subbase for the closed subsets of X L. It now follows that the

image under f' of each subbasic closed set in A L is a subbasic closed

set in X M and since f* is injective, f": is a closed map.

Thus f: satisfies all of the conditions of Definition 2.10, and

so f'* is an iseomorphism of X L onto X M.




PROPOSITION 5.12. If L, M, f, 0, and Y satisfy the hypotheses

of Theorem 5.11 and M is complete, then M is iseomorphic with X L via

an iseomorphism which makes the following d a:re'. commutative:




LL

L I
I

NK^ ^







PROOF. In view of Theorem 5.11, we need only show that

e,: M - X M is an iseomcrphism; i.e., that e. is surjective. If

e XYM, then by Proposition 4.5 (ii), 4 is centered. Since M is compact

and X consists of closed subsets of M, it follows that nf / 0. By

Proposition 2.16 (iv), nh is a singleton; i.e., xh. = {a) for some

a c M. It follows that e = e (a).

Thus if f* is the iseomorphism defined in Theorem 5.11, then
-1
S= e o f*': X L M is an iseomorphism with the property that

Soe, e 0 f" o ee = e oe o f = f.




THEOREM 5.13. If X is a topological lattice, then the completion

of X by cuts is iseomorphic with the superextension of X with respect to

the usual subbase T; i.e., X is iseomorphic with XTX.


PROOF. Clearly X and X are topological lattices and d is a

continuous lattice inclusion (5.4 and 5.8). T and T are admissible
-l
subbases for X and X, respectively (4.4 and 5.9),and {d- [A]: A e T} = TD

and {d [A]: A T} = T (5.7). Suppose that VI, V2 T such that

VI 2 V2 0. If V1 and V2 are both in TD or both in T then it
-1 -1
follows from the definition of T(5.7) that d- [V ] (1 d- [V] 2 0.

In the remaining case we may assume that VI eD and V2 TI. It now

follows from the definition of T (5.7) and Proposition 5.10 that

d-1Vl ] 1 d- V] 2 0.








Thus X, X, d, T, and T satisfy the hypotheses of Theorem 5.11,

and since X is complete, it follows'from Proposition 5.12 that X is

iseomorphic with XTX and the following diagram commutes.



eT

X ------ TX






ST





If we restrict our attention to the case where X is totally ordered,

then T becomes a weakly-normal subbase (4.12), and the completion by cuts

(X X) is Hausdorff (2.18 (ii)). The question arises as to how we can

characterize the locally compact, totally ordered spaces for which the

Alexandroff one-point compactification (wX) is the same as the completion

of X by cuts. Before answering this question we need a few preliminary

results.


c
DEFINITION 5.14. For each A c X, D(A)c = {x e X: D(A) c X (x)}.



PROPOSITION 5.15. For any subset A c X, D(A) and D(A) are

closed subsets of X. Furthermore, D(A) duallyy, D(A)c) is either void

or decreasing duallyy, increasing).








PROOF. Let A c X. That D(A) is closed follows from the fact

that D(A) = n({X (a): A c X (a)} (5.3). By Proposition 2.4 (i), D(A)

is either void or an ideal of X; i.e., D(A) is either void or decreasing.

To see that D(A)c is a closed subset of X, we will show that

D(A)c = r{X (a): a E A}. Since A c D(A), it follows from the definition

of D(A)c that for each a E A and each x E D(A)c, a < x so that x E X (a).

Thus D(A) c c {X (a): a E A). To see the other inclusion, suppose
c
that x E X such that x i D(A) Thus A X (x) so there is some

a E A with a0 i x. Hence x i X (a ) and so x i l{X (a): a E A};
c
therefore, D(A) = l{X (a): a E A} is a closed subset of X. It now

follows from Proposition 2.4 (i) that D(A)c is either void or increasing.



LEMMA 5.16. Let X be a totally ordered topological lattice.

(i) For each A c X, D(A) r D(A)c / 0 if and only if v A exists

in X; moreover, if v A exists in X, then D(A) r D(A)c = {v A}.

(ii) For each x e X, v D({x}) = x.

(iii) For each A c X, if vX D(A) exists, then AX D(A)c exists

and vX D(A) = AX D(A)C

(iv) For each A c X, if any of V A, v D(A), or A D(A)c exists

in X, then they all exist and are equal.

(v) For each A c X, if A d[A] = d(x) for some x e X, then

AA = x.


PROOF. (i) Let A c X. Suppose that b = v A exists in X.

Thus A c X (b) so that D(A) c X (b) and hence b E D(A)c. If a E X such

that A c X (a), then b = v A e X (a). Thus b E A{X (a): A c X (a)}

= D(A).









c c
Hence b E D(A) r D(A)c. If x E X such that x E D(A) n D(A)c, then

since x E D(A)c and b E D(A), it follows from the definition of D(A)c

that b E D(A) c X (x) and so b x. In the same way since b E D(A)c

and x e D(A), it follows that x E XD(b); therefore, x < b. Thus x = b

and D(A) n D(A)c = {v A}.

Conversely, suppose now that D(A) h D(A)c 0 and let

b E D(A) n D(A)c. Thus A c D(A) c X (b) so that a < b for each a E A.

If x E X such that x < b, then since b / XD(x) and b E D(A), it follows

from the definition of D(A) (5.3) that A X (x). Thus there is some

a E A such that a X XD(x); i.e., a 4 x. It now follows that b = v A.

Thus D(A) n D(A)c = {v A}.


(ii) Since for each x E X, D({x)) = X (x), it is clear that

x = v D({x}).


(iii) Let A c X such that v D(A) exists in X. Let b = v D(A).

Thus D(A) c X (b) and by the definition of D(A)c, b E D(A)c. Since

D(A)c is increasing (5.15), it follows that X (b) c D(A)c. Recall that

D(A) is closed in the interval topology (5.15) and hence in the order

topology [ 9; Theorem 12]; hence, b = v D(A) E D(A). It follows that

for each x e D(A)c, b < x and so x e X (b); therefore, D(A) c_ X (b).

Thus D(A)c = X (b) and so A D(A)c = b = v D(A).


(iv) Let A c X and suppose that v A exists in X. Let a = v A.

By (i) a E D(A)c so that D(A) c X (a). Also by (i), a E D(A) and since

D(A) is decreasing (5.15), it follows that X (a) c D(A). Thus D(A) = XD(a)

= D({a}). By (ii), v D(A) exists in X and v D(A) = a. Now using (iii)

we see that A D(A)c exists in X and A D(A)c = a.







Suppose now that v D(A) exists in X. By (iii), A D(A)c

exists and A D(A)c = v D(A). Thus we need only show that v A exists

and v A = v D(A). Let a = v D(A). Since D(A) is closed in the

interval topology (5.15), it is also closed in the order topology

[ 9; Theorem 12] and so a = v D(A) e D(A). Since D(A)c is also

closed in the interval topology (5.15) and hence in the order topology,

it follows that a = A D(A)c E D(A)c. Thus a E D(A) n D(A)c and by

(i) v A exists in X and a = v A.

Finally suppose that A D(A)c exists in X and let a = A D(A).
c c c
Since D(A)c is closed, it follows as before that a = A D(A) E D(A)C

Thus D(A) c X (a). Now if x E X such that A c X (x), then by the

definition of D(A), D(A) c X (x) so that x e D(A)c. Thus a : x and

so a E X (x). Hence a E O{X (x): Ac X (x)} = D(A). Thus

D(A) r D(A)c X 0. By (i) v A exists in X and a = v A. By a previous

case v D(A) exists and v D(A) = a = v A = A D(A)c


(v) Suppose A c X such that A d[A] = d(x) for some x e X.

For each a E A, d(x) 5 d(a) and so x a since d is a lattice inclusion.

If c E X such that x < c, then since d is injective, d(x) < d(c) and

there is some d(a) e d[A] such that d(x) 5 d(a) and d(c) A d(a). Since

X is totally ordered (4.12 and 5.13) it must be the case that d(a) < d(c).

Thus since d is injective, x 5 a < c. It follows that x = A A.




The following lemma is well known and so its proof will be


omitted.







LEMMA 5.17. If H is any Hausdorff locally compact space, and

if Y and Z are two Hausdorff compactifications of H with

IY HI = IZ HI = 1, then Z and Y are homeomorphic.



The following theorem characterizes those situations in which

the completion by cuts is the Alexandroff one-point compactification.




THEOREM 5.18. Suppose that X is a locally compact, totally

ordered topological lattice. Then in order for the completion by cuts

of X (X) to be homeomorphic with the Alexandroff one-point compactifi-

cation of X (wX) it is necessary and sufficient that there exists

exactly one subset A c X such that v A and A (X A) do not exist in X.


PROOF. Necessity: If there does not exist B c X such that v B

does not exist in X, then for each B c X, v B = v D(B) = b for some

b E X (5.16 (iv)). Thus D(B) / 0, and since D(B) is a closed ideal

of X (2.4 (i)), it follows that D(B) = d(b). Therefore X = X is compact,

and since X / wX, X i wX -- a contradiction. Thus there is some B c X

such that v B does not exist in X. Let A = D(B). Since A is decreasing

(5.15) and X is totally ordered, it follows-that X A = D(A)c. Since

D(A) = D(D(B)) = D(B), it follows that D(A)c = D(B)c. Thus by Lemma 5.16

(iv), v A and A (X A) do not exist in X and so at least one such subset

of X exists. Suppose now that there is another set C c X such that v C

and A (X C) do not exist in X. Recall that D(C) and A are members of

X and that X is homeomorphic with wX. If D(C) E d[X], then there is







some x E X such that D(C) = d(x). Since d(x) = D({x}), it follows

that v D(C) = x (5.16-(ii)). Hence v C = x (5.16 (iv)), but this

contradicts the fact that v C does not exist in X. Thus D(C) i d[X].

By a similar argument, using the fact that v A does not exist in X,

we see that A e d[X]. Now since IX d[X]I = owX XI = 1 and

D(C), A E X d[X], it follows that D(C) = A. If C / D(C), then since

C c D(C), there is some x E D(C) C, and since v D(C) does not exist

in X, d(x) < D(C) in X. Let G = d[X C]. Then d(x) E G so that G

is a non-void subset of X. Since X is complete, A G exists in X and

A G 5 d(x). Thus A G / D(C), so it must be the case that A G E d[X]

and there is some z E X such that A G = d(z). Hence z = A (X C)

(5.16 (v)); i.e., A (X C) exists in X--a contradiction. Therefore,

C = D(C) = A, and A is the unique subset of X for which v A and A (X A)

do not exist in X.


Sufficiency: Let A c X be the unique set such that v A and

A (X A) do not exist in X. Thus v D(A) does not exist in X (5.16 (iv)),

and since X D(A) = D(A)C, A (X D(A)) does not exist in X (5.16 (iv)).

By the uniqueness of A, A = D(A) and so D(A) / D({x}) for all x E X

(5.16 (ii)); i.e., D(A) = A E X d[X]. If B E X such that B j d[X],

then v B does not exist in X, for suppose that there is some b E X such

that v B = b. Then since B is a closed decreasing subset of X (5.15),

it follows that b E B and XD(b) = B; therefore, B = D({b}) = d(b) E d[X]

--a contradiction. Finally, since v B does not exist in X and B = D(B),

it follows that A (X B) = A D(B)c does not exist in X (5.16 (iv)).

Hence by the uniqueness of A, A = B.








Thus X d[X] = {A} and since X is a Hausdorff compactification

of the locally compact Hausdorff space X formed by adjoining one point

to.X, X is homeomorphic with wX (5.17).





In view of the fact that the open n-cell in Euclidean n-space

can be characterized as the product of n-copies of the open interval

(0, 1), it follows from Theorem 5.13 and the Product Theorem (3.4), that

the closed n-cell (i.e., the n-cell with boundary attached) is a de Groot

compactification of the open disc. Likewise, since Hilbert space can

be characterized as the countably infinite product of the open interval

(0, 1) [3], it follows that the Hilbert cube is a de Groot compactifi-

cation of Hilbert space. (See Example 8.3 for details.)













6. TOPOLOGICAL LATTICE COMPLETIONS



Recall that Theorem 5.13 provided a superextension characteri-

zation of the completion by cuts. The purpose of this chapter is to

obtain algebraic characterizations of other lattice superextensions.

This is done via the notion of topological lattice completions.

Throughout this chapter X will denote a topological lattice.

The topological lattice completions of X are complete topological

lattices having X as a sublattice and a subspace. We show that all

lattice superextensions of X are actually topological lattice comple-

tions of X (Theorem 6.22). Conversely, all topological lattice

completions of X which satisfy a certain density condition are shown

to be lattice superextensions of X (Theorem 6.20).

Throughout this chapter Y will denote a complete topological

lattice (i.e., a complete lattice equipped with the interval topology)

and all subbases considered are assumed to be subbases for closed sets.



-NOTATION 6.1. Let Z be a non-void subset of Y.

(i) Z = Z.

(ii) Y {z E Z: z y y), if {z E Z: z : y) 1 0, or
Zla = {y E Y: y = Z, if {z E Z: z 5 y} = 0


^ {z E Z: y z}, if {z E Z: y 5 z} i 0, or
Zb
= {y E Y: y = Z, if {z E Z: y : z} 0


Z1 = Zla Zlb.







(iii) Z2a = (Zla)lb

Z2b =(Zlb)la

2 2a 2b
z z n z

(iv) 00 la = b 2a 2b



It should be noted that Definition 6.1 can be easily extended

to obtain Za for each ordinal a, and if we define the "lattice closure

of Z in Y" to be Z where a is the least ordinal such that Z = Zl,

then the map which takes Z onto its lattice closure in Y can be shown

to be a Moore closure operator, or completion-operator on Y as in

Definition 5.1. See Appendix II for the details.



PROPOSITION 6.2. Suppose that A C B c Y. Then

(i) Ala cBla

(ii) Ab c Bb

(iii) A1 c B1
1 la lb 2
(iv) A c A c A U A c A2


PROOF. (i) If y E Ala, then {a e A: a < y} c {b E B: b y}

and y = v {a E A: a } y} v {b e B: b y} 5 y; i.e., y e Bla


(ii) Dualize the argument for (i).


(iii) This is immediate from (i) and (ii) above and the defini-

tions of A1 and B1 (6.1 (ii)).


(iv) The proof is immediate if A = 0 (6.1 (iv)); so assume

that A / 0.








That A c A follows from the facts that for each w E A,

w = v a E A: a w} e Al and w = {a E A: w a a} e Al. Clearly,
1 la Ib
A c A la Ab (6.1 (ii)).
la 2 la
To see that Al c A, let w Aa. Thus
la latlb 2a lbTh
w =A {z E A : w z} e (Al ) = A Finally, since A c A it
la (bla 2b
follows from (ii) above that w E A c (Al) = A ; therefore,
2a 2b 2 lIb 2
w E A A = A That A c A follows dually. Thus
1 la lb 2
A c A-c Aa Ab c A2



PROPOSITION 6.3. Suppose that A c Y.
la
(i) If y E A and A n YD(y) X 0, then Y (y) = n({Y(a): a E A

and a y}.

(ii) If y e A and A r Y (y) X 0, then YD(y) = ({YD(a): a E A

and y a a}.


PROOF. Both parts (i) and (ii) follow immediately from the

definitions.



PROPOSITION 6.4. Suppose that A c Y.

(i) If A contains the supreme of all non-void finite subsets

of A and y Ala such that A n Y (y) i 0, then y E (A n Y (y)).

(ii) If A contains the infima of all non-void finite subsets

of A and y E Alb such that A n Y (y) / 0, then y E (A n Y (y)).


PROOF. (i) Let A satisfy the hypotheses of (i) and suppose

that y E Ala such that A n YD(y) / 0. To see that y E (A A YD(y))

let W be any basic open set of Y containing y.







If W = Y U{Y (zi): i = 1, ***, n} for some finite set

{z1 ,*, z ) Y, then since for each i, z. y, it follows that

YD(y) n (U{YI(z.): i = 1, "**, n)) = 0 and so
(A f YD(y)) n (U{Yi(z.): i = 1, **, n}) = 0. Hence

A. n YD(y) C W and W n (A n YD(y)) / 0.

If W = Y U{YD(wj): j = 1, ***, m} for some finite set

{w ***, w } c Y, then for each j = 1, **', m, y 4 w.. Since
I- m]
y = v {a E A: a 5 y}, it is possible to pick,. for each j E {1, **', m},

some x. E A with the property that x. y and x. 4 w.. Let

a = v {x.: j = 1, ***, m}. By the hypotheses on A and since each

x. e A, it follows that a E A and a 5 y; i.e., a E A F YD(y). If

a i W, then there is some j E {l, ***, m} with a E YD (W); therefore,

x. a 5 w. and x. w.--a contradiction. Thus a E W and so

W n (A n YD(y)) $ 0.

In the remaining case W = U C V where

U = Y U{YD(w ): j = 1, ***, m} and V = Y U{Y (2z): i = 1, ***, n}

for some finite set {wl, '*, w, zl, *, zn }c Y. By an argument

analogous to the one above, there is some a E U n (A n YD(y)) and

since A n YD(y) c V, it follows that a E W n (A n Y (y)).

Since W was arbitrarily chosen, it follows that y E (A n YD(y)).


(ii) Dualize the argument for (i).



PROPOSITION 6.5. If i: X Y is a lattice inclusion and if

i[X]1 = Y, then v i[X] = v Y and A i[X] = A Y.








PROOF. Clearly v Y e Y = i[X]1 c i[X]la (6.2 (iv)). Thus

v Y = v {i(x) E i[X]: i(x) : v Y} = v {i(x) E i[X]} = v i[X]. The

other equality is proved dually.



NOTATION 6.6. If i: X Y is a lattice inclusion such that

Y = i[X]1, then we let

= {Y} u{YD(i(x)): x E X} U{Y (i(x)): x E X).



PROPOSITION 6.7. If i: X Y is a lattice inclusion such that
1 A
i[X] = Y, then T is an admissible subbase for Y.


PROOF. Clearly T consists of ideals and dual ideals of Y. Since

Y = i[X] it follows from Proposition 6.3 that each member of the usual
A A
subbase for Y is an intersection of members of T; therefore T is a

subbase for the closed subsets of Y.
A
To see that T is a T -subbase for Y, suppose that z E Y. Then

since z E i[XI X] la ni[X]lb, it follows that
A
){T e T: z E T} = (n{YD(i(x)): x E X and z i(x)}) )

(n{Y (i(x)): x E X and i(x) z}) n Y

= YD(z) n Y (z) (6.3) = {z}.
A
Finally suppose that z E Y and T E T such that z i T. Clearly

T / Y. If T E D, then there is some t E X such that T = YD(i(t)).

Since z i i(t) and z E i[X]1 c i[X]la (6.1 (ii)), there is some x E X
A
such that i(x) z and i(x) 4 i(t). Thus z e Y (i(x)), Y (i(x)) E T,

and YD(i(t)) ) Y (i(x)) = 0. If T E then a dual argument shows that







A
there is some YD(i(x)) E TD with z E YD(i(x)) and T fi YD(i(x)) = 0.
A A
By Definition 2.11 (ii), T is a T -subbase for Y. Hence T is an

admissible subbase for Y (4.2 (i)).



PROPOSITION 6.8. If i: X Y is a lattice inclusion such

that Y = i[X]1, then {il[T]: T E TD TD and {i-l T]: Te E } = TI

where T is the usual subbase for X.


-1 A -1 A
PROOF. Clearly TD c {i [T]: T E T } and T c {i [T]: T E T .

To see the other containments, recall that i is a lattice inclusion;

therefore, for each a E X, i-[Y (i(a))] = {x E X: i(x) 5 i(a)}

-l
= {x E X: x : a} = XD(a) e TD and dually, i-[Y (i(a))] = X (a) e T



PROPOSITION 6.9. Suppose i: X Y is a lattice inclusion such
1 A -1 -1
that i[X] = Y. If S, T E T, then i [S] n i [T] / 0 whenever S n T X 0.


A
PROOF. Suppose S, T E T such that S r T / 0. If S and T are
A A
both in TD or both in TI, then the conclusion is immediate from the defi-
A
nition of T (6.6) and the fact that i is a lattice inclusion. For the

remaining case we may assume that there exist t and s in X such that

T = Y (i(t)) and S = YD(i(s)). Since S f T / 0, we see that

i(t) 5 i(s) and since i is a lattice inclusion, t S s. Thus

i-l[S] n i-l[T] = XD(s) n Xi(t) t 0.



DEFINITION 6.10. Suppose that X and Y are topological lattices,

Y is complete, and i: X Y is a function. Y will be called a topological







lattice completion of X with respect to i whenever the following

two conditions are satisfied:-
1 2
(i) Y = i[X] or Y = i[X]

(ii) i is a continuous lattice inclusion.



PROPOSITION 6.11. If i: X Y is a lattice inclusion and

i[X] = Y, then Y is a topological lattice completion of X with respect

:to i; i.e., the map i is continuous.

^-1 A
PROOF. Since T is a subbase for Y (6.7), and {i-l[T]: T E T}

is the usual subbase for X (6.8), it follows immediately that i is

continuous.



PROPOSITION 6.12. If Y is a topological lattice completion of

X with respect to i: X Y, then i is an iseomorphism onto its image;

i.e., i is an embedding.


PROOF. In view of the definitions of topological lattice

completions (6.10) and lattice inclusions (2.2 (ii)), we need only show

that the restriction i: X i[X] is a closed map. For each a E X,

i[X (a)] = {i(x): x E X and x a}.

= {i(x): x E X and i(x) i(a)} (since i is a lattice inclusion).

= i[X] fYD(i(a)).

Dually, for each a E X, i[X (a)] = i[X] r)Y (i(a)). Thus the image

of every member of Tis a closed set in i[X], and since i is injective,

it follows that the image of each closed set in X is closed in i[X].








THEOREM 6.13. If i: X Y is a lattice inclusion and

i[X]1 = Y, then Y is iseomorphic with the completion of X by cuts.


PROOF. By Proposition 6.11, the function i: X Y is a
A
continuous lattice inclusion. T and T are admissible subbases for

X and Y, respectively, such that (i-1 [T: T E } = T and

{i-l[T]: T E } = T (6.7, 6.8, and 4.4). If T, S E T such that

T r S / 0, then i-lT] n i-l[S] / 0 (6.9). Thus X, Y, i, T, and

satisfy the hypotheses of Theorem 5.11, and since Y is compact it

follows from Proposition 5.12 that Y is iseomorphic with XTX, the

completion of X by cuts (5.13).




REMARK 6.14. It follows from Theorem 6.13 that for any topo-

logical lattice X, there is only one completion Y of X for which

Y = X. This occurs just when X is a complete lattice. In the case
2 1
where X c Y and X2, but not X is a complete sublattice of Y, the pos-

sibilities are not so limited. In fact, in this case a lattice inclusion

need not be continuous (see Example 8.6, Part I); however, for the

remainder of this chapter we will consider only topological lattice

completions of X.

In order to use Theorem 5.11 (as was done in the proof of

Theorem 6.13) to characterize one of these "larger" topological lattice

completions as a lattice superextension, it is necessary to obtain

a subbase which satisfies certain conditions. Namely, if Y is a







topological lattice completion of X with respect to i, then we must

choose some subbase E for Y such that

(i) E is an admissible subbase for Y,
-1
(ii) {i-[S]: S E E) is an admissible subbase for X, and

(iii) if S, T E E with S 0 T / 0, then i- SI] r i-l[T] / 0.

Keeping in mind conditions (i) and (ii) we make the following

definition.



NOTATION 6.15. If Y is a topological lattice completion of X

with respect to i, then

S() = {Y} U{YD(y): y E (iX]la i[X]I) U i[X]}

u{ (y): y e (i[x lb i[X]l) U i[X]}.



PROPOSITION 6.16. If Y is a topological lattice completion of

X with respect to i and Y = i[X1, then (i) T.


PROOF. This follows immediately from the fact that when

Y = i[X]1, then i[X]1 = i[X]la = i[X]b and so i[X]la i[X]1

i[X]Ib i[X]1 = 0.



PROPOSITION 6.17. If Y is a topological lattice completion

of X with respect to i, then

(i) (i) is an admissible subbase for Y, and

(ii) {i-l[S]: S E Z ()} is an admissible subbase for X.






PROOF. (i) Clearly Z consists of principal ideals and prin-

cipal dual ideals of Y. In order to see that E ) is a subbase for Y,

we need only show that all principal ideals of Y and principal dual
(i)
ideals of Y which are not in E are closed in the topology on Y

which has (i) as a subbase. If YD(y) is a principal ideal of Y and

Y (y) Z(i), then it follows from the definition of (i) that either

(a) y E i[X] or

(b) y E i[X]2 (i[X]la U i[X]lb).

Dually, if z E Y such that Y (z) I ) then either

(c) z e i[X] or

(d) z e i[X]2 (i[X]la U i[X]b).


(a) If y E i[X] then y E i[X] and so by Proposition 6.3 (ii),

YD(y) = n{YD(i(x)): x E X and y 5 i(x)}.


(c) If z E i[X] then z i[X]la, and by Proposition 6.3 (i),

Y (z) = n{Y (i(x)): x E X and i(x) < z}.


Thus for each w e i[X] Y (w) and YD(w) are closed in the topology

generated by Z( Hence in view of the definition of ) (6.15), for
la lb
each w E i[X] U i[X] Y (w) and Y (w) are closed in the topology
(iD
generated by E

2 2a 2a ab
(b) If y E i[X]2, then y E i[X]2a, and since i[X]2 = (i[X]la)b,

YD(y) = M{YD(w): w E i[X]la and y w} (6.3 (ii)).


(d) If z E i[X]2, then z E i[X]2b, so that since i[X]2b = (i[X]lb) l

Y (z) = n{Y (w): w E i[X]lb and w 5 z} (6.3 (i)).







It now follows from parts (b) and (d) that if q e i[X]2 (i[X]la U i[X]b),

then YD(q) and Y (q) are intersections of closed sets in the topology on
M (i)
Y generated by E and hence are closed in that topology. Thus E

is a subbase for the interval topology on Y.

(i)
To see that E is a T -subbase for Y, let z e Y and

= {SE E ): z E S}. Clearly R is linked. To see that P is a

maximal linked system of ), suppose that T () such that Ti {.
(i)
Then z i T. We will assume that T E D and note that a dual argu-

ment should be used when Te Thus there is some y E (i[X]la i[X)

U i[X] such that T = YD(y) and z 4 y. "We consider the possibilities

for z.

(e) z e (i[X]lb i[X]1) U i[X].

(f) z e i[X la

(g) z e i[X]2 (iCX] a i[X]b).

lb 1 (i)
(e) If z e (i[X] i[XI) U i[X], then S = Y (z) E E z e S, and

S n Y(y)= Y(z) n Y (y) = 0.


(f) If z e i[X]a, then since z 4 y, it is possible to pick some

w E i[X] such that w < z and w 4 y. Thus S = Y (w) E z e S, and

S ) YD(y) = YD(w) n YD(y) 0.


(g) If z e i[X]2 (i[X]la U i[X]lb), then since z e i[X]2b and

z 4 y, there is some q e i[X]l such that q.s z and q i y. If

q / i[X]l, then S = YI(q) E E(), z e S, and S = YDI(q) 0 YD() .

If q e i[X]l i[X]a, then since q 4 y, there is some w E i[X] such that
(i)
w q and w f y. Hence S = Y (w) E z E S since w < q < z, and

S n Y() Y = (w) n Y (y) = 0.








It follows from the arguments in (e), (f), and (g) that in all
(i)
cases there is some S E( such that z E S and S f T = 0. Clearly

S E H so that f U{T} is not linked. Thus f is a maximal linked system

of E By Proposition 2.13, E is a T -subbase for Y; therefore,

E() is an admissible subbase for Y (4.2 (i)).


(ii) Let E = {i-1 T]: T E (}. Since i is continuous, it

is clear that E is a collection of closed subsets of X. Also, since
-1 (i)
i is a lattice inclusion, it follows that {i [T]: T E E } is a

collection of ideals of X and {i- [T: T E E i)} is a collection of

dual ideals of X. Thus E c F. It follows from the definition of

E ) (6.15) that E contains all the principal ideals and principal

dual ideals of X. Hence by Proposition 4.3, E is an admissible subbase

for X.





In view of Proposition 6.17, the subbase () which we have

chosen for Y (where Y is a topological lattice completion of X with

respect to i) satisfies the first two conditions in Remark 6.14. It

is not true, however, that Zi) necessarily satisfies the third condi-
(i)
tion; i.e., it is not always true that if S, T E ) such that

S r T / 0, then i- [S] f i-l[T] X 0. An example of this can be seen

in 8.8. Thus we are lead to define the notion of "lattice density."







DEFINITION 6.18. If Y is a topological lattice completion of

X with respect to i: X Y, then we say X is lattice dense in Y when-

ever the following condition is satisfied:

If y e i[X]lb i[X] and z e i[X]la i[X]1 with the property

that y < z, then there is some x E X such that y < i(x) < z.



It should be noted that if Y = i[X] then X is lattice dense

in Y since i[X]la i[X]1 = i[X]b i[X1 = 0.



PROPOSITION 6.19. Suppose that Y is a topological lattice

completion of X with respect to i: X -+ Y and X is lattice dense in Y.

If S, T E (i) such that S n T / 0, then i-l[S] n i-1T ] 0.

-l
PROOF. If either T = Y or S = Y, then since i- [Y] = X, the
(i) (i)
proof is trivial. If T and S are both in Z or both in E then

since {i- [R]: R E } i)} is a collection of closed ideals of X and

dually {i [R]: R E } is a collection of closed dual ideals of X,

the proof again is trivial (4.5 (i)). In the remaining case we may

assume without loss of generality that there exist t E (i[X] i[X] )

U i[X] and s (i[X]lb i[X]1) U i[X] such that YD(t) = T and

Y (s) = S. Since S T / 0, 'it must be the case that s t. If

s e i[X], then s = i(x) for some x E X and x e i-l[Y (i(x))] 0 i- YD(t)]

= i-1[S] n i-l[T]. Dually, if t E i[X], then t = i(x) for some x E X

and x e i [Y (t)] ) i [Y (s)] = i [CS] i-l[T]. Thus suppose that

s, t i[X]. Since neither s nor t is in i[X] and s s t, it follows







that s < t. Now since X is lattice dense in Y, there is some x E X
-1 -1
with s < i(x) < t; therefore, x e i [Y (t)] n i [Y (s)]
-1 -1
Si-l[ES] i-lT].




THEOREM 6.20. Given that Y is a topological lattice completion

of X with respect to the inclusion map i: X Y, X is lattice dense in

Y, and E = {i -[TI: T E ) }, then Y is iseomorphic with X X via an

iseomorphism f: X X Y with the property that f o e = i.


PROOF. By the definition of topological lattice completions

(6.10), i is a continuous lattice inclusion. E and ) are admissible

subbases for X and Y, respectively, such that {i [S]: SE E( } =
-1 (i) (i)
and {i [CS: S E E(} = E (6.17). If S, T E E such that

S r T / 0, then i- [S] r i-l[T] / 0 (6.19). Thus X, Y, i, E, and

E) satisfy the hypotheses of Theorem 5.11. Since Y is complete,

there is an iseomorphism f: X X Y with the property that f o e = i

(5.12).





We now consider the converse question; i.e., which lattice

superextensions are topological lattice completions of X in which X is

lattice dense?



PROPOSITION 6.21. If E is an admissible subbase for X, then
for eac+ e E[x la + lb
for each T E % duallyy, T E C ), VT e [XI duallyy, A T E e [XI ).
I







PROOF. Suppose that T E D. Thus v T = v e [T] (4.10 (iv)).

It follows that v T = v {e (x): e (x) e e [X] and e (x) v T };

therefore, v T E e [Xla. A dual argument shows that if T e E, then
AT [X]lb
A T E e e.EXl




THEOREM 6.22. If E is an admissible subbase for X, then

X X is a topological lattice completion of X with respect to e : X X X.


PROOF. Let E be an admissible subbase for X. Recall that e

is a continuous lattice inclusion (2.16 (v) and 4.9) and that for each

e X X,

v {A S: S e : } = n = A {V S S e: ED} (4.10 (vii)).

Thus since for each S E Er ,i A S+ < X (4.10 (vi)) and

A S E e EX]b (6.21), it follows that
b2b
S= v { E XX: e[X]lb and E '} E (e [X]lb)la = e [X]2b

Dually, since for each S E f n D, v S+ E e [X]la (6.21) and < v S+

(4.10 (vi)), then

S= A { : XEX: p ee [X]la and < .} e (eJ[X] la ) = e [X]2a
2a 2b X2
Thus for each e E XX, E e X] 2 e EXI = e [XE ; i.e., EX = e X]

and so X X is a topological lattice completion of X with respect to

e X t X.




We conclude this chapter by exhibiting those lattice super-

extensions of X in which X is lattice dense.







DEFINITION 6.23. Suppose that E is an admissible subbase

for X. Then for each e E X, we let

X(D,f) = {x e X: e (x) : Z} and

X(I,f) = {x E X: X 5 e (x)}.



REMARK 6.24. It is easily seen from Definition 6.23 that for

each I e X,

v e [X(D,X)] 3r/ A e [X(I,e)].

Hence exactly one of the following statements is valid:

(i) v e [X(D,X)] = A = A e [X(I,.)]

(ii) v e [X(D,i)] = Z < A e [X(I,4)]

(iii) v e [X(D,.)] < z = A e [X(I,.)]

(iv) v e [X(D,Z)] < Z < A e [X(I,Z)].

Clearly for each E X EX, X(D,;) is a closed ideal of X which

may or may not be in E and X(I,X) is a closed dual ideal of X which may

or may not be in E.



DEFINITION 6.25. If E is an admissible subbase for X, then E

will be called thick if and only if for each pair , ) E X X such that

/ satisfies condition (ii) in Remark 6.24 and V satisfies condition (iii)

in Remark 6.24, it follows that 0 = E U{X(D,X), X(I,w)} is an admissible

subbase for X and there exists an iseomorphism f: X X X X such that

f o e = e .







THEOREM 6.26. If E is a thick, admissible subbase for X,

then X is lattice dense in A X.


PROOF. To see that X is lattice dense in A X, let ', t~l A X

such that e e [X]la e [X]1, 1 e X]lb e[X]l, and < <. Thus

o = v {e (x): x e X and e (x) 5f} = v e [X(D,Y)] (6.23).

Since Z I e [X] it must be the case that

< A {e (x): x E X and j e (x)} A e [X(I,f)] (6.23).

Hence X satisfies condition (ii) of Remark 6.24. By a dual argument

we see that A satisfies condition (iii) of Remark 6.24. Thus since E

is thick, X X is iseomorphic with X X where 0 = Z U{X(D,Z), X(I,N)} (6.25).

Let f: X X A X denote the iseomorphism. Since f is a lattice

isomorphism, f o e = e0 (6.25), and X(D,.) E 0D, it follows that

f(L) = f(v e [X(D,V)]) = v f[e [X(D,Y)]] (2.2)

= v e [X(D,)] = v X(D,f)+ (4.10 (iv)).

Dually, since X(I,M) E 0I,

fO() = A X(I,A)+.

Thus X(D,Z) e f(X) and X(I,n) e f(y) (4.10 (vi)). Since f preserves

the order of A X (2.5) and Yt < V, it follows that f(m) f(Z) and so

X(I,m) e f(m) n 01 c f(x) n OI (4.7);

therefore X(D,C), X(I,z) e f() and so X(D,;) 0 X(I,') / 0.

Let x E X(D, ;) X(I,n). Then

f(4) S e (x) 5 f( ).

Since f o e e we have f(M) f(e (x)) < f(Y), and since f is a lattice

isomorphism, yZs e (x) X. Finally, since i~ e [X] and X1 e [X], it

follows that M< e (x) < Z. Consequently, X is lattice dense in A X.
ZJ









PROPOSITION 6.27. If Y is a topological lattice completion

of X with respect to i: X Y and X is lattice dense in Y, then

E = {i-l[S]: S E (i)} is a thick, admissible subbase for X.


PROOF. Recall that E is an admissible subbase for X (6.17 (ii)).

By Theorem 6.22, Y is iseomorphic with A X. To see that E is a thick

subbase for X, suppose that f, ln E X X such that Z satisfies condition (ii)

,of Remark 6.24 and )4 satisfies condition (iii) of Remark 6.24. Since

v {e (x): x E X and e (x) 5-}

= < A {e (x): x E X and 5 .e (x)} (6.23, 6.24),
la 1 (i)
it follows that X E e X -e X. Thus {e X: } E ZD (6.15),

and so X(D,X) = i-l[E{ XEX: Z}] e ZD. Dually, tE e [X]lb e [X]1,
(i) -1
SE XE X: A :5 } e and so X(I,M) = i 1[{ EX X: Y W] e E .

Consequently, 0 = E U{X(D,); X(I,%)} = E is an admissible subbase for

X, and there is an iseomorphism f: A X X @X, namely, the identity map,

such that f o e = e Hence E is thick.



PROPOSITION 6.28. The usual subbase T and the subbase F are

both thick.


PROOF. To see that T is thick, we recall that X X is the

completion by cuts of X so that X X = e [X] Thus it follows that

every member of A X satisfies condition (i) in Remark 6.24. Hence T

vacuously satisfies the definition of a thick subbase.








To see that F is thick, suppose that , mY A X such that

Satisfies condition (ii) of Remark 6.24 and M satisfies condition (iii)

of Remark 6.24. Since X(D,/.) is a closed ideal of X and X(I,Y, is a

closed dual of X, then X(D,4) E FD and X(I,M E e (4.1 (ii)). Thus

0 = r U{X(D,;), X(I,,)} = Fis an admissible subbase for X (4.4) and

the identity map f: AX X A X is an iseomorphism such that f o e = e .



COROLLARY 6.29. X is lattice dense in A X and A X.


PROOF. Since both T and r are thick (6.28), admissible sub-

bases for X (4.4), the proof is immediate from Theorem 6.26.




All of the admissible subbases which are considered in the

examples (Chapter 8) are thick subbases for X. Whether or not every

admissible subbase for X is thick remains an open question.




THEOREM 6.30. Suppose that X is a topological lattice and

0 is an admissible subbase for X. Then X is lattice dense in A X

if and only if there exists a thick admissible subbase E for X such

that AnX is iseomorphic with AX via an iseomorphism f: A X AXX

with the property that f o e = e .


PROOF. Let 0 be an admissible subbase for X. Thus A X

is a topological lattice completion of X with respect to

eg: X A X (6.22).




67

-1 6
If. X is lattice dense in X X, then E = {e [S]: S e 0}

is a thick admissible subbase for X (6.27) and X X is iseomorphic

with X X via an iseomorphism f: X X A X with the property that

f o e = e (6.20).

Conversely, if there is some thick admissible subbase E for

X and an iseomorphism f: XX X X such that f o e = eg, then since

X is lattice dense in X X (6.26), it follows that X is lattice dense

in X X.















7. LARGEST AND SMALLEST LATTICE SUPEREXTENSIONS


In this chapter we will consider the possible existence of

certain maps between different lattice superextensions of the same

topological lattice. The results will then be used to establish the

existence of maximal and minimal lattice superextensions for all

topological lattices and in certain cases, largest and smallest lattice

superextenions (Theorem 7.15). Thus in the realm of lattice super-

extensions we have analogues to the Stone-Cech and Alexandroff compacti-

fications. Kaufman [12] obtained similar analogues for the totally

ordered case. Theorem 7.15 is a generalization of Kaufman's results.

In this chapter, as in the previous chapters, X will denote a

topological lattice. Also, to avoid confusion, we will sometimes use

the symbol "*" in place of "+" to denote the subbasic closed sets of

a lattice superextension; e.g., if 0 is an admissible subbase for X,

then for each T e 0,

T* = {f E X X: T e } and 0* = {T*: T e O}.



DEFINITION 7.1. Let 0 and E be admissible subbases for X.








(i) We say that X X algebraically dominates X X if and only

if there is a surjective, order-preserving function f: X X -+ X X such

that the diagram


X X






e I

Xx


commutes (i.e., f o e = e ) and such that f preserves the supreme

of ideals in X and the infima of dual ideals in X; i.e., if I is an

ideal of X and D is a dual ideal of X, then f(v e [I]) = v (f o e )[I]

and f(A e [D]) = A (f o e )[D]. Such a function is called a dominating

function of X X over X X.

(ii) We say that the superextension X dominates the super-

extension X X if and only if X X algebraically dominates X X via a

continuous dominating function.



It might be noted that although X X and X X are both lattices,

a dominating function f is not required to preserve all supreme and

infima of X X; i.e., f is not required to be a lattice homomorphism.

The reason for this is apparent from Example 8.9 in which a topological

lattice X is exhibited for which the lattice X X algebraically dominates

the lattice X X (and also for which the superextension X X dominates







the superextension X X), but for which there are X, E X FX such

that f( ) A f(nz) / f( A nt). (See 8.9 for details.) Example 8.7

shows that algebraic domination does not necessarily imply superexten-

sion domination, for in this example there exists a unique algebraic

dominating function f: A X X X which is not continuous.



PROPOSITION 7.2. Suppose that 0 and E are admissible subbases

for X such that X X algebraically dominates X X via a dominating

function f: XX X 0X. Then

(i) for each R E D duallyy, R E i), f(v R+) = v e[R]

duallyy, f(A R+) = A e [R]),

(ii) for each R EED OD duallyy, R E Z 0 I), f(vR+) = v R;

duallyy, f(A R+) = A R*), and

(iii) for each X E XEX, (t nO) c f(.).


PROOF. () Suppose that R E D. Then by the hypotheses on

f and since R is an ideal of X, it follows that

f(v R+) = f(v e [R]) (4.10 (iv))

= v (f o e )[R] = v e [R].

If R E I, then a dual argument shows that f(A R ) = A e [R].


(ii) Suppose that R E D n D. Then since f(v R+) = v e [R]

((i) above) and v e [R] = v R* (4.10 (iv)), we see that f(v R+) = v R*.

Dually, if-R E EI r 0 then f(A R ) = A R*.

(iii) Suppose that X E X X. Let R E (; n 0 ). Thus < v R+

(4.10 (vi)), and since f is order-preserving, f(G) < f(v R ) = v R*

((ii) above);i.e., R E f(Z) (4.10 (vi)). Hence (t fl 0D) < f(4), and by

a dual argument, (V n 0 ) c f()). Therefore, ( n 0) c f(A).







LEMMA 7.3. If E is an admissible subbase for X, then for

each a E X, v e [XD(a)] = e (a) and A e [X (a)] = e (a).


PROOF. Let a E X. Then for each x E XD(a), x : a so that

e (x) 5 e (a) since e is a lattice inclusion (4.9); therefore,

v e [XD(a)] < e (a). Since a E XD(a), it follows that e (a) = v e [XD(a)].

A dual argument shows that e (a) = A e [X (a)].



PROPOSITION 7.4. If 0 and E are admissible subbases for X such

that 0 c_ then A X algebraically dominates A X.


PROOF. We define f: A X A X by

f(i) = v {A e [S]: S (4 fE )} for each E C X.


foe e: Let x E X. For each S ee (x) n E x E S so
--L --
that A e [S] < e (x); therefore, f(e (x)) = v {A e [S]:

S E e (x) ) } < e (x). Now suppose that T E e (x) n 8E. Thus

x E T and since T e 0 I c it follows that T E e (x). Hence by

Proposition 4.10 (iv) and the definition of f, A T* = A e [T] < f(e (x));

i.e., T E f(e (x)) f 01 (4.10 (vi)) and so e (x) 5 f(e (x)) (4.7).

It follows that f o e (x) = e (x) for each x e X.


f preserves the order of A X: Suppose that Y, 4 E A X such that

/ r7. Thus (. E ) c () r ) (4.7) and so f(f) = v {A e [S]:

S E n Z } < v {A e [S]: S E 7 ,t E } = f(Y>).







For each E X X, ( t( 0) c f(f): Let e E X'X. If

T E t 01, then by the definition of f, A e [T] f(/). Since

A e [T] = A T* (4.10 (iv)), it follows that T E f(C) (4.10 (vi)),

and so (n A 0 ) c f(C). If T E ( r 0 D), then for each S E (.e 0r ),

T f S ? 0. Thus there is some x E S n T and so A e [S] 5 e (x) 5 v e [TI.

It follows that

f(R) = v {A e [S]: S E (X n zI)} V e [T] = v T* (4.10 (iv));

therefore, T E f(f) (4.10 (vi)) and so ( n 0 ) c f(4).


f is surjective: If PZE @0X, then Yis a linked system of E

and hence is contained in some maximal linked system ) of E. Since

/1 c_ (Y n 0) c f(e), it follows from the maximality of M that f(/) = r.


f preserves supreme of ideals and infima of dual ideals of X:

Suppose that I is an ideal of X and Z = v e [I]. Since f preserves the

order of X, we see that for each x e I, f(e (x)) = e (x) 5 f(;) and

so v e [I] 5 f(). If T E (v e [I] n 0 ), then for each x E I,

T e e (x) (4.8) and so x e T; consequently, I c T. Since T E ED, it

follows that T E v e [I] = Z (4.8). Thus T E (P n 0) c f(4) and so

v e [I] 0D c_ f(1). By the definition of the order on X X (4.7),

f(.) v e [I]; therefore f(W) = f(v e [I]) = V e [I] = v (f o e )[I].

A dual argument shows that f preserves the infima of dual ideals of X.


In view of the above arguments, X X algebraically dominates X X.







PROPOSITION 7.5. Suppose that 0 and E are admissible subbases

for X such that 0 c E.

(i) If for each f EA X, (f 0 0 ) is contained in a unique

maximal linked system of 0, then there is a unique dominating function

f: XX X+ X. Moreover, f may be defined by f(e) = {T 0:

(r n o) U{T} is linked} for each X E X X.

(ii) If for each t E XEX,(t n 0) is a maximal linked system

of 0, then there is a unique and continuous dominating function

f: XAX A X.


PROOF. (i) Since 0 c E, there is a dominating function

f: A X + X (7.4). Suppose that g: A X A X is also a dominating

function. Thus for each E Xe X, (f n 0) c g(t) and (' n e) c f(.C)

(7.2 (iii)). Since (9 n 0) is contained in a unique maximal linked

system of 0, it follows that g(V) = f(X) for each Z E X; i.e., f = g

and f is unique.

To-see that f can be defined as stated, let X E X X and

S= {T e 0: ( i n 0) u{T} is linked}. To see that f is linked, let

S, T H. Then (.t n ) U{S) is linked and ( 0n ) U{T) is linked.

Since ( A 0) is contained in a unique maximal linked system ) of 0,

it. follows that ( n 0) u{S} and ( n 0) u{T} are both subsets of -.

Thus S n T / 0 and ffis linked. To see that f is a maximal linked

system of 0, suppose that T E 0 such that 0 u{T) is linked. Thus

(0 0) U{T) is linked, and so by the definition of f, T E Sf.

Since both ff and f(Z) are maximal linked systems of which

contain ( An 0) and (X 0 0) is contained in a unique maximal linked

system of 0, it follows that f(d) = f.








(ii) If for each e XA X, (4 ( 0) is a maximal linked

system of 0, then (. 0 O) is contained in a unique maximal linked

system of 0. By (i) above, there is a unique dominating function

f: A X X X defined by f(t) = {T E 0: (x n O) U{T} is linked} for

each E A X. Since (i n 0) is a maximal linked system of 0, it follows

that f(1) = ( nl 0). Thus for each T e 0,
-l
f-l[T* = {X e X: f(4) e T*} = { e XX: T E f()}

= E X X: T E } = T+ E

Since 0* is a subbase for the closed sets of X X, it follows that f

is continuous.



DEFINITION 7.6. If 0 and Z are admissible subbases for X,

then we say that 0 is equivalent to E whenever X X is iseomorphic

with XX via an iseomorphism f: X X X such that f o e = e .



PROPOSITION 7.7. If 0 and E are admissible subbases for X such

that each of AOX and X X algebraically dominates the other, then 0 and

E are equivalent. Specifically, we will show that if f: XAX XAX and

g: X X X X are dominating functions, then

(i) g o f is the identity on X X and f o g is the identity on X X

(ii) f and g are injective

(iii) f and g are closed maps

(iv) f and g are homeomorphisms

(v) f and g are lattice isomorphisms

(vi) X X is iseomorphic with X X.
i, ---- --- --- 9







PROOF. The hypotheses describe the following diagram:


EX




g


where g o e = e and f o e = e .


(i) Let R E D. Thus R is an ideal of X. By the properties

of dominating functions (7.1 (i)) and Proposition 4.10 (iv),

g 6 f(v R ) = g o f(v e ER]) = g(v f o e[R]) =

g( v e [R]) = v go e [R] = V e [R] = v R+.

Dually, for each Re c g o f(A R+) = A R+. Now let c e X X. If

R E x E ZD, then Z v R (4.10 (vi)), and so f(Z) < f(v R ) since f

is order-preserving. Thus g o f(Y) : g o f(v R+) = v R since g is

order-preserving. Hence R e g o f(L) (4.10 (vi)); i.e.,Z n ED c g o f(g).

Dually, f E c g o f(0) so that = g o f(Z) and g o f is the identity
I-
on A X.

A similar argument shows that f o g is the identity on X X.


(ii) If t and YIare elements of X X such that f(Y) = f(7), then

by (i) above t = g o f(Z) = g o f(lz = Y. Thus f is injective.

A similar argument shows that g is injective.







(iii) Let R E D. If Z E R+, then R E and so X v Rt (4.10 (vi)).

Since f is order-preserving, f(;) f(v R+) = v e [R] (7.2 (i)). Thus

f[Rt] c Q E X0X: < V e [R]}.

Conversely, if a E X X such that I v e [R], then since f is surjective,

there is some o E X X with f( ) = .. Now since g is order-preserving

and g o f is the identity on X X, it follows that / = g o f(4) =

g(Y) 5 g(v e [R]). Thus by the properties of g (7.1 (i)),

S5 g(V e [R]) = v (g o eQ)[R] = v e [R] = v R+ (4.10 (iv)).

Hence R E Z (4.10 (vi)) and so $ ER : therefore

) E XX: g v e0[R]} = fE[R] = g-l[R+].

A dual argument shows that if Re E then

g-[1R+] = fE[R] = { e gX: A eO[R] :}.

Since E is a subbase for the closed sets of X X, it follows from the

above qualities that f-1 = g is continuous. Hence f is a closed map.

A similar argument shows that g is a closed map.


(iv) Since each of f and g is closed, continuous, and bijective,

each must be a homeomorphism.


(v) Suppose that C is a non-void subset of X X and = v C.

Since f preserves the order of X X, it is clear that /Z= v {f( ):

E C} I f(). To see that f(4) 179 let T X n E D. Thus M7Z< v T*

(4.10 (vi)) so that for each ) E C, f( ) v T*. Since g is order-

preserving and g o f is the identity on X X, it follows that for each

9E C,


) = g o f(O) < g(v T*) = v e [T] (7.2 (i)).







Thus Z = v {: 9 e C)} v e [T]. Finally, since f is a dominating

function (7.1 (i)), it follows that

f(X) 5 f(v e [T]) = v f o e [T] =v e T] = v T* (4.10 (iv)).

Hence, T E f(4) (4.10 (vi)). Therefore, N OD c f('-) so that

M1= f()); i.e., f(v C) = v f[C]. A dual argument shows that

f(A C) = A f[C] and so since f is bijective, f is a lattice isomorphism.

A similar argument shows that g is a lattice isomorphism.


(vi) That f and g are iseomorphisms follows immediately from

(iv) and (v) above.



COROLLARY 7.8. If 0 and Eare admissible subbases for X such

that 0 c E and X X algebraically dominates XAX, then 0 is equivalent

to E.


PROOF. Since 0 c E, it follows that X X algebraically dominates

SX (7.4). By the hypotheses, XoX algebraically dominates X X. Thus

0 is equivalent with E(7.7).



PROPOSITION 7.9. Suppose that 0 is an admissible subbase for X

and E = 0 U T. Then

(i) E is an admissible subbase for X, and for each f- X X,

(Zf 0) is a maximal linked system of 0, and

(ii) for each Z E XX and each a E X, if ( fl 0) e e (a),

then X (a) e 1, and dually, if e (a) (t 0 0), then X (a) e X.
--- D -- --- --0 --- I







PROOF. (i) Since T c Z= 0 U T c F, E is an admissible sub-

base for X (4.3). Let C EX X. Clearly ( r 0) is linked. To see

that (X r 0) is a maximal linked system of 0, let T E 0D such that

(c. n 0) U{T} is linked. If T i X then there is some U EE I r

such that T n U = 0. Since I = 0 U T and {T} U ( n 0I ) is

linked, it must be the case that U E T ; i.e., U = X (u) for some

u E X. Since u / T and 0 is a T -subbase for X, there is some S E 0

such that u E S and S ( T = 0. It follows that X (u) c S, and

S E (. n 0); therefore, (. 0) U{T} is not linked -a contradiction.

Thus T E X. Dually, if T E 0 such that ( ?n ) U{T} is linked, then

T e X, and hence (X F 0) is a maximal linked system of 0.


(ii) By (i) above and Proposition 7.5 (i) there is a unique

dominating function f: X X X aX which may be defined by f(X) = (X n 0)

for each E E X X. Suppose that X e E X and a E X such that

(X n 0) = f() < e0(a). For each S E (X r E ), A S+ .(4.10 (vi))

so that f(A S ) + f(P) since f is order-preserving. If S E 0 then

f(A S+) = A S* (7.2 (ii)) and since f(X) < e (a), it follows that

A S* < e (a); therefore a E S so that S X XD(a) / 0. If S Oi ,

then there is some b E X such that S = X (b). In this case

f(A S+) = A e [S] (7.2 (i)) = A e [X (b)] = e (b) (7.3). Thus

e (b) f(Z) < e (a); therefore b a since e0 is a lattice inclusion

(4.9) and S n XD(a) = X (b) A XD(a) / 0. It follows that t U{XD(a)}

is-linked and so XD(a) E 1. A dual argument shows the other statement.








PROPOSITION 7.10. If 0 is an admissible subbase for X, then

0 is equivalent to 0 VU T.


PROOF. Let E = 0 U T. Then E is an admissible subbase for

X (7.9 (i)). For each e X X, (X fC 0) is a maximal linked system

of 0 (7.9 (i)) so that X X algebraically dominates X X via a unique,

continuous function f: X X X X which may be defined by

f(X) = (K n 0) for each 4,e X X (7.5). To see that f is an iseomorphism,

we need only show that f is a closed, injective lattice homomorphism (2.10).


f is injective: Suppose that -, Y) X X such that / Y,7 Then

we may assume, without loss of generality, that there exist T E (f n Z) ,

and S E (M rn E D) such that T n S = 0. If both T and S are in 0, then

clearly f(g) = (V n 0) X (t'n 0) = f(). Suppose that T 1 0 ;

i.e., there is some b E X such that T = X (b). Note that e (b) % f(N,

for if e (b) 5 f(;), then T = X (b) E P(7.9 (ii))--a contradiction.

Thus e (b) i f(/) and there is some T' e 0 such that T' E e (b) and

T' I f(W) = (>? n 0) (4.7). It follows that X (b) c T',

T' E (i n 0) = f(Z), and hence f(X) / f(/y). In the case that T E 0I

and S 1 0D, a dual argument shows that there is some S' E 0D such that

S c S', S' E f(M)), S' i (f n 0) = f(Y-), and hence f(X) / f0f).


f is a lattice homomorphism: Suppose C is a non-void subset

of X X. Since f is order-preserving, it follows that for each e C,

f(Y) 5 f(v C); therefore, v f[C] : f(v C). If T E v f[C] n D, then








for each E C, T E f(t) = (t 0) (4.8); hence T E/ for each E C

so that T E V C (4.8)'. Thus T E (v C n 0) = f(v C); i.e.,

(v f[C] r D ) C (f(v C) n 0 ). It follows that f(v C) 5 v f[C],

and so f(v C) = v f[C]. A dual argument shows that f(A C) = A f[C].


f is a closed map: Let R E E. If R E 0, then

f[R+] = {f(f): R E } = {f(Y): R e (Y ) 0)}

= {f(): R E f = {f(C): f(4) E R*}

= R* since f is surjective.

If R i 0, then R E E Oc T. Suppose that R E TD; i.e., there is some

a E X such that R = XD(a). Thus

f[R]+ = {f(c): t : e (a)} (4.10 (vi), (iv) and 7.3)

= {f(t): (X n 0) = f(c ) 5 f o e (a) = e (a)} (7.9 (ii) and

since f is order-preserving)

= {) EA X: e (a)} (since f is surjective).

Dually, if R = X (a) for some a E X, then

f[R+] = { E XX: e (a) ~}.

Thus the image under f of each member of + is a closed subset of XOX,

and since f is injective, it follows that f is a closed map.


It follows from the preceding arguments that there is an

iseomorphism f: X X X X such that f o e = eg. Thus 0 is equivalent

with 9 U T = E (cf., 7.6).




In view of Proposition 7.10 we may now assume without loss of

generality that,for our purposes, all admissible subbases contain T.

The following theorem now follows immediately from Propositions 7.4 and 7.10.







THEOREM 7.11. If E is any admissible subbase for X, then

X X algebraically dominates XTX and is algebraically dominated by X.





We consider now some situations in which one lattice super-

extension of X dominates another lattice superextension of X not only

algebraically, but also as a superextension.



PROPOSITION 7.12. If 0 and E are admissible subbases for X such

that 0 c E and 0 is weakly-normal, then

(i) for each o E X X, (G 0 0) is contained in a unique maximal

linked system of 0, and

(ii) the superextension X X dominates the superextension X X.


PROOF. (i) Suppose that E X X and ( n 0) is not contained

in a unique maximal linked system of 0. Then there exist IW, >7 e X X

such that M / A (f ) 0) c ; and (O r 0) c Y. Since Y/ Y?, there

exist T E g and S E Ysuch that S n T = 0 (2.16 (ii)). By the weak-

normality of 0, there is a finite screen {T1, T2, **, Tn } c of S and

T (2.11 (iii)). Since B X = X X, it follows that X is prime (2.18 (iii))

and hence there is some j {1, 2, -**, n} such that T. E I. Therefore,

T r T. / 0 and S n T. 0--a contradiction to the fact that

{T1, ', Tn} is a screen for S and T. Thus (X n 0) is contained in

a unique maximal linked system of 0.







(ii) By (i) above and Proposition 7.5 (i), there is a unique

dominating function f: X XX which may be defined by

f(4) = {T E 0: (o t 0) U{T} is linked} for each E e X X.

To see that f is continuous, let T E 0 and E X X such that

2 f-1[T^]. Thus T i f( ) so that {T} u (C n 0) is not linked.

Hence there is some S E (f r 0) such that T r) S = 0. By the weak-

normality of 0, there is a finite screen c 0 of S and T. Let

C = U{W : W E T and W r S = 0}. Clearly C is a closed subset of

X X. Since S E c and S n W = 0 implies that S+ FW+ = 0 (2.16 (i)),

it follows that W+ for each W E T with W n S = 0; i.e., d C.

If y is any member of f-[T*], then since E X X = U{W+: W c }, there

is some We such that e V; i.e., e. Since W e 0, we see that

S(9 0) c f(9) (7.2 (iii)), and since T E f(g), it follows that

T n W f 0; therefore, S W = 0. Thus a e U {W+: We T and

W S = 0} = C. It follows that f-i [T* cC and X / C. Thus since

Swas an arbitrary point not in f- [T*], it follows that f- [T*] is

a closed subset of X and so f is continuous.





If T is a non-void collection of subsets of X, then Tn denotes

the collection of arbitrary intersections of members of Y.



PROPOSITION 7.13. If 0 is an admissible subbase for X and
S= then
r = T UT then
D I'







(i) for each Xe E XX, (V n T) is a maximal linked system of

T, and

(ii) the superextension X X dominates the superextension X X.


PROOF. (i) Let E X X. Clearly ( r T) is a linked system

of T. If U E TD such that (V n T) U{U} is linked, then either

U = X E (~ T) or there is some a E X such that U = XD(a). To see

that XD(a) E e, let T E n oe By the hypotheses, T = M{V E TI:

T c V}. For each V E TI such that T c V, it follows that V E (X n T),

V n XD(a) X 0, and a E V; therefore a E T so that T n XD(a) f 0.

Since T was an arbitrary member of (.t 0 i), it follows that

(V n O I) U{XD(a)} is linked and hence U{XD(a)} is linked.

Consequently, XD(a) E ( n T). Dually, if U E TI and (X \ T) U{U}

is linked, then U e (X n T). Thus (.f A T) is a maximal linked

system of T.


(ii) That the superextension X X dominates the superextension

X X follows immediately from (i) above and Proposition 7.5 (ii).




We say that the lattice superextension X X is "larger" than

the lattice superextension X X (or, equivalently, X X is "smaller"

than X X) if and only if the superextension X X dominates the super-

extension X X.

In the following theorem we sum up the preceding propositions

obtaining two situations in which X X is actually the smallest lattice

superextension of X. The proof is immediate from Propositions 7.12

and 7.13.







n n
THEOREM 7.14. If T is weakly-normal or if F = TD U TI, then

XTX is the smallest lattice superextension of X; specifically, if 0 is

any admissible subbase, then the superextension X X dominates the super-

extension X X.




In view of the relationship between T, F and any admissible

subbase E, namely, T c E c F,it is natural to ask if X X is always

the smallest or if X X is always the largest lattice superextension of

X. In general the answer is negative to both questions. As noted

previously there is a topological lattice X for which the unique dominating

function f: X X -- X X is not continuous; i.e., X X is not smaller than

X X and X X is not larger than X X (8.7). Even under the restriction
n n
that r = T U T X X is not necessarily the largest. In Example (8.6,

Part II) there is a topological lattice which satisfies this restriction,

but for which there is an admissible subbase Z such that there is no

continuous dominating function f: X -X X; i.e., X X is not the largest

superextension of X.

The theorem below follows immediately from Proposition 7.12 (ii)

and Theorem 7.14. It is stated here for completeness.




THEOREM 7.15. If all admissible subbases for X are weakly-normal,

then X X is the largest lattice superextension of X and X X is the

smallest lattice superextension of X.








DEFINITION 7.16. (Kaufman [12]) Y is called an orderable

compactification of a totally ordered topological lattice X if Y is

a topological compactification of X and the order on X can be extended

to Y so as to yield the given topology on Y as the order (interval)

topology.



COROLLARY 7.17. (Kaufman) If X is totally, then X has a

largest and a smallest orderable compactification.


PROOF. Suppose that X is totally ordered. Thus each admissible

subbase for X is weakly-normal (4.12). By Theorem 7.15, FX is the

largest and X TX is the smallest lattice superextension of X. It is

shown in Appendix I that Kaufman's orderable compactifications of X are

precisely the lattice superextensions of X; therefore, X X is the largest

orderable compactification of X and X X is the smallest orderable compacti-

fication of X.




In view of Kaufman's method for obtaining orderable compactifi-

cations, we may characterize the lattice superextensions of a totally

ordered space X as the topological closures of hE[X] in I where I is

the closed interval [0,1], E is any subset of {f: f is a continuous

increasing map of X into I such that A f[X] = 0 and v f[X] = 1} which
E
separates the points of X, and h : X -* I is defined by nf(hE(x)) = f(x)

for each f E E and each x e X.















8. EXAMPLES


1 1
EXAMPLE 8.1. Let X = (0, ) U (-,1) with the usual ordering.

Then the completion of X by cuts, [0,1] with its usual ordering, is

iseomorphic with XTX (5.13), and the largest superextension X X is
1 3
iseomorphic with [0,-] U [ ,1] with its usual ordering. (The maximal
1 1
linked system of r defined by {(0,-)} U {(0,a]: < a < 1} U
'2 2
1 1
{[a,l): 0 < a < -} corresponds to the point in A X, and the one defined
2 4 F
by {(-,1)} U {(0,a]: < a < 1} U {[a,l): 0 < a < -} corresponds to
2 2 2
in X X.)
4 F





EXAMPLE 8.2. Let X = (0,1) with its usual ordering. Then

T = r and consequently every lattice superextension of X is iseomorphic

with [0,1] with its usual ordering.





EXAMPLE 8.3. Let n be a positive integer. For each

j e (1, **, n}, let I. = (0,1) with its usual ordering. Thus each

'I. is a topological lattice.

Let Y be the topological product of {I.: j = 1, ***, n}, and

for each j = 1, ***, n, let E. = T. (Note that for each I., T = F.)







-1
Thus E = {W. CT]: T E. j E {1, ', n}} is a T -subbase for Y (3.1),
] ] 1
and the superextension (de Groot compactification) of the product Y

with respect to E, X Y (B Y), is homeomorphic with the product of the

superextensions {X I : j = 1, .**, n} (3.3) (de Groot compactifications


{8 I.: j = 1, ***, n} (3.4)).

Since each I. = (0,1), it follows from Example 8.2 that X Y

is homeomorphic with a closed n-cell in Euclidean n-space; therefore,

a de Groot compactification of an open n-cell in Euclidean n-space is

the n-cell with boundary attached. Note also that since the results in

Chapter 3 (Products of Superextensions) were given for arbitrary products

of T1-spaces, we can generalize this example to the case that Y is the

product of an arbitrary collection of copies of (0,1). In particular,

in view of Anderson's result [3], a de Groot compactification of Hilbert

space is the Hilbert cube.


EXAMPLE 8.4. Let I = [0,1] with its usual

X = {x e I: x is rational}. Then under the usual c

logical lattice, and since I is the completion of X

iseomorphic with I (5.13).

For each irrational y E I, let y. and yu b

in X which correspond to y. Let

Y = X UI{y: y e I and y is irrational} U

{yu: y I and y is irrational}.


Ordering, and let

ordering X is a topo-

by cuts, X X is



e distinct points not








We extend the ordering on X to one on Y in the following way:


(a) If y c I and y is irrational, then for every x e X

and x' e X such that x < y < x', we let x < yp < y < x'.

(b) If y, z e I are both irrational with y < z, then we

let y <' z .


We prove the following statements:



(i) Y is a complete (totally ordered) lattice, and X ip a

sublattice of Y.


PROOF. It is easily seen from the definition of the ordering

on Y that X is a sublattice of Y, and since I is totally ordered, it

follows that Y is also totally ordered. To see that Y is complete,

suppose that C is a non-void subset of Y. Let C' = (C n X) U{y e I:

y is irrational and either yg E C or y e C}. Thus C' is a non-void

subset of the complete lattice I so that there is some c E I such

that c = v C'. If c E X, it follows from the definitions that c = v C
I Y
and. c E Y. If c is irrational and cu C, then again from the definitions

we see that c = v C Y. Finally, if c is irrational and c i C,
u Y u
then v C = c CY Y. A dual argument shows that A C exists in Y.



(ii) iAX is iseomorphic with Y.


PROOF. Define i: X Y by i(x) = x for each x e X; i.e., i is

the injection map of X into Y. Since X is a sublattice of Y, i is a







lattice inclusion. The qualities below follow readily from the defi-

nitions and some properties of the real numbers:


(a) If y E I is irrational, then

i[YD(y)] = i- YD(Y)] = n{X (x): x E X and y I x} and

i-lCY (y)] =i-1[Y (y ) = n{X I(x): x E X and x I y).
-[ -l
(b) If x E X, then i EY(i(x))] = XD(x) and i [Y (i(x))] = X (x).


It follows from (a) and (b) that i is continuous; i.e., i is a continuous

lattice inclusion.


(c) i[X] = i[X], i[X]la = {y : y I and y is irrational} UiEX],
lb
and i[X] = {y : y E I and y is irrational} Ui[XI.


It follows from (c) that Y = i[X]la Ui[X]b c i[X]2 c Y, i.e., Y = i[X]2

Thus Y is a topological lattice completion of X with respect to i (6.10).

To see that X is lattice dense in Y, suppose that v, w e Y with

v iX]la i[X], w i[X]b i[X]1, and w
(c) that there exist y, z e I such that y and z are irrational with

w = yu and v = z Since yu < z it must be the case that y < z

and so there is some rational number x E X such that y < x < z;

i.e., w = y
It now follows that Y is iseomorphic with X X where E = {i- [S]:
E (i 20). To see that Y is iseomorphic with X, we need only
S e 1 } (6.20). To see that Y is iseomorphic with A X, we need only








show that E = F. Recall that

Y) = {Y} U{Y D(i(x)): x e X} U {YI(i(x)): x e X}

{YD(yD): y e I and y is irrational} U

{Y (yu): y e I and y is irrational} (6.15).
-l
If T is a closed ideal of X, then either T = XD(x) = i- Y (i(x))] for

some x E X or T = {x e X: x < y} = i- [YD ()] for some irrational

y e I. In either case it is clear that T E D. A dual argument shows

that T. E E whenever T is a closed dual ideal of X. Thus E = r and Y

is iseomorphic with X X.





EXAMPLE 8.5. Let A, B, C be subsets of the plane defined as

follows:

A = [0,1) x {0}
2 2
B = {(x,y): (x-1) + y = 1, 0 < x < 2, 0 5 y < 1}

C = (3,4] x {0}.

Since A, B, and C are copies of half-open intervals we may assign to

each of them a natural (total) ordering from left to right. Also since

A and B are both copies of [0,1), there is a natural lattice isomorphism

p: A B. Let X = A U B u C. X becomes a lattice by defining a

supremum and an infimum for each pair of elements of X. Let z, w E X.

We let

Sp(w) v z, if w E A and z e B

z V w = w z = w V z if w, z e A, w, z e B, or w, z C

z if w E A U B and z e C







-1
Sw A p (z), if w E A and z e B

z A W = W A w A z, if w, z E A, w, z c B, or w, z E C

w, if w c A U B and z E C.

(i) The lattice described above can be visualized as follows:


(0,0) A (1,0)


(2,0)


0
(3,--) (4,0)
(3,0) C (4,0)


(ii) A X can be visualized as follows:
T


point in A X which is

not in eT[X].


eT[A] eT[C]








(iii) If 0 = T U T, then 0 = T U{AUB, C} and X X can be
visliD Ias
visualized as follows:


points in X X which are

not in e [X]


e [B



e -----A]
e@[A]


e EC]


(iv) F = T U{A, AUB, C} and X X can be visualized as follows:



e [B] points in A X which are

S/ not in eF[X]





er[A] e [C]




From (ii), (iii), and (iv) we see that three quite natural sub-

bases yield different lattice superextensions.





EXAMPLE 8.6. Suppose that L and M are lattices. We define the

lattice direct product, X, of L and M to be the set L x M with the




93


ordering defined by (a,b)
It is easily seen that the lattice direct product is indeed a lattice.




Part I


Let X be the lattice direct product (0,1) x (0,1) where (0,1)

is the open interval with the usual ordering. Let Y1 be the lattice

direct product [0,1] x [0,1],.where.[0,1] is the closed interval with

the usual ordering. X may be visualized as follows:


(0,1) (1,1)










(0,0) (1,0)

X


(i) X is a sublattice of Y, and Yi = = X2a X2b but the

inclusion map i: X Y is not continuous. Thus Y is not a topological

lattice completion of X with respect to i.


PROOF. It is easily seen from the definition of lattice direct


~la lb 2 2 2a 2b
product that X is a sublattice of YI, and it follows readily that

Y C XlaU Xlb C X2 i Y so that Y, = X2 (= X2arX2b).
I -- 1




Full Text

PAGE 1

NATURAL COMPACTIFICATIONS OF LATTICES By LINDA JOE WAHL SMITH A DISSERTATION PRESENTED TO THE GRADUATE COUNCIL OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY UNIVERSITY OF FLORIDA 1969

PAGE 2

UNIVERSITY OF FLORIDA 3 1262 08552 2935

PAGE 3

To Nevins

PAGE 4

ACKNOWLEDGMENTS The author wishes to extend her sincere and grateful appreciation to Dr. G. E. Strecker and Dr. G. A. Jensen for their helpful suggestions in the preparation of this paper, and to acknowledge Dr. J. de Groot , Dr. W. E. Clark, and Dr. Gale Nevill for serving on her Supervisory Committee. To her husband, Nevins, and her parents, Mr. and Mrs. J. K. Wahl and Mrs. Nevins Smith, she extends many thanks for all the love and moral support they have given over the years and especially during her work on this paper. For typing assistance, she is very much indebted to Mrs . Margaret Parramore . Ill

PAGE 5

TABLE OF CONTENTS Page ACKNOWLEDGMENTS iii KEY TO SYMBOLS v 1. INTRODUCTION 1 2. PRELIMINARIES 5 3. PRODUCTS OF SUPEREXTENSIONS 15 4. LATTICE SUPEREXTENSIONS 20 5. THE COMPLETION BY CUTS 29 6. TOPOLOGICAL LATTICE COMPLETIONS 48 7. LARGEST AND SMALLEST LATTICE SUPEREXTENSIONS 68 8. EXAMPLES 85 APPENDIX I KAUFMAN'S ORDERABLE COMPACTIFICATIONS HI APPENDIX II LATTICE CLOSURES 113 BIBLIOGRAPHY 117 BIOGRAPHICAL SKETCH 119 IV

PAGE 6

KEY OF SYMBOLS X a topological lattice (2.8 (iii)) Xj^(a) = {x-e X: X < a} and X^(a) = {x e X: a < x} r„ (2.3 (ii)) the collection of all closed ideals of X Fj (2.3 (ii)) the collection of all closed dual ideals of X r = F^ u r^ Tj^ (2.3 (iii)) the collection of all principal ideals of X together with X T, (2.3 (iii)) the collection of all principal dual ideals of X together with X T = Tj^ u Tj.. (4.1 (iii)) is the usual subbase for X I an arbitrary admissible subbase for X (4.2 (i)) AX (2.15) the superextension of X with respect to E S (or sometimes S" to avoid confusion) ... (2.14) the set of all maximal linked systems of Z which contain S 1 (or sometimes E-) = {S"*": S e E}. (2.14) e (2.16 (v)) the natural embedding of X into A X 6_X (2.17) the topological closure of e [X] in AX (called the de Groot compactif ication of X with respect to E) 4* the collection of arbitrary intersections of V 2 the power set of X

PAGE 7

D(A) = n{Xp(a): A = X^(a)} where A e 2^ (5.3) X (5.2) the completion of X by cuts d: X->X (5.4) the map defined by a ^ D({a}) T = {Xjj(d(a)): a e X} [j{X^(d{a)): a e X} u{X} (5.7) ooX the Alexandroff one-point compact if icat ion of X z^ = z^^ n z^^ (6.1) z' z'^ n z^^ (5.1) T = {Yj^(i(a)): a e X} ij{Yj(i(a)): a e X} u{Y} (where Y = iCX]-"-) (6.6) Z^"-^ (6.15) vi

PAGE 8

1. INTRODUCTION Armed only with the definition of compactification, and asked to compactify the open disc in the plane, a novice topologist would most likely simply add its boundary. However, among the "standard" methods which have been developed for obtaining compactif ications of spaces (Stone-Cech, Alexandroff, Freudenthal , Wallman, etc.) not one 1 V yields this seemingly "natural" compactification. The Stone-Cech compactification (although it has nice mapping properties) adds so many points that the disc's "intrinsic character" is not preserved, and it is in some sense "lost" in the new space. At the other extreme, the Alexandroff compactification does not add enough. What we seek, then, is a method for producing compact extensions which yields "natural" compactif ications in the sense that the closed interval [a, b] "naturally" compactif ies the open interval (a, b), the extended real line "naturally" compactif ies the rationals, the closed disc "naturally" compactifies the open disc, and the Hilbert cube "naturally" compactif ies Hilbert space. In this dissertation we define and investigate such a method. Frink [8] has developed a Wallman-type method for obtaining many compact if ications for a given space. It has been shown by Alo and Shapiro [2] that the compactification of the open disc by the closed disc is of Wallman-type.

PAGE 9

The method which we use is due to de Groot . For a given space , this method produces compact extensions (called superextensions) and compactifications (called de Groot compactifications) which are dependent upon both the space and a chosep subbase for its closed sets. Hence, in general, for each space there may be many superextensions and de Groot compactifications. (See Chapter 2 for details.) In particular, for locally compact Hausdorff spaces, it is possible (by astutely choosing the subbases) to obtain both the Stone-Cech and the Alexandroff compactifications as de Groot compactifications. Another aspect of the versatility of the de Groot compactifications is that they (unlike any of the "standard" compactifications mentioned above) are productive. Specifically, we will show in Chapter 3 that the topological product of any collection of de Groot compactifications : (respectively, superextensions) of a family of spaces is homeomorphic with a natural de Groot compactification (respectively, superextension) of the topological product of the family. In order to obtain the desired "natural" compactifications, we consider only topological products of topological lattices (i.e., lattices with the interval topology) and restrict our attention, when dealing with lattices, to those superextensions which arise with respect to certain "admissible" subbases. This restriction is not overly severe since, as can be seen in Chapter 8, for a given topological lattice there are, in general, many "lattice superextensions" (i.e., superextensions with respect to admissible subbases). In Chapter 4 we will

PAGE 10

show how the order of the original lattj.ce can be extended to a partial ordering of any lattice superextension in such a way that the superextension becomes a complete lattice (in the interval topology). Thus each lattice superextension of a topological lattice turns out to be not only a topological compact ificat ion, but also a lattice completion--hence a very "natural" type of compactification. In Chapter 5 we consider the algebraic structure of a topological lattice X and show that its completion by cuts can be realized as a particular lattice superextension. In Chapter 6 we :attempt to algebraically characterize other lattice superextensions of X. This is done using the notion of topological lattice completions of X, these being certain complete lattices containing X as a sublattice, We show that all lattice superextensions of X are in fact topological lattice completions of X and that those topological lattice completions of X satisfying a particular "density" condition are lattice superextensions of X. In [12] Kaufman considers (totally) orderable compact ificat ions for totally ordered spaces X; i.e., compactifications Y of X for which the given topology on Y is the same as the order (interval) topology on Y. Among other things, he establishes the existence of a "largest" and a "smallest" orderable compactification for each totally ordered space. In Chapter 7 we consider the notion of "domination" for lattice superextensions of topological lattices and generalize Kaufman's results to these compactifications of lattices. Hence each topological lattice

PAGE 11

has a lattice compactif ication which is maximal relative to mapping properties (analogous to the Stone-Cech compactification for more general spaces) and a lattice compactification which is minimalrelative to mapping properties (analogous to the Alexandroff compactification for locally compact Hausdorff spaces). Chapter 8 consists of examples of lattice superextensions of various kinds of lattices. Also, examples are given which show the necessity of certain restrictions which are made in the text.

PAGE 12

2. PRELIMINARIES The purpose of this chapter is to state most of the basic definitions, propositions, and theorems which will be needed in the following chapters. Algebraic terms which are not specifically defined here can be found in Birkhoff [5] and topological terms in Kei"ley [13]. DEFINITION 2.1. (i) A set X is called partially ordered if there is a binary relation < on X which is reflexive, antisymmetric, and transitive. (ii) If X is a partially ordered set with more than one element in which each pair of elements a and b have a supremum (denoted by a V b) and have an infimum (denoted by a a b), then X is called a lattice . (iii) If X is a lattice and each subset C of X has a supremum 2 xn X and an mfimum m X, then X is said to be complete . (iv) If X is a lattice and a, b e X, then a and b are said to be comparable if either a < b or b < a. (v) If X is a lattice and a, b e X, then a < b will mean that a 2 b and a ?^ b. We consider only non-trivial lattices in order to avoid the possibility that vx = aX . 2 Throughout we will use the standard conventions that for a lattice X, a0 = vX and v0 = aX, and for a topological space X, 00 = X, 5

PAGE 13

(vi) X is said to be totally ordered if for every pair of elements x and y of X , x and y are comparable; i.e.,x Y is a function. (i) The map i is called a lattice homomorphism whenever i satisfies the following: For each non-void subset C of X , if v C X exists in X, then v i[C] exists in Y and i(v C) = v i[C]. Dually, if 1 A 1 AC exists in X, then a i[C] exists in Y and i(A C) = a i[C]. A I X Y (ii) The map i is called a lattice inclusion and X is called a sub lattice of Y if i is an injective lattice homomorphism. (iii) The map i is called a lattice isomorphism and X is said to be lattice isomorphic with Y if i is a surjective lattice inclusion. DEFINITION 2.3. Let X be a lattice and I a non-void subset of X. (i) I is' increasing if a e I whenever a £ X and i < a for some i e I . lis decreasing if a e I whenever a e X and a ^ i for some i e I . (ii) I is an ideal of X if I is decreasing and a v b £ I for each pair a, b e I. I is a dual ideal if I is increasing and a a b £ I 3 for each pair a , b £ I . 3 We do not consider to be an ideal or dual ideal in order to simplify the statements of proofs.

PAGE 14

(iii) I is a principal ideal of X if there is some fixed a e X such that I={x£X:x
PAGE 15

each I. is decreasing, and hence 1^0. By (i) above, I is an ideal of X. A dual argument shows that the intersection of a finite collection of dual ideals of X is a dual ideal of X. ( iii ) Since X ^ 5^ , for each F e 3^ there exists x e X F. Let X = v{x : F e V? }. If x e [J^, then x e F for some F e3^ , and since F is decreasing and x < x, we see that x e F--a contradiction. Thus X e X uj^and so u>?^ X. A dual argument proves the other statement, PROPOSITION 2.5. If X and Y are lattices and i : X -> Y is a lattice homomorphism , then i preserves the order of X . PROOF. Let i: X ^ Y be a lattice homomorphism and suppose X, y e X such that x < y. Thus y = x v y and by the definition of lattice homomorphism i(y) = i(x v y) = i(x) v i(y); therefore A I i(x) 2 i(y) and so i preserves the order of X. DEFINITION 2.6. Let X be a topological space and A c X. The topological closure of A in X will be denoted by A or A when no confusion seems likely. DEFINITION 2.7. Suppose X and Y are topological spaces and f : X -> Y is a function. (i) f is called an embedding of X into Y if f is injective, continuous, and a closed map onto its image. (ii) f is called a dense embedding of X into Y if f is an Y embedding and f[5f] = Y.

PAGE 16

(iii) f is called a homeomorphism and X is said to be homeomorphic with Y whenever f is a surjective embedding. (iv) Y is called a compact if icat ion of X if Y is compact and f is a dense embedding. DEFINITION 2.8. Suppose that X is a lattice. (i) A closed interval of X is a subset A of X such that either A is a principal ideal of X, A is a principal dual ideal of X , or A is the non-void intersection of a principal ideal of X and a principal dual ideal of X. (In the latter case, there exist a, b e X such that A={xeX:a Y is a function, f is called an iseomorphism and X is said to be iseomorphic with Y if f is both a lattice isomorphism and a homeomorphism .

PAGE 17

10 The remainder of this chapter contains definitions and some fundamental results concerning superextensions of T -spaces. These results are due to de Groot , Jensen, and Verbeek [ 7 ]. It is within the framework of superextensions that we will study lattices in the following chapters . DEFINITION 2.11. Suppose that X is a T -space and E is a subbase for the closed subsets of X. (i) If A and B are subsets of X, then A and B are said to be screened by a subcollection ^F of E and ^ is called a screen for A and B if U 'i' = X and no member of "i* meets both A and B. Two elements x and y are screened by 'i' if {x} and {y} are screened by ^, and an element X e X and a subset A c^ X are screened by H* if A and {x} are screened by ^. (ii) E is called a T subbase for X if for each x e X, n{S 6 E: x e S} = {x}, and whenever x e X and S £ E with x ^ S, there is some T e E with x e T and T H S = 0. (iii) E is said to be weakly-normal if each pair of disjoint members of E can be screened by a finite collection from E. (iv) A non-void subcollection ^ of E is said to be linked if each pair of elements of i^ has a non-void intersection; ^ is called a maximal linked system of E if -< is maximal in E with respect to the property of being linked. The next two propositions are found in [7 ]. Their proofs are included only to acquaint the reader with the terminology.

PAGE 18

11 PROPOSITION 2.12. A space X is_ T if_ and_ only if there is a T subbase for the closed sets of X . PROOF. Suppose that X is a T -space and let Z be the collection of all closed sets of X. Clearly E is a subbase for the closed subsets of X. Let X e X. Since X is T , {x} e Z, and it follows that {x} c_ n{S e E: X e S} £ {x}; i.e., n{S e E: x e S} = {x}. If x e X and S e E with x ^ S, then clearly {x} e E, x e {x}, and {x} Pi S = 0. Hence E is a T -subbase for X (2.11 (ii)). Conversely, suppose that E is a T -subbase for X. Let x e X. To see that {x} is closed, let y e X such that y i {x}. Hence y i ri{Se E: xe S} since E is a T -subbase for X. Thus there is some closed set S £ E with {x} c^ S and y ^ S. It follows that {x} is a closed subset of X. PROPOSITION 2.13. If. ^ is_ a_ subbase for the closed subsets of a_ T space X , then E is_ a_ T subbase for X if_ and only if for each x e X , (Se E: xe S} is a maximal linked system of E . PROOF. Suppose that E is a T -subbase for X and x e X. Let ^={SeE:xeS}. Clearly ff is a linked system of E. If T e E and T ^ fi, then x ^ T. Thus since E is a T -subbase for X there is some S e E such that x e S and T n S = 0. It follows that S e if and so ^ U {T} is not linked; therefore /(' is a maximal linked system of E. Conversely, suppose now that E is a subbase for the closed subsets of a T -space X such that {Se E: xe S} is a maximal linked system of E for each x e X. Let x be a fixed but arbitrary point of X,

PAGE 19

12 Clearly {x} c_ n{S e E: x e S}. If y e X and y ?^ x, then it follows from the fact that X is a T -space that there is some T e E such that X e T and y ^ T. Therefore y i r\{S e 1: x e S} and so n{s e Z: X 6 S} = {x}. Finally, let x e X and T e E with x i T . Thus T ^ {S € E: X € S} and since {S e E: x e S} is a maximal. linked system of E, there is some S e E such that x e S and S H T = 0. Hence by Definition 2.11 (ii), E is a T -subbase for X. NOTATION 2.14. Let E be a T -subbase for a space X. XX will denote { j^ c^ E : j;^ is a maximal linked system of E}. For each S £ E, S will denote {i£ e XX: S € if } , and E will denote {S : S e e}. DEFINITION 2.15. Suppose that E is a T -subbase for a space X. The set XX equipped with the topology which has E as a subbase for the closed sets is called the superextension of X with respect to E. The following proposition states some results found in [ 7 3 which follow readily from the definitions. The proof is thus omitted. PROPOSITION 2.15. Suppose that ^ is_ a T subbase for the space X . (i) If S, T e E with S D T = 0, then S"^ H t"^ = . (ii) I£^ e X X and S e E such that ^ U {S} is_ linked , then S e ^ (iii) If_ ^ is_ a_ linked system of E , then ^ is_ contained in some maximal linked system of E. (iv) I£ Jif e XX and H/ ^ 0, then HjC = {x} for some x £ X.

PAGE 20

13 (v) I£ e : X * A X is_ defined by^e(x) = {S e Z: x e S}, then e is an embedding of X into XX, and e [S] = S H e [X] for each S € Z. (vi) The space X^X is T, . — * L 1 (vii) The space XX is_ compact . (viii) I£ S and T are_ in_ E and_ S U T = X, then s"^ U t"^ = X^X, DEFINITION 2.17. Let Z be a T -subbase for a space X. (i) Let B^X denote the topological closure of e [X] i^ XX; this will be called the de Groot compact if icat ion of X with respect to E. Note that {B X nS:SeE}isa subbase for the closed subsets of Bj-X. (ii) If ;^ c E, then ,^ is said to be prime if and only if whenever S, , S^ , • • • , S e E such that U{S.: i = l,2,''',n} = X, there 1 2 n 1 exists some j e {l,2,''',n} such that S. e. }(. . (iii) A subset
PAGE 21

14 (i) If / is_ a_ finite subcollection of E , then [JJ^ X if and only if Ule^XHT"*": T e^} = 3j.X. (ii) If E is weaklynormal , then 3„X is Hausdorff . (iii) ^ J? e ^yX , then ^ e 3 X if_ and only if ^ is_ prime . (iv) If j^ is any maximal centered system of E , then ;^ is_ prime . ( V ) _If_ E _is_ weakly-normal and ^^ £ AX, then ;^ e 3 „X if_ and only if ^ contains a maximal centered system of E . The proof of Proposition 2.18 can be found in [ 7 ] Chapter 1, Proposition 1.5 through Proposition 1.9. Propositions 2.16 and 2.18 together with the fact that a T -space is completely regular if and only if it can be embedded in a compact Hausdorff space outline a proof for the following statement from [ 6 ] : A T -space X is completely regular if and only if there exists a weakly-normal T -subbase for the closed subsets of X. EXAMPLE 2.19. Suppose that X is a completely regular T -space and E is the collection of all non-void zero sets of X. Then E is a weakly-normal T -subbase for X, and 3„X is the Stone-Cech compactification of X. (This follows from Proposition 2.16 (i) and Theorem 6.5 of [lO].)

PAGE 22

PRODUCTS OF SUPEREXTENSIONS In this chapter we investigate the productivity of superextensions of T -spaces and show that a natural superextension of the product of an arbitrary family of T -spaces is the product of the superextensions of the family. We also show that the corresponding de Groot compactification of the product is the product of the de Groot compactifications . Throughout this chapter we let {Y } be a family of T -spaces, Z a a€ A X a be a T -subbase for the closed sets of Y , Y be the topological product of the family {Y } , and for each a e A, it : Y -> Y be the projection 3. 3€A * 3. 3. function. Also without loss of generality we may assume that for each a e A, the set Y is itself a member of the subbase E . a a PROPOSITION 3.1. Z = {tt "''[S]: S e Z , a e A} is a T., subbase a. 3. X for the closed sets of Y. PROOF. It is clear that Z is a subbase for the closed sets of the product space Y. Suppose that y, z e Y and z e H { tt [T] : ir [T] e Z 3, 3 and y e TT~ [T]}. Thus for each a e A, it (z) e n{T: T e Z and ir (y) e T}. -^ a a a a Since Z is a T -subbase for Y , n {T e Z : tt (y) e T} = {tt (y)}; a 1 a aa-' a-' therefore it (z) = ti (y) for each a e A and hence z = y. Now let y e Y 3 3 and TT~ [S] e Z with y <' tt~ [S]. If follows that tt (y) «' S, and 3 3 3 15

PAGE 23

16 since E is a T, -subbase for Y , there is some T e E such that a 1 a a TT (y) e T and S n T J= 0. Therefore, tt~ [T] e Z, y e it" [T] and 3 3 3 iT~ [S] n TT~ [T] = 0. Hence E is a T -subbase for Y. (cf. Definition 3 3 J. 2.11 (ii)). In view of the above proposition we may consider the superextension of Y with respect to E. PROPOSITION 3.2. Let_ P be_ the topological product of the family {AY} . For each (5^ ) . e P, the set ^ = {Q e E: for each L a a€A a acA — ^^— • a a e A, IT [Q] e Jj } is a maximal linked system of E. a a ^ PROOF. Suppose that Q, R e ^. For each a e A , it [Q], tt [R] e ;C 3 3 3 so that TT [Q] n TT [R] /! 0; therefore, Q H R i^ so that ff is linked, a a Suppose now that tt [S] e E and f{ U{tt~ [S]} is linked. If a e A and a ^ b, then tt [tt"-'-[S]] = Y . If a = b, then tt [tt"-''[S]] = S. Thus for a D a a D each a e A it follows that tt [tt" [S]] e / and so tt~ [S] e f< ; i.e., f{ a D a b is a maximal linked system of E. THEOREM 3.3. There is a_ homeomorphism f from the product P of the superextensions { X Y } to the superextension X Y of the product L 3 36A — '" " ' — '• L ' — — — — — a Y. Furthermore , if g is_ the unique continuous map which makes each of the diagrams commute.

PAGE 24

then the diagram commutes, 17 PROOF. By Proposition 3.2 there is a function f : P -> X Y defined by f((^ ) J = {Q e Z: for each a e A, it [Q] e iC } . a aeA a a We will show that f » g = e . For each y e Y, f(g(y)) = {Q e E: for each a € A, TT [Q] e tt (g(y))} = {Q e E: for each a e A, tt [Q] e e (tt (y))} 3i 3. 3. li 3. a = {Q e E: for each a e A, tt (y) e tt [Q]} = {Q e Z: y € Q} = e (y). d, 3, Li To see that f is injective, let and ^ be members of P such that f ^ f< . Thus there must be some b e A such that 't, (/) ^ tt (/^) , and since t^^^S) and it (^) are maximal linked systems of E , it follows that there exist T e ^u^(^) and S e -n (/t) such that S H T = 0. Hence TT, [S] n TT, [T] = and so tt, [S] and tt, [T] cannot be contained in the b b b b same maximal linked system of E. Recall that if a e A and a ?* b , then TT [tt~ [T]] = Y , and if a = b, then tt [tt~ [T]] = T. It follows that for a D a a b each a e A, TT [tt. [T]] e tt (4). Hence by the definition of f, a b a / TT~ [T] e f(Q). By a similar argument we see that tt~ [S] e f (^) . Thus f(p i fW. To see that f is surjective, let ^ e X Y. For each aeA, let / ={TT[Q]:QeJC}. For each a e A , ;C is clearly linked, and if 3 3. 3 S e E such that I U{S} is linked, then it follows that >£. u{tt~ [S]} 3 ^^ 3

PAGE 25

18 is linked and hence tt [S] e i^ . Therefore S = it [it [S]] e ^ so that 3. 3 3 d. ;C is a maximal linked system of E . Thus (^ ) . e P and by the defi3 3 3 3G A nition of f, f((^ ) „) = {Q e Z: for each a e A, tt [Q] e ^ } * a aeA a a = {Q e E: Q € j^} = <^. The continuity of f is an immediate consequence of the fact that for each b e A and each T e E^ , f ~"''[(Tr"-'"[T])'''] {{t ) . e P: b b a aeA f^^^a^aeA^ ^ (^"V])^} = {(^)^^, c P: ."^CT] e f ( C^-!^),^,)} = ^(^^acA ^ ^•' ^°^ ^^^h ^ ^ ^' V\'tT" ^ ^a^ = ^^^a^aeA ^ ^ = T(= tTj^[tt^-^[T]]) £/j^} = ^^^[{^e A^ Y^ : T e^ }] = ^'^Ct'^] which is a b closed subset of P. Finally we show that f is a closed map. Using the fact that f is injective and applying f to the above equality we know that for each a e A and T € E , f[^"-^[T"^]] = f [f"^[(TT"-'-[T])'^]] = (tt'-^CT])"^. It follows a a a a from the fact that {tt [T]:aeA,T eE}isa subbase for the closed a a sets of P and that f is bijective,that f is a closed map. Consequently, X Y is homeomorphic with P. Let Q be the topological product of the family (6 Y : a e A}. u 3 a Thus Q is a subspace of the product space P. COROLLARY 3.4. There is a_ homeomorphism h from the product Q of the compactifications (3 Y } .to the de Groot compactif ication i 3 3CA ~~'~~ ' ' "— ' ' a BY of the product Y.

PAGE 26

19 PROOF. Let h be the restriction of f in Theorem 3.3 to Q. Since f is a homeomorphism we need only show that h[Q] = 6 Y. To see that BY c h[Q],let ^e Y. Recall that in the proof ' of Theorem 3.3 we showed that for each aeA,/ ={TT[S]:Sei^}is a maximal linked system of Z and that f ( ( / ) ,) = iC. To see that a a acA (Jf ) e Q we will show that for each a e A, / is prime and Hence /. € 6 Y (of.. Proposition 2.18 (iii)). Let a e A and suppose that a {T. , T„, •••, T } c z such that U{T.: i = 1, 2, •••, n} = Y . It 12 n — a 1 a follows that {tt~ [T ], 7r'"'"[T ], •••, tt~"'"[T ]} c E and 3. J3 z 3 n U{-n~ [T.]: i = 1, 2, •••, n} = Y. Since iS is in 6_Y and hence is a 1 L prime, there is some j e {1, 2, •••, n} such that tt [T.] eJ^. Thus ttCtt [T.]] = T. e^ and so ^ is prime. aa] ja a^ To see that h[Q] c by, let (X ) . e Q. Thus for each a e A, — Z a aeA / e B Y and so X is prime. Suppose now that f is a finite subset a Zj a a a of I such that U*f = Y. Recall that each member of 4" is of the form TT [T] for some aeA and T e E . Let B = {aeA: there is some T e E a a a with Ti [T] e 4'}. Suppose that for each b e B thei^e is some yj^ e Yj^ U{T € E^: ti" [T] e f}. Then we can pick y e Y such that ^u(y) = Yv, for each b e B. It follows that for each tt" [T] e ^, b b a y i -n [T] and so y e Y UH* — a contradiction. Hence there must be 3 some b e B such that U (T e E, : tt" [T] e 4"} = Y^ . Since ii.^ is prime, b b b b there is some S e E, such that tt^ [S] e Y and S € £^ . It follows from b b b the definition of f that tt"''"[S] e f((;f ) J. Thus fiit) J is prime b a aeA a aeA ^ and so h((;f ) .) = f((<}^ ) .) e B_X. Consequently, h is a homeomorphi 3 3£A 3 36A L from Q onto B_Y. sm

PAGE 27

4. LATTICE SUPEREXTENSIONS For the remainder of the paper we restrict our attention to superextensions of topological lattices with respect to T^-subbases which consist of ideals and dual ideals. For these "admissible" subbases we show that the superextensions are actually de Groot compactifications (Theorem 4.6). Moreover, (with respect to a "naturally induced" ordering), they are shown to be complete lattice extensions of the original lattice (Theorem 4.11). NOTATION 4.1. (i) For each a e X, X (a) = {x e X: x < a} and X (a) = {x e X: a < x}. (D denotes decreasing and I denotes increasing.) (ii) r = {C c X: C is a closed ideal of X}, T^ = {C c X: C is a closed dual ideal of X}, and T = r U T . (iii) T = {X} u {Xp(a): a e X} u {Xj(a): a e X}. (We refer to T as the usual subbase for X.) (iv) If E is a subbase for the closed sets of X and E £ T, then Ej = E n Tj and Ep = E ^ r^ . >. DEFINITION 4.2. (i) If E is a T -subbase for the closed subsets of X and E £ r,then E is called an admissible subbase for X. (ii) A lattice superextension of X is any superextension of the form A X where E is an admissible subbase for X. I 20

PAGE 28

21 . . PROPOSITION 4.3. IfT^Ecr, then E is_ an admissible subbase for x: PROOF. It is clear from the definition of the interval topology (2.8 (ii)) that T is a subbase for the closed subsets of X. It is also clear that any collection of closed subsets of X containing T is a subbase for the closed sets of X. Hence Z is a subbase for X. To see that Z is a T -subbase for X, let a e X. Thus since T c^ Z , {a} c n{S £ Z: a e S} c n{U e T: a e U} £X (a) H X (a) = {a}; i.e., r\{S e Z: a e S} = {a}. Finally, suppose a e X and S e Z such that a j^ S. If S e Z^^, then X^Ca) e Z , a £ X (a) and X (a) n S = 0. Dually, if S £ Z_, then X^(a) £ Z^^, a £ X^(a) and X^(a) H S = 0. Thus Z is a T -subbase for X (cf . , Definition 2.11 (ii)) and hence is an admissible subbase for X (4.2 (i)). COROLLARY 4.4. T and T are admissible subbases for X. '^ By the above corollary XX and XX are lattice superextensions of X. Examples of these can be found in Chapter 8. Next we establish that all lattice superextensions of X are in fact compactifications of X which can be considered as complete lattices having X as a sublattice. PROPOSITION 4.5. Let Z be an admissible subbase for X. (i) Z and Z are centered systems of Z. (ii) Every linked system of Z is centered .

PAGE 29

22 PROOF. (i) is immediate from Proposition 2.4 (ii). (ii) Suppose that "J^ is a linked system of E and F , F , •••, F e i^. If all of the F. belong to Y. , then by (i) above n{F.: i = 1, 2, •••, n} ^ 0. Similarly, if all of the F. belong to Z , then n{F.: i = 1, 2, •••, n} ^ 0. In the remaining case we may assume, by at most rearranging the subscripts, that there is some integer m such that l
PAGE 30

23 In view of Theorem 4.5 every lattice superextension of X is actually a topological compact if ication of X. To see that it is also a complete lattice with the interval topology and that it has X as a sublattice we continue as follows. DEFINITION 4.7. Suppose that E is an admissible subbase for X and X., Me XX. We say that ^< /^whenever 7yir\ 'l^ '^ ^ r\ Z , or, dually, whenever <>t r\ 1 ^ mC^ 1 . Since containment is a partial ordering, it follows readily that 4.7 defines a partial ordering on the elements of AX. Throughout the remainder of this paper, AX will be understood to be partially ordered according to Definition 4.7 whenever E is an admissible subbase for X. The following theorem shows that XX is a complete lattice whenever Z is admissible. THEOREM 4.8. Suppose that Z is_ an admissible subbase for X . Every non-void subset of AX has a_ supremum in AX and an infimum in AX. Specifically , if ?! C £ A^X and_ if_ C=n{<^nE^:<^eC} and_ Cj = r\{^ r\ Zj-. ^ e C},

PAGE 31

24 then V C = C U{S e E : C U{S} is linked } and A C = C U{S £ E : C U{S} is linked }. PROOF. It is easily seen that ^ = C^^ U{S e E^: C^ U{S} is linked} is a linked system of E. Suppose that T e E and ^ U{T} is linked. Since iC r\ ^j E. ^ ^ ^r ^°^ each ^e C, it follows that ^ U{T} is linked for each A X defined by x'-^{S£E:x£ S} is_ a^ lattice inclusion ; thus X is_ a_ sublattice of XX. PROOF. That e is injective follows from the fact that it is a topological embedding (Proposition 2.15 (v)). To see that it

PAGE 32

25 preserves suprema, suppose that C ^X such that v c exists in X. If S e e_(v C) n E , then v C e S, and since S is decreasing it follows that C £ S. Therefore S e e (c) O E for each c c C and by Proposition 4.8, S € V e [C]. Hence by Definition 4.7, v e [C] < e (v c). Conversely, if S e V ej.[C] D E^, then S e e^(c) O E^ and c e S for each c e C. Thus C £ S and, since S is a closed ideal of X , it follows from Frink [ 9; Theorem 12] that v^ C e S. Hence S e e^iv C), and by Definition 4.7, ej.(v^ C) < V e^[C]; i.e., e (v C) = v e [C]. A dual argument shows that if C c X and a c exists in X, then e (a C) = a e„[C]. By X EXE Definition 2.2 (ii), e is a lattice inclusion. PROPOSITION 4.10. For an^ admissible subbase E of X, the following hold : (Recall that for each S e E , s"*" = {;C e XX: S e/}. ) (i) If S e E^ and T e E^, then s"^ = { jf e A X : J& < v s"*"} and t"*" = {/ e Xj,X: a t"^ < ^} . (ii) For each /rte X^X, {^ e X^X: ^ < «t} = nis'^: S e m. n E } and {/ e XX: m < £.] = n{s"^: S e m r\ 'i^J • (iii) For each me X^X, e'-^lU e \^X: cf < >»l}] = H (^^ H E^^) and e^-^[{^ e X^X: ^ < ^}] = oiftD E^). (iv) For each T e E , v t"*" = v e [T], and for each T e E , A t"^ = A ej.[T]. (v) If. T e E and S e E , then S T ?! i£ and only if A S* < V t!^. (vi) If ;^ e X X, T e E , and S e E , then T e ;f if_ and only if JC 5 V T , and S e <;C i£ and only if a s"*" < ^. (vii) For each iC e X^X, v {a s"^: S e.iC H ^^t) = if = A {V s"^: S e je HEjj}.

PAGE 33

25 PROOF. (i_) Note first that by Proposition 4.8, v S e AX and S £ V s . Clearly, if ;£ e S , then ^ < v S . Conversely, if Z < V s"*", then SeVS D Y. ^ '£ D Z and so ^ e S . The second equality is proved dually. (ii) Suppose that yn, e XX. It follows from Definition 4.7 that {:L € X X: ? H E^}. The second equality is proved dually. ( iii ) Recall from Proposition 2.16 (v) that for each T e E, T n e [X] = e [T]. Thus since e is in j active, it follows that for each T e E, T = e^^[e^[T]] = e~^[T"^ n e^[X]] = e^^[T^]. Let /^ e X^X. By the above remark and (ii) vfe see. that e {_{^ e X Y.: £ < /t}'\ = e~^[ niS^: S em D E^^}] = n{e^-^[S^]: S e ^ H E^} = n{S: Se/>inE^} = Diy^i n ^) ' The other equality is proved similarly, ( iv ) Let T 6 E . Again using Proposition 2.16 (v) we see that e [T] £ t"^ and so v e [T] < v t"^. If S e v e [T] D E , then it follows from Proposition 4.8 that T ^ S and so S £ v T •, therefore by Definition 4.7, v T < v e [T] and hence v t"*" = v e [T]. A dual argument shows that if T £ E , then a t"^ = a e [T]. (v_) If there is some x £ T H S, then by (iv) above A S < e (x) < V T . Conversely, suppose that j£=aS
PAGE 34

27 (vi) Suppose that J^ e X X and T e Z . By (i) above ^ < \f l'^ if and only if iC e T , and by definition, £ e 1 if and only if T e ^. Thus T £ ^ if and only if ^ < v t"*". Dually, if S e Z , then ^ S>'^ < it if and only if S e X. ( vii ) Suppose that ^ e \ X. Let (2 = v (a s"^: S e ;(f n E } , and '^ =A{vT'*":TeJ^n E }. By (vi) above, for each S e jf fl E and each T e ^if n E , a s"*" < <^ < v t"*" so that ^ < £ < X. Let R e ^ n Sj^. Thus by (vi), ^ < v r"^. Since for each S e ..f H E AS < jZ , it follows from (v) that S n R ?^ for each S e / n E ; i.e., (/ n E ) U{R} is linked. Therefore^ U {R} is linked, and so R e ^. Hence by (v) and the definition of :^ R e c^ By Definition 4.7, ^ < ^ and hence Q = % £>. THEOREM U.ll. Any lattice superextens ion of X is_ a_ complete topological lattice having X as_ a_ sublattice . PROOF. By Theorem 4.8 and Proposition 4.9 any lattice superextension of X is a complete lattice having X as a sublattice. By Proposition 4.10 (i), the given subbase for a lattice superextension XX is a collection of principal ideals and principal dual ideals of XX, and by Proposition 4.10 (ii) each principal ideal and principal dual ideal of XX is closed in the given topology on XX. Thus the topology on XX generated by E is the interval topology on XX; i.e., XX is a topological lattice.

PAGE 35

28 We can conclude this chapter by showing that in the case where X is totally ordered, all of the lattice superextensions of X are also totally ordered. PROPOSITION 4 . 12 . If. X is_ totally ordered and E is_ an_ admissible subbase for X, then T, is_ weakly-normal and AX is_ totally ordered . PROOF. To see that I is weakly -normal , let S, T e E such that S n T = 0. Either S u T = X , in which case {S,T} is a screen for S and T, or there is some xeX-(S U T). In the latter case, since I is a T -subbase for X we may pick S', T' € Z such that x e S' H t', S n S' =0, and T H T' = 0. It now follows from the facts that X is totally ordered and that S and T are not both in E or E that X = X (x) u X (x) c_ S' u T'. Thus {S',T'} is a screen for S and T. Hence by Definition 2.11 (iii), E is weakly-normal . To show that XX is totally ordered, suppose that ^ and Cp^ are members of AX with ^ ^ d. Then thereexist T e 0, H E and S e X H E such that S H T = 0. Since X is totally ordered it must be the case that t < s for each t £ T and seS. IfReXHE, then it follows from R H S ^ that T c_ R, R e ^ n E , and thus ^ ^ ^-

PAGE 36

5. THE COMPLETION BY CUTS In this chapter we characterize the completion by cuts of a topological lattice as a particular lattice superextension (Theorem 5.13). In doing so we prove a theorem (5.11) which is used extensively in proving the main results in this and the succeeding chapter. Throughout this chapter we let X denote a topological lattice. VJe first characterize the completion of X by cuts as a particular complete sublattice of the power set, 2 , of X which is ordered by inclusion. Then using the theorem mentioned above we show that the completion of X by cuts is iseomorphic with the superextension of X with respect to the usual subbase T. We conclude the chapter by considering the case where X is totally ordered and investigating the conditions on X under which the completion by cuts is homeomorphic with the Alexandroff one-point compactification. DEFINITION 5.1. A completion operator on a lattice L is a map L L : 2 -> 2 which satisfies the following conditions: CI. A c
PAGE 37

30 REMARK 5.2. It follows from Ward [15] that if $ is a completionoperator on L, then A = {A e 2 : A = ^(A)} is a complete lattice, ordered by inclusion. Moreover, for each 4* c^ A, (i) A 4' = n 4* and (ii) V H" = $( Uf), and the mapping from L to A defined by x > ^({x}) is an injective, order-preserving map. Next, we consider a specific completion-operator on X. PROPOSITION 5.3. The ma£ D: 2^ ^2^ defined by A^> n{X(a): Ac_x(a)}is_a completion-operator on X. PROOF. C]^: Clearly, for each A e 2^, A c n{X (a): A £X (a)}. £2^: For each A e 2^, by CI, D(A) £ d(D(A)). Conversely, if A £ X (a) for some a e X, then by definition, D(A) ^^r^Ca); therefore, D(D(A)) c D(A). C3_: Suppose that A c B c_ X. If D(B) = X, then clearly D(A) c D(B), If, however, B £ ^n(^) f°i^ some a e X, then, since A c^ B c X (a), it follows that D(A) c D(B). It should be noted that for each a e X, D({a}) = X(a) and that D(A) is the set of all predecessors of all successors of all members of A; i.e., D(A) = {x e X: x < a for all a e X with the property that b < a for all b e A}.

PAGE 38

31 It follows from Proposition 5.3 and Remark 5.2 that X = {A c^ X: D(A) = A} (when partially ordered by inclusion) is a complete lattice and that the map d: X -» X defined by a > D({a}) is injective and order-preserving. In view of the equivalence for D(A) noted in the preceding paragraph and Birkhoff [ 5; Chapter V, §9], X is the completion of X by cuts. PROPOSITION 5.4. The_ injection d : X -»X defined b^ x + D({x}) is a_ lattice inclusion . PROOF. Since d is injective, we need only show that d is a lattice homomorphism (2.2 (i)). Suppose that C is a non-void subset of X such that v c exists in X. It follows from Remark 5.2 that V d[C] = D(u{d(x): X e C}) = D(C). Let c = v C. Thus C c X (c) X X U and hence D(C) < D({c}); i.e., v d[C] < d(v c). Finally, if a e X X ^ such that C c X (a), then a is an upper bound of C; hence c = v C < a U A so that c e X (a). It follows that d(c) = d(v c) < D(C) = v d[C]. " ^ X Thus d(v C) = V d[C]. A dual argument shows that if ?? C <= x such ^ X that AC exists, then d(A C) = a dCC]. Thus d is a lattice X ^ X homomorphism (2.2 (i)). Next, we demonstrate that X (when it is equipped with its interval topology) is also a compact if icat ion of X.

PAGE 39

32 PROPOSITION 5.5. For each A e X (i) A {d(a): A < d(a)} = A = v {d(a): a e A}, (ii) Xp(A) = n{Xj^(d(a)): A < d(a)}, , . (iii) Xj(A) = n{Xj(d(a)): a e A}. PROOF. Let A e X. (i_) Since a e A if and only if d(a) ^ A, it is clear that V {d(a): a 6 A} < A < A {d(a)j A < d(a)}. To see that v {d(a): a e A} = A, let B e X such that B < A. Since X is ordered by inclusion there is some a e A such that a ^ B; therefore X (a) ^ B so that d(a) i B. Thus v {d(a): a e A} = A. To see that a {d(a): A ^ d(a)} = A, suppose that B e X such that A < B. Since D(A) ^ D(B), it follows from the definition of D that there is some a e X such that A <=_ X_(a) and B i_ X (a). Thus A < d(a) and B i d(a). It follows that A = a {d(a): A < d(a)}. (ii) Since A < d(a) implies that X (A) £X (d(a)), it is clear that Xp(A) c n{Xj^(d(a)): A < d(a)}. To see the other inclusion, suppose that B e X such that B e r\{X (d(a)): A < d(a)}. Thus B < d(a) for each a e X such that A < d(a). By (i), A = a {d(a): A < d(a)} and it follows that B < A; i.e., B e X (A). ( iii ) Since a e A if and only if d(a)
PAGE 40

33 PROPOSITION 5.6. For each a e X, d[Xj^(a)] = X (d(a)) and d[Xj(a)] = Xj(d(a)). PROOF. Let a € X. Since d preserves the order of X, it is clear that d[X (a)] £X (d(a)), and since X (d(a)) is closed. d[X (a)] £ X (d(a)). To see the other inclusion, let A e X (d(a)) and let W be a basic open set containing A. If W is of the form X U{X^(B.): i = 1, •••, n} for some I 1 finite set {B, , •••, B } c X, then since A e W, it must be the case 1 n — that for each i, a X < B .. . Since, by Proposition 5.5 (i), A X = A {d(x): A X < d(x)}, it follows that for each i e {l, •••, n} there is some b. e X such that a X < d(b.) and B. ^ d(b.). Let 1 111 c = (a {b.: i = 1, •••, n}) a a. Then c £ X, c < a, and since d is a lattice inclusion, d(c) < d(a). Thus d(c) e d[X (a)]. Finally, note that d(c) e W, for if not, then d(c) e X (B.) for some i € {1, •••, n}, and so B. < d(c) < d(b.) —a contradiction. Hence 1 1 W n d[Xj^(a)] i 0. Consider now the case that W has the form X u{X(C.): j = 1, for some finite set (C, , •••, C } c X. Since for each j = 1, •••, m, 1 m — J 5 5 5 A ^ C, and since A < d(a), it follows that for each j, d(a) j^ C . . Thus d(-a) € W n d[Xp(a)]. Hence W n d[X^(a)] i0. Finally, we consider the case that W = V H U where V = X U{X (C): j = 1, •••, m} and U = X U{X (3.): i = 1, •••, n} for some finite set {B, , •••, B , C, , •••, C}c X. Since A e V, it must 1 n 1 m — be the case that A r^ a X and so A ?^ 0. Since A ^ C. and A = v {d(x): x e (5.5 (i)), it follows that for each j e {1, •••, m} there is some

PAGE 41

34 X. e A such that d(x.) i C. Let x = v {x.: j = 1, •••, m}. Since d is a lattice inclusion, it follows that for each j, d(x.) <. d(x). Also d(x) < A, and since A < d(a), d(x) e d[X (a)]. Note that d(x) e V, for if not, then there is some j e {1, •••, m} with d(x) e X„(C.) and thus d(x.) < d(x) < C. — a contradiction. Note D D D D also that d(x) e U, for if not, then there is some i e {1, •*•, n} with d(x) e X (B.); thus B. < d(x) < A so that A i U — again a contradiction. Consequently, d(x) e W H dCX (a)]. Hence, in each of the possible cases, W H d[X (a)] ^ 0, so that, since W was arbitrarily chosen, A e d[X (a)]. A dual argument shows that d[X (a)] = X (d(a)). LEMMA 5.7. The collection T = {X (d(a)): a e X} U {Xj(d(a)): a e X} y {X} is a subbase for X, {d~"'"[A] : A e T } = T , and {d' [A]: A e T^} = T^. PROOF. Clearly T is a collection of closed subsets of X and T is contained in the usual subbase for X. Thus to show that T generates the interval topology on X we need only verify that all principal ideals of X and principal dual ideals of X, which are not in T, are closed in the topology generated by T. This follows immediately from Proposition 5.5 parts (ii) and (iii). To see that {d"-''[A]: A e Tj^} = Tj^ and {d~-^[A] : A e T^} = T^, note that d [X] = X and since d is a lattice inclusion-, d"-'"[Xj^(d(a))] = X^ia) and d"-'-[Xj(d(a) )] = X (a) for each a e X.

PAGE 42

35 PROPOSITION 5.8. The completion of X b^ cuts is a compactification of X ; specifically , the mapping d : X -» X is_ a_ topologically dense embedding . PROOF. Since X is complete, we know that it is compact (2.9). That d is continuous follows immediately from the facts that T is a subbase for the closed subsets of X and {d [A]: AeT}=T(5.7)isa collection of closed subsets of X. That d is injective follows from the fact that d is a lattice inclusion (2.2 (ii)). To see that d is an embedding, we need only show that d is closed onto its image. Recall that d is a lattice inclusion and is (consequently) injective. Thus for each a e X, d[X (a)] = Xp(d(a)) n d[X] and d[X^(a)] = Xj(d(a)) H d[X]; therefore, d[Xj^(a)] and d[X^(a)] are closed in d[X]. Hence d is an embedding since T is a subbase for the closed sets of X. Finally, to see that d[X] is topologically dense in X, recall that we require that X ^ 0. Let AcX. IfAX
PAGE 43

35 PROPOSITION 5.9. T is an admissible subbase for X. PROOF. By Lemma 5.7, T is a subbase for X. To see that T is a T -subbase for X let A £ X. By the definition of T, n{U £ T: A e U} [n{Xp(d(a)): A < d(a)}] H [n{Xj(d(a)): d(a) < A}] H X. Since n{X^(d(a)): A < d(a)} = Xj^(A) (5.5 (ii)) and n{X^(d(a)): d(a) < A} = n{X (d(a)): a £ A} = X (A) (5.5 (iii)), it follows that n{U e T: A £ U} = Xj^(A) H X^(A) = {A}. Finally, suppose. that A £ X and U £ T such that A i^ U. We consider first the case that U = X (d(a)) for some a £ X. Since h i H, it follows that A i d(a). Since A = v {d(x): x £ A} (5.5 (i)) there is some X £ A such that d(x) ^ d(a). Thus X (d(x)) £ T and it follows that A £ X (d(a)) and X (d(a)) H X (d(a)) = 0. Since x £ A if and only if d(x) < A, we may dualize the above argument in the case that U £ Tj. Hehce T is a T -subbase for X (2.11 (ii)). Since T consists of ideals and dual ideals of X, it is an admissible subbase for X (4.2 (i)). PROPOSITION 5.10. I£a, b £ X, then X (d(a)) H X (d(b)) = if and only if X (a) n X (b) = 0. PROOF. Since d[X (a)] £X (d(a)) and d[X (b)] c X (d(b)), it is easily seen that X (d(a)) n X (d(b)) = implies that X (a) n X (b) = 0. Conversely, if X (a) n X (b) = 0, then since b ^ a and d is a lattice inclusion, it follows that d(b) ^ d(a). Hence X^(d(a)) n Xj(d(b)) = 0.

PAGE 44

37 The following theorem will have several applications in this chapter and Chapters 6 and 8. THEOREM 5.11. Suppose that L and M are topological lattices which satisfy the following conditions : ( i ) f : L ^ M is_ a continuous lattice inclusion . (ii) and 4* are admissible subbases for L and M, respectively, such that Op = {f""^[V]: V e "V^} and = {f"-'-[V]: V e ^ }. (iii) If. V , V 6 ^isuch that V H V ?^ , then f"^Cv^] n f ^[v^] ?f 0. Then X.L is_ iseomorphic with AM via an iseomorphism f" which is an extension of f ; i.e. , such that the following diagram commutes. -^ AqL A^M PROOF. Definition of f " : Suppose that ^ e A L. Clearly, £.' = {y € "V: f~"^[V] € ;^} is a linked system of H. If V e V and ^ {V} uz' is linked, then by conditions (ii) and (iii), if u{f~ [V]} is linked. Since
PAGE 45

38 f* is an extension of f; i.e., e q f = f''' ° e : In vjj.aw of condition (ii) we see that for each x e X, f" o e (x) = f*{{M e 0: X e W}) = f>H{f"^[V]: x e f"-'-[V], V e 4-}) = {V e 4*: f(x) e V} = e (f(x)). Thus the diagram commutes. f is surjective : For each Trie AM it follows from (ii) and (iii) that {f [V]: V e W is a maximal linked system of 0. By the definition of f ''« , f»'K{f [V]: V e W) = ^; therefore, f" is surjective, f* is a lattice inclusion : To see that f»'« is injective suppose J , -^ e XL such that fl ^ ot Thus it follows from (ii) that there are V, V e 4* with f""'"[V] ej), f~"''[V'] e C^, and f~"'"[V] O f""^[V'] = 0. By condition (iii), V H v' = and since, by the definition of f ''' , V e f''(.p and V e f*(:fi), it follows that f*{p i i-i^K) . To see that f-* is a lattice homomorphism , suppose that C is a non-void subset of X^L, and let /. = v C. Let V e 4* . By the definition of f'"', V e f"(^p) if and only if f [V] e jf . Since ^ = V C, f""'"[V] 6 iC if and only if f~"'"CV] e^ for each te C (4.8). For each iC e C, it follows from the definition of f" that f [V] e .C if and only if V e f"(;^). Finally, by Proposition 4.8, V e f"(if) for each
PAGE 46

39 f" is continuous : It follows from the definition of f" that for each f [V] e and each £. e XL, f [V] £ «if if and only if V € f"(;^). Thus for each V e 4, f-'^Cv"*"] = {£. e AgL: f^^'C^) e V^} = {^ e XL: V e f''(^)} (2.14) = U e AgL: f"^[V] e jf } = (f"^[V])^ which is a member of the subbase for the closed sets of A L. Since {V : V e v} is a subbase for the closed subsets of A M , it follows that f" is continuous. f " is a closed map : For each V e m , it follows from the above equality and the injectivity of f" that f''=C(f"^[v])^] = f^cLf^v-^cv"*"]] = V^. By condition (ii), {f~"^[V]: V e H*} = and so { (f'^^CV] )'*' : V g v} = 0"^ is' a subbase for the closed subsets of A L. It now follows that the image under f" of each subbasic closed set in A L is a subbasic closed set in AM and since f" is in j active, f" is a closed map. Thus f" satisfies all of the conditions of Definition 2.10, and so f" is an iseomorphism of A L onto AM. V PROPOSITION 5.12. I£L, M, f, 0, and_ V satisfy the hypotheses of Theorem 5 .11 and M is_ complete , then M is_ iseomorphic with A L via an iseomorphism r which makes the following diagram commutative : ^0^ ^0.

PAGE 47

40 PROOF. In view of Theorem 5.11, we need only show that e : M -^ AM is an iseomorphism; i.e., that e is surjective. If ^e ^vyM, then by Proposition 4.5 (ii), ^ is centered. Since M is compact and ^ consists of closed subsets of M, ix follows that HJf ^ 0. By Proposition 2.16 (iv), HJ^ is a singleton; i.e., H^ = {a} for some a e M. It follows that ^ e (a). Thus if f" is the iseomorphism defined in Theorem 5.11, then r = e ° f " : XL -^ M is an iseomorphism with the property that -hop = P o -h** op — P op oT=r. THEOREM 5 .13 . I^ X is_ a topological lattice , then the completion of X b^ cuts is iseomorphic with the superextension of X with respect to the usual subbase T; i.e., X is iseomorphic with A^X. PROOF. Clearly X and X are topological lattices and d is a continuous lattice inclusion (5.4 and 5.8). T and T are admissible ~ -1 subbases for X and X, respectively (4.4 and 5.9), and {d [A]: A e T } = T and {d~ [A]: A e T_} = T (5.7). Suppose that V V e T such that V, n V^ ?^ 0. If V, and V^ are both in T„ or both in T^ , then it 12 12 D I follows from the definition of T (5.7) that d~ [Vj fl d" [V^] / 0. 1 z In the remaining case we may assume that V £ T and V e T . It now follows from the definition of T (5.7) and Proposition 5.10 that d~^[V^] n d'^EV^] li 0.

PAGE 48

41 Thus X, X, d, T, and T satisfy the hypotheses of Theorem 5.11, and since X is complete, it follows from E*roposition 5.12 that X is iseomorphic with XX and the following diagram commutes. X -*A^X d* X -«S — *X X e«r -^ T If we restrict our attention to the case where X is totally ordered, then T becomes a weakly -normal subbase (4.12), and the completion by cuts (XX) is Hausdorff (2.18 (ii)). The question arises as to how we can characterize the locally compact, totally ordered spaces for which the Alexandroff one-point compactification (wX) is the same as the completion of X by cuts. Before answering this question we need a few preliminary results. DEFINITION 5.m. For each A c X, D(A)^ = {x e X: D(A) 5.Xj^(x)}, PROPOSITION 5.15. For an^(r^ subset A c x, D(A) and_ D(A)'^ are closed subsets of X. Furthermore , D(A) ( dually , D(A) ) is either void or decreasing ( dually , increasing ) ,

PAGE 49

42 PROOF. Let A c_x. That D(A) is closed follows from the fact that D(A) = n{X (a): A e Xj^(a)} (5.3). By Proposition 2.4 (i), D(A) is either void or an ideal of X; i.e., D(A) is either void or decreasing. To see that D(A) is a closed subset of X, we will show that D(A) = ri{X (a): a e A} . Since A c_D(A), it follows from the definition c c of D(A) that for each a e A and each x e D(A) , a < x so that x e X (a). Thus D(A) £ n{X (a): a e A}. To see the other inclusion, suppose that X € X such that x «' D(A) . Thus A ^ ^n^^^ ^° there is some a e A with a i x. Hence x ^ X (a ) and so x ^ n{X (a): a e A}; therefore, D(A) = n{X (a): a e A} is a closed subset of X. It now follows from Proposition 2.4 (i) that D(A) is either void or increasing. LEMMA 5.16. Let X be_ a_ totally ordered topological lattice , (i) For each AcX, D(A) H D(A)'^ ^ 0if and only if_ v A exists in X; moreover , if_ v A exists in X, then D(A) O D(A)^ = {v A}. (ii) For each x e X, v D({x}) = x. (iii) For each A c X, if v D(A) exists , then A D{A)^ exists X A and v^ D(A) = a^ D(A)^. (iv) For each A £ X , if_ any of v A, v D(A), or A D(A)^ exists in X , then they all exist and are equal . (v) For each A £ X, i£ a d[A] = d(x) for some x e X, then A A = X. PROOF. (i) Let A £ X. Suppose that b = v A exists in X. Thus A £ Xj^(b) so that D(A) £ X (b) and hence b e D(A)^ . If a e X such that A £Xj^(a), then b = v A e Xp(a). Thus b e n{X (a): A £X (a)} = D(A).

PAGE 50

U3 Hence b e D(A) n D(A)*^. If x e X such that x e D(A) O D(A)^, then c c since x c D(A) and b e D(A), it follows from the definition of D(A) that b € D(A) c^ ^n^^) ^^'^ so b 5 x. In the same way since b e D(A) and X e D(A), it follows that x e X (b); therefore, x < b. Thus x = b and D(A) n D(A)^ = {v A}. Conversely, suppose now that D(A) H D(A) i and let b e D(A) n D(A)^. Thus A c_ D(A) c X (b) so that a < b for each a e A. If X e X such that x < b, then since b { '"^n^^^ ^"^^ ^ ^ D(A), it follows from the definition of D(A) (5.3) that A ^X (x). Thus there is some a e A such that a i X (x); i.e., a ^ x. It now follows that b = v A. Thus D(A) n D(A)'^ = {v A}. (ii) Since for each x e X, D({x}) = X (x), it is clear that X = V D({x}). ( iii ) Let A c X such that v D(A) exists in X. Let b = v D(A). Thus D(A) c X (b) and by the definition of D(A)^, b e D(A)*^. Since D(A)^ is increasing (5.15), it follows that X (b) £ D(A)^. Recall that D(A) is closed in the interval topology (5.15) and hence in the order topology [ 9; Theorem 12]; hence, b = v D(A) e D(A). It follows that c c for each x e D(A) , b < x and so x e X (b); therefore, D(A) c_ X (b). Thus D(A)'^ = X (b) and so A D(A)^ = b = v D(A). (iv) Let A £ X and suppose that v A exists in X. Let a = v A. By (i) a e D(A)^ so that D(A) ^X (a). Also by (i), a e D(A) and since D(A) is decreasing (5.15), it follows that X (a) c D(A). Thus D(A) = Xj^(a) = D({a}). By (ii), v D(A) exists in X and v D(A) = a. Now using (iii) c c we see that a D(A) exists in X and a D(A) = a.

PAGE 51

44 Suppose now that v d(A) exists in X. By (iii), a d(A) c exists and a D(A) = v D(A). Thus we need only show that v A exists and V A = V D(A). Let a = v D(A). Since D(A) is closed in the interval topology (5.15), it is also closed in the order topology [ 9 ; Theorem 12] and so a = v D(A) e D(A). Since D(A) is also closed in the interval topology (5.15) and hence in the order topology, it follows that a = a D(A)*^ e D(A)^. Thus a e D(A) H D(A)^ and by (i) V A exists in X and a = v A. c c Finally suppose that a D(A) exists in X and let a = A D(A) . c c c since D(A) is closed, it follows as before that a = a D(A) e D(A) . Thus D(A) £ ^n^a)Now if x 6 X such that A c_ X (x), then by the definition of D(A), D(A) 5. X (x) so that x e D(A)^. Thus a < x and so a e X (x). Hence a e n{X (x): A ^X (x)} = D(A). Thus D(A) n D(A)^ t 'Ji. By (i) v A exists in X and a = v a. By a previous case V D(A) exists and v D(A) = a = v A = a D(A)'^. (v_) Suppose A c^ X such that a d[A] = d(x) for some x e X. For each a e A, d(x) < d(a) and so x < a since d is a lattice inclusion. If c e X such that x < c, then since d is infective, d(x) < d(c) and there is some d(a) e d[A] such that d(x) < d(a) and d(c) -^ d(a). Since "X is totally ordered (4.12 and 5.13) it must be the case that d(a) < d(c) Thus since d is injective, x < a < c. It follows that x = a A. The following lemma is well known and so its proof will be omitted.

PAGE 52

45 LEMMA 5 . 17 . If_ H is_ any Hausdorff locally compact space , and if Y and Z are two Hausdorff compact if icat ions of H with |Y-H| = |Z-H| =1, then Z and Y are homeomorphic . The following theorem characterizes those situations in which the completion by cuts is the Alexandroff one-point compact if icat ion. THEOREM 5.18. Suppose that X is_ a locally compact , totally ordered topological lattice . Then in order for the completion by cuts of X ( X ) to_ be_ homeomorphic with the Alexandroff one-point compactification of X ( uX ) i^ i^ necessary and sufficient that there exists exactly one subset A £ X such that v A and a (X A) do_ not exist in X. PROOF. Necessity : If there does not exist B c_ X such that v B does not exist in X, then for each Bc_X,vB = vD(B) = b for some b e X (5.16 (iv)). Thus D(B) ^ 0, and since D(B) is a closed ideal of X (2.4 (i)), it follows that D(B) = d(b). Therefore X = X is compact, and since X ^ ujX, X ^ wX -a contradiction. Thus there is some B c_ X such that V B does not exist in X. Let A = D(B). Since A is decreasing (5.15) and X is totally ordered, it follows that X A = D(A)^. Since D(A) = D(D(B)) = D(B), it follows that D(A)^ = D(B)^. Thus by Lemma 5.16 (iv), V A and a (X A) do not exist in X and so at least one such subset of X exists. Suppose now that there is another set C £ X such that v C and A (X C) do not exist in X. Recall that D(C) and A are members of X and that X is homeomorphic with wX. If D(C) e d[X], then there is

PAGE 53

46 some X e X such that D(C) = d(x). Since d(x) = D({x}), it follows that V D(C) = X (5.16' (ii)). Hence v C = x (5.16 (iv)), but this contradicts the fact that v C does not exist in X. Thus D(C) i d[X]. By a similar argument, using the fact that v A does not exist in X, we see that A i d[X]. Now since |x d[X]| = |wX x| =1 and D(C), A e X d[X], it follows that D(C) = A. If C ;^ D(C), then since C c D(C), there is some x e D(C) C, and since v D(C) does not exist in X, d(x) < D(C) in X. Let G = d[X C]. Then d(x) e G so that G is a non-void subset of X. Since X is complete, A G exists in X and A G < d(x). Thus A G i^ D(C), so it must be the case that A G e d[X] and there is some z e X such that a G = d(z). Hence z = A (X C) (5.16 (v)); i.e., A (X C) exists in X — a contradiction. Therefore, C = D(C) = A, and A is the unique subset of X for which V A and A (X A) do not exist in X. Sufficiency : Let A £ X be the unique set such that v A and A (X A) do not exist in X. Thus v D(A) does not exist in X (5.16 (iv)), and since X D(A) = D(A)^, a (X D(A)) does not exist in X (5.16 (iv)). By the uniqueness of A, A = D(A) and so D(A) i D({x}) for all x e X • (5.16 (ii)); i.e., D(A) = A e X dCX]. If B e X such that B i d[X], then V B does not exist in X, for suppose that there is some b e X such that V B = b. Then since B is a closed decreasing subset of X (5.15), it follows that b e B and Xp(b) = B; therefore, B = D({b}) = d(b) e d[X] ^--a contradiction. Finally, since v B does not exist in X and B = D(B), it follows that a (X B) = A D(B)^ does not exist in X (5.16 (iv)). Hence by the uniqueness of A, A = B.

PAGE 54

47 Thus X d[X] = {a} and since X is a Hausdorff compactification of the locally compact Hausdorff space X formed by adjoining one point to X , X is homeomorphic with wX (5.17). In view of the fact that the open n-cell in Euclidean n-space can be characterized as the product of n-copies of the open interval (0, 1), it follows from Theorem 5.13 and the Product Theorem (3.4), that the closed n-cell (i.e., the n-cell with boundary attached) is a de Groot compactification of the open disc. Likewise, since Hilbert space can be characterized as the countably infinite product of the open interval (0, 1) [3], it follows that the Hilbert cube is a de Groot compactification of Hilbert space. (See Example 8.3 for details.)

PAGE 55

6. TOPOLOGICAL LATTICE COMPLETIONS Recall that Theorem 5.13 provided a superextension characterization of the completion by cuts. The purpose of this chapter is to obtain algebraic characterizations of other lattice superextensions. This is done via the notion of topological lattice completions. Throughout this chapter X will denote a topological lattice. The topological lattice completions of X are complete topological lattices having X as a sublattice and a subspace. We show that all lattice superextensions of X are actually topological lattice completions of X (Theorem 6.22). Conversely, all topological lattice completions of X which satisfy a certain density condition are shown to be lattice superextensions of X (Theorem 6.20). Throughout this chapter Y will denote a complete topological lattice (i.e., a complete lattice equipped with the interval topology) and all subbases considered are assumed to be subbases for closed sets. NOTATION 6.1. Let Z be a non-void subset of Y. (i) Z° = Z. (ii) \^Y^^ € Z: z < y}, if (z e Z : z ^ y} i^ 0, or 2^^ = ^y ^ Y: y = |^^2, if {z € Z: z < y} = > iAylz e Z: y < z}, if {z e Z : y < z} i^ 0, or V Z, if {Z . Z: y . z} = ^ 7} z^^ n z^^ 48

PAGE 56

49 (iii) Z^^ = (Z^^^^ Z^^ = (Z^^^^ z^ = 7?^ n z2^ (iv) 0° = 0^^ = 0^^ = 0^^ = 02^ It should be noted that Definition 6.1 can be easily extended ct to obtain Z for each ordinal a, and if we define the "lattice closure a . Qt, a+1 of Z in Y" to be Z where a is the least ordinal such that Z = Z , then the map which takes Z onto its lattice closure in Y can be shown to be a Moore closure operator, or completion-operator on Y as in Definition 5.1. See Appendix II for the details. PROPOSITION 6.2. Suppose that A ^ B ^ Y. Then , . . .la „la (i) A £ B ,.., ,1b „lb (ii) A c_ B (iii) A^ c B^ (iv) A c A-"c A^^ U A^^ c A^. PROOF. {i) If y e A"""^, then {a e A: a < y } £ {b e B : b < y} and y = V {a e A: a < y} < V {b e B: b < y} < y; i.e., y e B . (ii) Dualize the argument for (i). ( iii ) This is immediate from (i) and (ii) above and the definitions of A and B (6.1 (ii)). (iv) The proof is immediate if A = (6.1 (iv)); so assume that A i"! 0.

PAGE 57

50 That A c_ A follows from the facts that for each w c A, w=v{aeA:a
PAGE 58

51 IfW = YU{Y(z.): i = l, •••,n} for some finite set (z,,***, z } c Y, then since for each i, z. i v, it follows that 1 n — 1 Yp(y) (\ (U{Yj(z.): i = 1, •••, n}) = and so (A n Y^(y)) n (U{Yj(z^): i = 1, •••, n}) = 0. Hence A. n Yj^(y) c W and W n (A n Y (y)) ^ 0. If W = Y U{Y (w.): j = 1, •••, m} for some finite set {w. , •••, w } c Y, then for each j=l, •••,m, y^w.. Since 1 m— --'''jj y = V {a e A: a < y}, it is possible to pick, for each j e {1, •••, m}, some X. e A with the property that x. < y and x. ^ w.. Let a = V {x. : j = 1, •••, m}. By the hypotheses on A and since each X. e A, it follows that a e A and a < y; i.e., a e A H Y (y). If a ^ W, then there is some j e {l, •••, m} with a e Y (w,); therefoije , X. s a s w. and x. < w.--a contradiction. Thus a e W and so W r\ (A n Yj^(y)) i0. In the remaining case W = U H V where U = Y u{Yj^(w.): j = 1, •••, m} and V = Y u^Y^Cz.): i = 1, •••, n} for some finite set {w, , •••, w , z^ , •••, z } c Y. By an argument 1 ml n — analogous to the one above, there is some aeU H (A nY(y)) and since A C\ Y (y) £ V, it follows that a e W O (A H Y (y)). Since W was arbitrarily chosen, it follows that y e (A H Y (y)) (ii) Dualize the argument for (i). PROPOSITION 6.5. I£i:X->-Yis_a lattice inclusion and if iCX]-"= Y, then v i[x] = v Y and a i[X] = a Y.

PAGE 59

52 PROOF. Clearly v Y e Y = iCX]"'" ^ iCX]"''^ (6.2 (iv)). Thus V Y = V {i(x) £ i[X]: i(x) < v Y} = v {i(x) e iCX]} = v i[X]. The other equality is proved dually. NOTATION 6,6. Ifi:X-+Yisa lattice inclusion such that Y = i[X]"^, then we let 'r = {Y} u{Yp(i(x)): X € X} u{Yj(i(x)): x e X}. PROPOSITION 6.7. Ifi:X->Yisa lattice inclusion such that i[X] = Y, then T is_ an_ admissible subbase for Y. PROOF. Clearly T consists of ideals and dual ideals of Y. Since Y = i[X] , it follows from Proposition 5.3 that each member of the usual subbase for Y is an intersection of members of T; therefore T is a subbase for the closed subsets of Y. To see that T is a T -subbase for Y, suppose that z e Y. Then since z e iCX]""" = iCX]"""^ niCX]"""^, it follows that n{T e T: z e T} = (n{Y (i(x)): x e X and z < i(x)}) O (n{Yj(i(x)): x e X and i(x) < z}) D Y = Yp(z) n Y^(z) (6.3) = {z}. Finally suppose that z e Y and T e T such that z i 1 . Clearly T ?! Y. If T 6 'r , then there is some t e X such that T = Y (i(t)). 1 la Since z i i(t) and z e i[X] £ iCX] (6.1 (ii)), there is some x e X such that i(x) < z and i(x) ^ i(t). Thus z e Y (i(x)), Y (i(x)) e T , and Y (i(t)) O Y (i(x)) =0. If T e ^ , then a dual argument shows that

PAGE 60

53 there is some Yj^(i(x)) e T^ with z e Y (i(x)) and T H Y (i(x)) = 0. By Definition 2.11 (ii), T is a T -subbase for Y. Hence T is an admissible subbase for Y (4.2 (i)). PROPOSITION 6.8. I£i:X->Yisa lattice inclusion such that Y = i[X]^, then {i'^[T]: T e T^} = T^ and {i"^[T]: T e T^ } = T^ where T is_ the usual subbase for X. PROOF. Clearly T^ £ {i"-'-[T]: T e T^} and T^ c {i'-'-CT]: T e T }. To see the other containments, recall that i is a lattice inclusion; therefore, for each a e X, i"-'"[Yp(i(a) )] = {x e X: i(x) < i(a)} = {x e X: X < a} = X^Ca) € T^ and dually, i"^[Yj(i(a) )] = X^Ca) e T^.. PROPOSITION 6.9. Suppose i: X -> Y is_ a lattice inclusion such that i[X] = Y. If S, T e T, then i~"'"[S] H i""^[T] i whenever S H T ;>! , PROOF. Suppose S, T e T such that S T j^ 0. If S and T are both in Tp or both in Tj , then the conclusion is immediate from the definition of T (6.6) and the fact that i is a lattice inclusion. For the remaining case we may assume that there exist t and s in X such that T = Yj(i(t)) and S = Yj^(i(s)). Since S H T ?^ , we see that i(t) < i(s) and since i is a lattice inclusion, t ^ s. Thus i'"^[S] n i"^[T] = Xj^(s) n Xj(t) i 0. DEFINITION 6.10. Suppose that X and Y are topological lattices, Y is complete, and i: X -> Y is a function. Y will be called a topological

PAGE 61

54 lattice completion of X with respect to i whenever the following two conditions are satisfied: (i) Y iCX]^ or Y = i[X]^. (ii) i is a continuous lattice Inclusion. PROPOSITION ,6.11. If i: X ^Y is a lattice inclusion and i[X] = Y, then Y is_ a_ topological lattice completion of X with respect to_ i ; i • e • J the map i is_ continuous . PROOF. Since ^ is a subbase for Y (6.7), and {i"''"[T]: T e T} is the usual subbase for X (6.8), it follows immediately that i is continuous. PROPOSITION 6.12. If Y is a topological lattice completion of X with respect to i : X ->Y , then i is_ an_ iseomorphism onto its image ; i.e., i is_ an_ embedding . PROOF. In view of the definitions of topological lattice completions (6.10) and lattice inclusions (2.2 (ii)), we need only show that the restriction i: X ^ i[X] is a closed map. For each a e X, i[Xp(a)] = {i(x): x e X and x < a}. = {i(x): X e X and i(x) < i(a)} (since i is a lattice inclusion), = i[X] nYj^(i(a)). "" Dually, for each a e X, i[Xj(a)] = i[X] nYj(i(a)). Thus the image of every member of T is a closed set in i[X] , and since i is injective, it follows that the image of each closed set in X is closed in i[X].

PAGE 62

55 THEOREM 6.13. Ili:X->Yis_a lattice inclusion and i[X] = Y, then Y is_ iseomorphic with the completion of_ X by cuts, PROOF. By Proposition 6.11, the function i: X -» Y is a continuous lattice inclusion. T and T are admissible subbases for X and Y, respectively, such that {i~ [T] : T e *? } = T and {i"-"-:!]: T e '?j} T^ (6.7, 6.8, and 4.4). If T, S e '? such that T n S ^ 0, then i"^[T] n i"^CS] i (6.9). Thus X, Y, i, T, and h satisfy the hypotheses of Theorem 5.11, and since Y is compact it follows from Proposition 5.12 that Y is iseomorphic with AX, the completion of X by cuts (5.13). REMARK 6.14. It follows from Theorem 6.13 that for any topological lattice X, there is only one completion Y of X for which Y = X . This occurs just when X is a complete lattice. In the case 2 1 where X £ Y and X , but not X , is a complete sublattice of Y, the possibilities are not so limited. In fact, in this case a lattice inclusion need not be continuous (see Example 8.6, Part I); however, for the remainder of this chapter we will consider only topological lattice completions of X. In order to use Theorem 5.11 (as was done in the proof of Theorem 6.13) to characterize one of these "larger" topological lattice completions as a lattice superextension, it is necessary to obtain a subbase which satisfies certain conditions. Namely, if Y is a

PAGE 63

56 topological lattice completion of X with respect to i, then we must choose some subbase E for Y such that (i) E is an admissible subbase for Y, (ii) {i [S]: S e E} is an admissible subbase for X, and (iii) if S, T e E with S H T ?! , then i'^^CS] H i""^[T] i 0. Keeping in mind conditions (i) and (ii) we make the following definition. NOTATION 6.15. If Y is a topological lattice completion of X with respect to i, then E^^^ = {Y} U{Yj^(y): y e (i[X]^^ i[X]^) u i[X]} utYj(y): y e (i[X]^^ i[X]^) u iCX]}. PROPOSITION 6.16. If Y is a topological lattice completion of 1 (i) ^ X with respect to i and Y = i[X] , then E = T. PROOF. This follows immediately from the fact that when Y = i[X]^, then iCX]-"= i[X]^^ i[X]^^ and so i[X]^^ i[X]^ iCX]^^ i[X]^ = 0. PROPOSITION 6.17. I£ Y is_ a topological lattice completion of X with respect to i , then (i) E is_ an_ admissible subbase for Y, and (ii) {i [S]:SeE }isan admissible subbase for X.

PAGE 64

57 PROOF. (i) Clearly Z consists of principal ideals and principal dual ideals of Y. In order to see that E is a subbase for Y, we need only show that all principal ideals of Y and principal dual ideals of Y which are not in Y. are closed in the topology on Y which has E as a subbase. If Y (y) is a principal ideal of Y and Y (y) ^ E , then it follows from the definition of E that either (a) y e i[X] , or (b) y e i[X]^ (i[X]^^ u i[X]^^). Dually, if z 6 Y such that Y (z) I E , then either (c) z e i[X] , or (d) z 6 i[X]^ (iCX]^^ U iCX]^^). (a^) If y e i[X] , then y e iCX] , and so by Proposition 6.3 (ii), Yjj(y) = n{Yp(i(x)): X e X and y < i(x)}. (c_) If z g i[X] , then z e i[X] , and by Proposition 6.3 (i), Yj(z) = n{Yj(i(x)): X e X and i(x) < z}. Thus for each w e i[X] , Y (w) and Y (w) are closed in the topology generated by E . Hence in view of the definition of E (6.15), for each w e i[X] u i[X] , Y (w) and Y (w) are closed in the topology generated by E (b) If y € i[X]^, then y e i[X]^^, and since i[X]^^ = (iCX]^^)^^, Yj^(y) = n{Yj^(w): w e i[X]^^ and y < w} (6.3 (ii)). (d) If z e i[X]^, then z e i[X]^^, so that since i[X]^^ = (iCX]""-^)^^, Yj(z) = n{Yj(w): w e iCX]''"^ and w < z} (5.3 (i)).

PAGE 65

58 It now follows from parts (b) and (d) that if q e i[X]^ (i[X]"''^ u iCX]"""^), then Y (q) and Y (q) are intersections of closed sets in the topology on Y generated by E , and hence are closed in that topology. Thus Z is a subbase for the interval topology on Y. To see that Z is a T -subbase for Y, let z e Y and ^= {S e E: z e S}. Clearly ^ is linked. To see that ^ is a maximal linked system of E , suppose that T e E such that T d H. Then z ^ T . We will assume that T e E^"*" and note that a dual argument should be used when T e E-. . Thus there is some y e (i[X] i[X] ) U i[X] such that T = Y (y) and z j^ y. We consider the possibilities for z. (e) z £ (i[X]^^ i[X]^) u i[X]. (f) z € i[X]^^. (g) z e i[X]^ (i[X]'^^ u iCX]^^). (e) If z e (i[X]^^ i[X]^) u i[X], then S = Y^Cz) e Z^^\ z e S, and S n Y^(y) = Yj(z) n Yj^(y) = 0. (f_) If z € i[X] , then since z ^ y, it is possible to pick some w e i[X] such that w < z and w ^ y. Thus S = Y (w) e E , z e S, and ^S n Y^iy) = Y^(w) n Yj^(y) = 0. (g) If z e i[X]^ (iEX]-"-^ u i[X]^^), then since z e i[X]^^ and z i Y, there is some q e i[X] such that q.< z and q ^ y. If q i i[X]"^, then S = Y^Cq) e E^^ , z e S , and S H Y^(y) = Y^(q) H Y^Cy) = 0. If q e i[X] £ i[X] , then since q j^ y, there is some w e i[X] such that w < q and w ^ y. Hence S = Y (w) e E "^ , z e S since w < q < z, and S n Y^(y) = Yj(w) n Y^(y) = 0.

PAGE 66

59 It follows from the arguments in (e), (f), and (g) that in all cases there is some S e E_ such that z e S and S H T = 0. Clearly S e ff so that //* U{T} is not linked. Thus ^ is a maximal linked system of Z .By Proposition 2.13, E is a T -subbase for Y; therefore, E is an admissible subbase for Y (4.2 (i)). (ii) Let E = {i~ [T]: T e E }. Since i is continuous, it is clear that E is a collection of closed subsets of X. Also, since i is a lattice inclusion, it follows that {i [T]: T e E_^ } is a collection of ideals of X and {i [T] : T e E^""" } is a collection of dual ideals of X. Thus E £_ F . It follows from the definition of E (6.15) that E contains all the principal ideals and principal dual ideals of X. Hence by Proposition 4.3, E is an admissible subbase for X. In view of Proposition 6.17, the subbase E which we have chosen for Y (where Y is a topological lattice completion of X with respect to i) satisfies the first two conditions in Remark 6.m. It is not true, however, that E necessarily satisfies the third condition; i.e., it is not always true that if S, T e E such that S D 1^0, then i~ [S] H i~ [T] ^ 0. An example of this can be seen in 8.8. Thus we are lead to define the notion of "lattice density."

PAGE 67

60 DEFINITION 6.18. If Y is a topological lattice completion of X with respect to i: X -> Y, then we say X is lattice dense in Y whenever the following condition is satisfied: If y e iCX]"*"^ i[X]^ and z e i[X]"^^ i[X]^ with the property that y < z, then there is some x e X such that y < i(x) < z. It should be noted that if Y = i[X] , then X is lattice dense in Y since i[X]^^ i[X]^ = i[X]^^ iCX]"^ = 0. PROPOSITION 6.19. Suppose that Y is_ a_ topological lattice completion of X with respect to i : X ->Y and X is_ lattice dense in Y. If S, T e E^^^ such that S H T ?^ , then i'"^[S] H i""'"[T] i 0. PROOF. If either T = Y or S = Y, then since i~ [Y] = X, the proof is trivial. If T and S are both in 1 or both in E_ , then since {i [R] : R e T^t^ s is a collection of closed ideals of X and dually {i~ [R] : R e zJ' } is a collection of closed dual ideals of X, the proof again is trivial (4-. 5 (i)). In the remaining case we may assume without loss of generality that there exist t e (i[X] i[X] ) U i[X] and s e (i[X]^^ i[X]^) U i[X] such that Yj^(t) = T and Y (s) = S. Since S H T ^ , it must be the case that s < t. If s e i[X], then s = i(x) for some x € X and x e i~ [Y (i(x))] H i [Y (t)] = i""'"[S] n i""''[T]. Dually, if t e i[X], then t = i(x) for some x e X and X € i~^[Yj^(t)] O i'^CY^Cs)] = i"^[S] H i~^[T] . Thus suppose that s , t «^ i[X]. Since neither s nor t is in iCX] and s < t, it follows

PAGE 68

61 that s < t. Now since X is lattice dense in Y, there is some x e X with s < i(x) < t; therefore, x € i" [Y (t)] H i" [Y (s)] = i"^[s] n i"^[T]. THEOREM 6.20. Given that Y ij_ a topological lattice completion of X with respect to the inclusion map i : X -> Y , X is^ lattice dense in Y, and E={i [TjiTeZ }, then Y is_ iseomorphic with XX via an iseomorphism f : XX ->• Y with the property that f o e = i. PROOF. By the definition of topological lattice completions (6.10), i is a continuous lattice inclusion. E and Z are admissible subbases for X and Y, respectively, such that {i [S]: Se E """ } E and {i""'"[S]: S e E^ ^ = E^ (6.17). If S, T e E^^^ such that S n T ?i 0, then i~"^[S] H i""^[T] ^ (6.19). Thus X, Y, i, E, and E satisfy the hypotheses of Theorem 5.11. Since Y is complete, there is an iseomorphism f : X X -v Y with the property that f » e = i (5.12). We now consider the converse question; i.e., which lattice superextensions are topological lattice completions of X in which X is lattice dense? PROPOSITION 6.21. If E is an admissible subbase for X, then for each T e E^^ ( dually , T e E ), v t"^ e e [X]"'"^ ( dually , a t"*" e ej.CX]''"^),

PAGE 69

62 PROOF. Suppose that T e E . Thus v t"*" = v e [T] (4.10 (iv)). It follows that V t"*" = V {e (x) : e (x) e e [X] and e (x) < v t"*" } ; Li Li Li Li therefore, v T £ e [X] . A dual arg 'nent shows that if T e E , then A t"^ e e^[X]^^. THEOREM 6.22. If_ E is_ an_ admissible subbase for X, then X X is_ a_ topological lattice completion of X with respect to e : X ->XX. PROOF. Let E be an admissible subbase for X. Recall that e is a continuous lattice inclusion (2.16 (v) and 4.9) and that for each ;€€ A^X, V {A s"*": S £ ^ n Ej} = ^ = A {v S"^: S £ ;€ n E^} (4.10 (vii)). Thus since for each S e ^ n E-., a s"*" < ;d (4.10 (vi)) and A S"^ £ e [X]"'"^ (6.21), it follows that ^ = ^ {^^ ^5;^= ^e ^e"^^^^^ ^^^ S -^^ ^ (e^[X]^^)^^ = e^CX]^^. Dually, since for each S £ ;f H E„ , v s"*" £ e [X]"""^ (6.21) and ^ < v s"*" (4.10 (vi)), then I = A {^£ X^X: ^ e e^[X]^^ and £


PAGE 70

63 DEFINITION 6.23. Suppose that Z is an admissible subbase for X. Then for each id e XX, we let XiD,iE) = {x e X: e^(x) < £} and X(I,^) = {x e X: .^ ^ e^(x)}. REMARK 6.24. It is easily seen from Definition 6.23 that for each ;£ e X„X, V e^[X(D,:f)] < Jf < A e^[X(I,;^)]. Hence exactly one of the following statements is valid : (i) V ej.[X(D,:£)] = ;^ = A e^[X(I,£)] (ii) V e^[X(D,;f)] = / < A e^[X(I,i6)] (iii) V e^[X(D,;d)] < £ = a e^CX(I,;t)] (iv) V e^CX(D,;C)] < :d < A ej,[X(I,;d)]. Clearly for each £ e AX, X(D,;^) is a closed ideal of X which may or may not be in E and X(l,t) is a closed dual ideal of X which may or may not be in E. DEFINITION 6.25. If E is an admissible subbase for X, then E will be called thick if and only if for each pair /, W e AX such that i. satisfies condition (ii) in Remark 6.24 and ^ satisfies condition (iii) in Remark 6.24, it follows that G = E U{X(D,jf), X(I,?r)} is an admissible subbase for X and there exists an iseomorphism f : A X -> AX such that ^ ° ^E = ^0-

PAGE 71

54 THEOREM 5.26. If. ^ i£ a thick , admissible subbase for X, then X is_ lattice dense in AX. PROOF. To see that X is lattice dense in XX, let J^, /? e AX such that I e e^CX]^^ e^[X]^,;>fe ^^^'^^^ " ^z""^^^' ^""^ ^^ ^^ ^^"^^ X = V {e^(x): X e X and e^(x) < X.} = v e^CX(D,Jd)] (5.23). Since £. i e [X] , it must be the case that ^ < A {e^(x): X e X and j2 < e^(x)} = A e^[X(I,^)] (6.23). Hence •£ satisfies condition (ii) of Remark 5.2M-. By a dual argument we see that "fli satisfies condition (iii) of Remark 5.24. Thus since E is thick, AX is iseomorphic with AX where = E U{X(D,^), X(I,/<)} (5.25) Let f: AX-)AX denote the iseomorphism. Since f is a lattice isomorphism, f o e = e (5.25), and X(D,;^) e , it follows that fC^) = f(v e^[X(D,/)]) = V f[e^[X(D,i)]] (2.2) V eQ[X(D,sd)] = V X(D,X)^ (4.10 (iv)). Dually, since X(I,;^) e , f(>?j) = A X(I,%)^. Thus X(D,;^) e f(jf) and X(I,/^) e f(/Y) (4.10 (vi)). Since f preserves the order of AX (2.5) and /^ < vd, it follows that f{n) < f(if) and so x(i,;7z) e fim) n 0^ c f (^) n 0^ (4.7); therefore X(D,;f), Xil,Pt) e f(it) and so X(D,:d) H X(I,^) ^ 0. Let X e X(D,;d) D Xil, 7n). Then f(^) < e^ix) < f(jd). Since f o e = e , we have fiH) < f(.e (x)) < f(iC), and since f is a lattice isomorphism, yn< e (x) < id. Finally, since X'i^ e [X] and Jd I e [X], it follows that /? < e (x) < *f . Consequently, X is lattice dense in A X.

PAGE 72

65 PROPOSITION 6.27. If Y is_ a topological lattice completion of X with respect to i : X -* Y and X is_ lattice dense in Y , then I = {i [S]: SeZ ^i^^ thick , admissible subbase for X. PROOF. Recall that Z is an admissible subbase for X (6.17 (ii)). By Theorem 6.22, Y is iseomorphic with XX. To see that Z is a thick subbase for X, suppose that i£ , M e AX such that t satisfies condition (ii) ,of Remark 6.24and ^ satisfies condition (iii) of Remark 6.24. Since V {ej,(x): X e X and e^(x) <^} = X. < A {e^(x): X e X and ^ <.e^(x)} (6.23, 6.24), it follows that £ e e^CX]"*"^ e^[X]"^. Thus {Q e XY: C < £] e E^^^ (6.15), L L ' L ' D and so X(D,^) = i"^[{^e A^X: ^ < t)'] e l^. Dually, /Tie e^[X]^^ e^CX]^, {jl e Aj,X: yn.< p e l^^^\ and so X(I,»z) = i~^[{ ^ e A^X: )^ < ^}] £ l^. Consequently, = Z u{X(D,Jl), X(I,»£)} = E is an admissible subbase for X, and there is an iseomorphism f : A X ->• AX, namely, the identity map. such that f o e = e . Hence I is thick, PROPOSITION 6.28. Th£. usual subbase T and the_ subbase V are both thick. PROOF. To see that T is thick, we recall that AX is the completion by cuts of X so that AX = e [X] . Thus it follows that every member of AX satisfies condition (i) in Remark 6.24. Hence T vacuously satisfies the definition of a thick subbase.

PAGE 73

55 To see that r is thick, suppose that S^, />^e AX such that «£ satisfies condition (ii) of Remark 5.2'+ and Z'? satisfies condition (iii) of Remark 5.24. Since X(D,/) is a closed ideal of X and X(l,Pt) is a closed dual of X, then X(D,;t) e r and X(I,;>^ e r (4.1 (ii)). Thus = r u{X(D,;C), X(I,/»)} = ris an admissible subbase for X (4.4-) and the identity map f : XX -^ AX is an iseomorphism such that f o e = e . COROLLARY 6.29. X is lattice dense in AX and AX. PROOF. Since both T and r are thick (5.28), admissible subbases for X (4.4), the proof is immediate from Theorem 5.25. All of the admissible subbases which are considered in the examples (Chapter 8) are thick subbases for X. Whether or not every admissible subbase for X is thick remains an open question. THEOREM 6.30. Suppose that X is a topological lattice and is an admissible subbase for X. Then X is lattice dense in AX if and only if there exists a thick admissible subbase E for X such that Ax is iseomorphic with Ax via an iseomorphism f : A„X -> A^X with the property that f o e = e . PROOF. Let be an admissible subbase for X. Thus AX is a topological lattice completion of X with respect to e^: X ^ AgX (5.22).

PAGE 74

67 -1 ^^0^ IfX is lattice dense in X^X, then Z = {e^ [S] : S e L } is a thick admissible subbase for X (6.27) and XX is iseomorphic with XX via an iseomorphism f : X X ^ XX with the property that f o e^. = eg (6.20). Conversely, if there is some thick admissible subbase E for X and an iseomorphism f : XX -> XX such that f ° e = e , then since X is lattice dense in XX (6.26), it follows that X is lattice dense in XX.

PAGE 75

7. LARGEST AND SMALLEST LATTICE SUPEREXTENSIONS In this chapter we will consider the possible existence of certain maps between different lattice superextensions of the same topological lattice. The results will then be used to establish the existence of maximal and minimal lattice superextensions for all topological lattices and in certain cases, largest and smallest lattice superextenions (Theorem 7.15). Thus in the realm of lattice superextensions we have analogues to the Stone-Cech and Alexandroff compactifications. Kaufman [12] obtained similar analogues for the totally ordered case. Theorem 7.15 is a generalization of Kaufman's results. In this chapter , as in the previous chapters , X will denote a topological lattice. Also, to avoid confusion, we will sometimes use the symbol """ in place of "+" to denote the subbasic closed sets of a lattice superextension; e.g., if is an admissible subbase for X, then for each T e 0, T''« = {t e XX: lei.) and 0''« = {P-: T e 0}. DEFINITION 7.1. Let and E be admissible subbases for X. 68

PAGE 76

69 (i) We say that XX algebraically dominates XX if and only if there is a surjective, order-preserving function f: XX -> XX such that the diagram \^ ^0^ commutes (i.e., f » e = e ) and such that f preserves the suprema of ideals in X and the infima of dual ideals in X; i.e., if I is an ideal of X and D is a dual ideal of X, then f(v e [I]) = v (f o e^)[I] and f(A e [D]) = a (f o e )[D]. Such a function is called a dominating function of XX over XX. ' ' Li ' W (ii) We say that the superextension XX dominates the superextension XX if and only if XX algebraically dominates XX via a continuous dominating function. It might be noted that although XX and XX are both lattices, a dominating function f is not required to preserve all suprema and infima of XX; i.e., f is not required to be a lattice homomorphism. The reason for this is apparent from Example 8.9 in which a topological lattice X is exhibited for which the lattice XX algebraically dominates the lattice XX (and also for which the superextension XX dominates

PAGE 77

70 the superextension AX), but for which there are i^, W e A X such that f(:^) A f (/%) i f{£. AfK). (See 8.9 for details.) Example 8.7 shows that algebraic domination does not necessarily imply superextension domination, for in this example there exists a unique algebraic dominating function f: AX * AX which is not continuous. PROPOSITION 7.2. Suppose that and E are admissible subbases for X such that AX algebraically dominates AX via a_ dominating function f : AX -> AX. Then (i) for each R e ^^ ( dually , R e E ), f( v r"*") = v e [R] ( dually , f(AR^) = aSqCR]), (ii) for each R e E Pi ( dually , R e E H ) , f ( v r"*") = v R>'( dually , f ( A R ) = A R--V), and (iii) for each ;C e A^X, iX. O 0) c f (jC) . PROOF. (i_) Suppose that R e E . Then by the hypotheses on f and since R is an ideal of X, it follows that f(vR^) = f(ve^[R]) (U.IO (iv)) = V (f o e^)[R] = V eg[R]. If R e ^j5 then a dual argument shows that f ( a R ) = a e [R]. (ii) Suppose that R e E H . Then since f ( v r"*") = v e [R] ((i) above) and v e [R] = v R" (i+.lO (iv)), we see that f(v r"^) = v R" . Dually, if.R e E Pi , then f(A r"*") = a R. ( iii ) Suppose that it e AX. Let Re {£ n 0j.). Thus jjif < v r"^ (4.10 (vi)), and since f is order-preserving, f(;^) < f(v R ) = v R" (Cii) above) ;i.e.. Re fU) (4.10 (vi)). Hence (^ H ) < f(/), and by a dual argument, (,^ n 0^) c f(^). Therefore, (;f r\ 0) c_f(/).

PAGE 78

71 LEMMA 7.3. ^ T. ^ an_ admissible subbase for X , then for each a e X, V e^^CXj^Ca)] = e^(a) and a e^CX^Ca)] = e^(a). PROOF. Let a e X. Then for each x e ^n^^)' x < a so that e (x) < e (a) since e is a lattice inclusion (4.9); therefore, V ej.[Xj^(a)] < e^(a). Since a e Xj^(a), it follows that e^(a) = v e^[Xj^(a)] A dual argument shows that e (a) = a e [X (a)], PROPOSITION 7.4. If_ and I are admissible subbases for X such that c^ E, then XX algebraically dominates AX. PROOF. We define f : A^X ^ X^X by f(;£) = V {a e [S]: S € (jC H Z )} for each £ e X^X. f ° e„ = e : Let x e X. For each See(x) H Z, xeSso that A e [S] < e (x); therefore, f(ej,(x)) = v {a e^ES]: S e e (x) n E } < e (x). Now suppose that T e e (x) H , Thus X e T and since T e £ E , it follows that T e e (x). Hence by Proposition 4.10 (iv) and the definition of f, a Ta e [T] < f(e (x))i i.e., T e f(e^(x)) H 0^ (4.10 (vi)) and so e^ix) < f(e^(x)) (4.7). It follows that f o e (x) = ^^^(x) for each x e X. f preserves the order of X X: Suppose that if, X"? e XX such that / < /r. Thus (;t n E ) ^ {rnr\ z^) (4.7) and so f(Af) = V {a eg[S]: S e £. n Ej} < V {A e^CS]: S e ;^ H E^} = fO^j).

PAGE 79

72 For each I e X ^ X, (;( D Q) E. Ht) : Let if e X^X. If T £ Jf n , then by the definition of f, A e [T] < f (/) . Since A e [T] A T" (4.10 (iv)), it follows that T e fOf) (4.10 (vi)), and so ( Jf H ) c f(iS). If T e (;f H 0^), then for each S e (/ H 0^), T n S ?! 0. Thus there is some x e S H T and so a e^ES] < e^Cx) < v e^CT] It follows that fU) = V {a e^CS]: S e (/ n E^)} < v e^CT] = v T'"' (4.10 (iv)); therefore, T e f(^) (4.10 (vi)) and so (^ H 0^^) c f (^) . f is surjective : If /"Z e ^(^> "then I'^is a linked system of E and hence is contained in some maximal linked system J^of E. Since fli<^ ii C\ 0) c f(^), it follows from the maximality of ^ that f(af) = /?! f preserves suprema of ideals and infima of dual ideals of X : Suppose that I is an ideal of X and ^ = v e„[I]. Since f preserves the order of XX, we see that for each x e I, f(e (x)) = e (x) ^ f('^) and so V e [I] < f(;f). If T e (V e [I] H ), then for each x e I, T e e (x) (4.8) and so x e T ; consequently, I £T. Since T e E , it follows that T e V e^[I] = ^ (4.8). Thus T e (/ H 0) c f{'£) and so V e [I] (^ 0^ £ f(jS). By the definition of the order on AX (4.7), f(;e) < V eg[I]; therefore f(^) = f(v e^[I]) = v SgCl] = v (f o e^)[I]. A dual argument shows that f preserves the infima of dual ideals of X. In view of the above arguments, XX algebraically dominates XX.

PAGE 80

73 PROPOSITION 7.5. Suppose that and E are admissible subbases for X such that c_ E. (i) If_ for each iteX^X, {£0 0) is_ contained in a unique maximal linked system of 0, then there is a_ unique dominating function f : XX -> XX. Moreover , f may be defined by f(i^) = {T e 0: (r n 0) u {T} is_ linked } for each / e XX. (ii) If for each ^ g X X,(Jf H 0) is a maximal linked system of , then there is a_ unique and continuous dominating function f: X^X . XgX. PROOF, (i^) Since c_ E , there is a dominating function f: XX -> XX (7.4). Suppose that g: XX -> XX is also a dominating Li \J tt \J function. Thus for each / e XX, (;^ H 0) c g(sd) and (if D 0) £ f(sJ.) (7.2 (iii)). Since (/ O 0) is contained in a unique maximal linked system of 0, it follows that g(;^) = fiat) for each Z e XX; i.e., f = g and f is unique. To 'See that f can be defined as stated, let / c XX and ^= {T e 0: (:tn 0) [j{T} is linked}. To see that ff is linked, let S, T e ^. Then U H 0) u{S} is linked and (if H 0) u{T} is linked. Since (Jf (^ 0) is contained in a unique maximal linked system y of Q, it' follows that (/HO) u{S} and (Jf H 0) u{T} are both subsets of ^ Thus S n T i' and ^ is linked. To see that ^ is a maximal linked system of 0, suppose that T e such that ^ u{T} is linked. Thus (it n 0) u(T} is linked, and so by the definition of ^, T e if. Since both ^ and f(J^) are maximal linked systems of which contain (Jf O 0) and (Jf H 0) is contained in a unique maximal linked system of 0, it follows that f(s^) = tf.

PAGE 81

74 (ii) If for each £e AX, (^ ^ 0) is a maximal linked system of 0, then (Jf H 0) is contained in a unique maximal linked system of 0. By (i) above, there is a unique dominating function f: X^X -> AgX defined by f(; XX such that f o e = e . ^ 2j y h \j PROPOSITION 7.7. If and E are admissible subbases for X such that each of XX and XX algebraically dominates the other , then and E are equivalent . Specifically , we will show that if f : XX -> XX and g: XX -> XX are dominating functions , then (i) g o f is_ the identity on_ X X and f o g is_ the identity on XX (ii) f and g are inject ive (iii ) f and g are closed maps (iv) f and g are homeomorphisms (v) f and g are lattice isomorphisms (vi) X„X is iseomorphic with X^X. L '

PAGE 82

75 PROOF. The hypotheses describe the following diagram: ^E^ 'O I XgX where g » e_ = e and f " e = e 0' (i^) Let R e I . Thus R is an ideal of X. By the properties of dominating functions (7.1 (i)) and Proposition 4.10 (iv), g 6 f(v r"^) = g o f(v ej,[R]) = g(v f o e^[R]) = g( V eg[R]) = V g o ggCR] = V ej,[R] = v R . Dually, for each ReL, g o fCAp"*") = ar , Now let ^^ e X^X. If R € X. C) Z , then / < v r'^ (4.10 (vi)), and so f(/) < f(v r"^) since f is' order-preserving. Thus g o f(^)
PAGE 83

75 ( iii ) Let R € E . If .;^ e r"^, then R e t and so t < v r"^ (4.10 (vi)) Since f is order-preserving, f(;£) < f(v r"*") = v e^ER] (7,2 (i)). Thus fCR"*"] c {j e AgX: ^ < ^ eg[R]}. Conversely, if ^ e XX such that fl. < v e [R], then since f is surjective, there is some Z e XX with f (i?) = ^ . Now since g is order-preserving and g ° f is the identity on XX, it follows that ^ = g o f(^) = g(jl) ^ g(v e [R]). Thus by the properties of g (7.1 (i)), 'i. < g(v eg[R]) = V (g o eg)[R] = v e^[R] = v r"^ (4.10 (iv)). Hence R e £ (4.10 (vi)) and so ;£ e R : therefore {J e XgX: ^ < V eg[R]} = fCR"^] = g"^CR"^]. A dual argument shows that if R e E , then -lr„+ g' [R^] f[R^] = {/ e XgX: A e^CR] < f }. e'since Z is a subbase for the closed sets of XX, it follows from the above equalities that f = g is continuous. Hence f is a closed map, A similar argument shows that g is a closed map. (iv) Since each of f and g is closed, continuous, and bijective, each must be a homeomorphism. (v_) Suppose that C is a non-void subset of XX and '^ ^ C . Since f preserves the order of XX, it is clear that iK^ {f(J): jl e C} < f(jd). To see that f(;(.) < />J let T e />Z H Thus >7? < v T='' (4.10 (vi)) so that for each j! e C , f{i) < v T". Since g is orderpreserving and g o f is the identity on XX, it follows that for each ^ = g o fip < g(v T'":) = V e^[T] (7.2 (i))

PAGE 84

77 Thus J^=v{ft:^eC}^v e [T]. Finally, since f is a dominating function (7.1 (i)), it follows that f(/) < f(v ej,CT]) = V f o e^m = v eg[T] = v T'*' (4.10 (iv)). Hence, T e f()t) (4.10 (vi)). Therefore, W H c_ f (it) so that ^= f(i^); i.e., f(v C) = v f[C]. A dual argument shows that f(A c) = A f[C] and so since f is bijective, f is a lattice isomorphism. A similar argument shows that g is a lattice isomorphism. (vi) That f and g are iseomorphisms follows immediately from (iv) and (v) above. COROLLARY 7.8. If_ and Y, are admissible subbases for X such that c^ E and AX algebraically dominates AX, then is_ equivalent to E. PROOF. Since £ E , it follows that AX algebraically dominates AX (7.4). By the hypotheses, AX algebraically dominates AX. Thus is equivalent with Z (7.7). PROPOSITION 7.9. Suppose that is_ an admissible subbase for X and E = u T. Then (i) E is_ an_ admissible subbase for X, and for each £ e AX, (JC n 0) is_ a^ maximal linked system of 0, and (ii) for each ii e AX and each a e X, i^ (:f H 0) < e (a), then X (a) e ii, and dually , i£ ^ (a) <. U H 0) , then X^(a) e .^.

PAGE 85

78 PROOF. (i_) Since Tc_Z=0 U T£r, Eisan admissible subbase for X (4.3). Let ^ e X X. Clearly i:C C) Q) is linked. To see that (<^ n 0) is a maximal linked system of 0, let T e such that CC n 0) u{t} is linked. If T ^ ;£ then there is some U e E H / such that T H U = 0. Since E = u T and {T} ij { XX which may be defined by f(jC) = (/ H 0) for each ^ e XX. Suppose that .^ e X X and a e X such that (.!f n 0) = fit) < CgCa). For each S e (X 1^ E^), a S^ < if (4.10 (vi)) then so that f(A s ) < f(J^) since f is order-preserving. If S e , f(A s"^) = A S" (7.2 (ii)) and since f(jC) < eg(a) , it follows that A S" < e (a); therefore a e S so that S H X (a) ^ 0. If S ii^ , then there is some b e X such that S = X (b). In this case f(A S^) = A SgLS] (7.2 (i)) = A eg[X^(b)] = e^ih) (7.3). Thus e (b) S fiX,) < e (a); therefore b < a since e is a lattice inclusion (4.9) and S H X (a) = X (b) H X (a) 7^ 0. It follows that X. U^X (a)} is linked and so X (a) e ^. A dual argument shows the other statement.

PAGE 86

79 PROPOSITION 7 . 10 . If_ is_ an admissible subbase for X , then is_ equivalent to Q \j T . PROOF. Let E = 9 U T. Then I is an admissible subbase for X (7.9 (i)). For each 1^ e XX, (:(. O 0) is a maximal linked system of (7.9 (i)) so that XX algebraically dominates XX via a unique, continuous function f : XX -> XX which may be defined by f(/) = CJi ^ Q) for each J6e XX (7.5). To see that f is an iseomorphism , we need only show that f is a closed, injective lattice homomorphism (2.10), f is injective : Suppose that *, ^'Ze XX such that * ^ /^ Then we may assume, without loss of generality, that there exist T e (^ H E ) ^ and S e (/T H e ) such that T H S = 0. If both T and S are in 0, then clearly fC^) = (rf H 0) M:^n 0) = fC^). Suppose that T ^ Q ; i.e., there is some b e X such that T = X (b). Note that e (b) ^ f(.^, for if eg(b) < f («?) , then T = X (b) e X>2(7.9 (ii)) — a contradiction. Thus e (b) i f(^) and there is some T' e such that T' e e (b) and T' ^ fO^) = (X7 n 0) (4.7). It follows that X (b) £ T', T' e (/ n 0) = f{it), and hence f{£) i f{Pl). In the case that T e 0^ and S i Q , a dual argument shows that there is some S' e such that S c S', S' e f(/7), SW (;f ^ 0) = f(lt), and hence f(J() i f(^. f is a lattice homomorphism : Suppose C is a non-void subset of XX. Since f is order-preserving, it follows that for each £e C, fCt) <. f(v C)i therefore, v f[c] < f(v C). If T e v f[C] D 0^, then

PAGE 87

80 for each ^ e C, T e f{;t) = (^ !^ 0) (4.8); hence T e
PAGE 88

81 THEOREM 7 . 11 . If. J^ is_ any admissible subbase for X , then XX algebraically dominates XX and is algebraically dominated by X„X, We consider now some situations in which one lattice superextension of X dominates another lattice superextension of X not only algebraically, but also as a superextension. PROPOSITION 7.12. If and Z are admissible subbases for X such that Q £_ I and is_ weakly-normal , then (i) for each £ e X^X, (Jl H 0) is_ contained in a_ unique maximal linked system of 0, and (ii) the superextension XX dominates the superextension XX. PROOF. (i^) Suppose that £ e XX and (.£ H 0) is not contained in a unique maximal linked system of 0. Then there exist /?7, /? e XX such that ^ ^ n, CC n 0) c /'Tand (it H 0) c /?. Since «?< /?, there exist T ei^tand S e /? such that S O T = (2.16 (ii)). By the weaknormality of 0, there is a finite screen {T,, T^, •••, T } <= of S and 1 z n — T (2.11 (iii)). Since B^X = X^X, it follows that It is prime (2.18 (iii)) and hence there is some j e {1, 2, •••, n} such that T. e ^. Therefore, T n T. i' and S r\ T . ^ 0—a contradiction to the fact that {T , •••, T } is a screen for S and T. Thus {£ H 0) is contained in a unique maximal linked system of 0.

PAGE 89

82 (ii) .By (i) above and Proposition 7.5 (i), there is a unique dominating function f: AX-*XX which may be defined by f(j£) = {T e 0: (^ n 0) y{T} is linked} for each il e XX. To see that f is continuous, let T e and ^ e XX such that i. i f"-^[T-"]. Thus T i fW so that {T} u (i^ '^ 0) is not linked. Hence there is some S e (^ H 0) such that T H s = 0. By the weaknormality of 0, there is a finite screen H* c^ of S and T. Let C = u{W : W e y and W H s = 0}. Clearly C is a closed subset of XX. Since S e ,1^ and S H w = implies that s"^ H w"^ = (2.16 (i)), it follows that ^ i^ w"*" for each W e "i" with \^<~^S = 0; i.e.,X^C. If C. is any member of f [T"] , then since 6 e \ X ulWiWe'i'}, there is some W e H* such that QeW;i.e.,Wed. Since W e , we see that i5 e (0 n 0) c f(^) (7.2 (iii)), and since T e f(p-), it follows that T n W i^ 0; therefore, S H W = 0. Thus ^ £ i)"^ £ u iw"*" : W e H* and wns = 0} = C. It follows that f~"^CT''0 c c and ^ »^ C . Thus since » -1 -1 X was an arbitrary point not in f [T"], it follows that f [T"] is a closed subset of XX and so f is continuous. If f is a non-void collection of subsets of X, then 4* denotes the collection of arbitrary intersections of members of H* . PROPOSITION 7.13. If_ is_ an admissible subbase for X and r = tJ u tJ, then

PAGE 90

83 (i) for each ;C e XX, (^C C\ T) is_ a maximal linked system of T, and (ii) the superextension AX dominates the superextension X X. PROOF. (i) Let ;^ e X^X. Clearly (:f O T) is a linked system of T. If U e Tp such that (/ O T) \J {U} is linked, then either U = X € (f. n T) or there is some a e X such that U = ^r,^^)To see that Xj^(a) e ;^, let T e ;t n 0^. By the hypotheses , T = n{V e T : T c V}. For each V e Tj such that T c V, it follows that V e i£ C) T) , V n Xj^(a) ^ 0, and a e V; therefore a e T so that T D X (a) ^ 0. Since T was an arbitrary member of (Jf O Q ) , it follows that (<^ n 0j) u{>^Q(a)} is linked and hence :C uiX^Ca)} is linked. Consequently, X^Ca) e (^ n T). Dually, if U e T, and (^ O T) U{U} is linked, then U e CC H T). Thus (;f H t) is a maximal linked system of T. (ii) That the superextension AX dominates the superextension AX follows immediately from (i) above and Proposition 7.5 (ii). We say that the lattice superextension AX is "larger" than the lattice superextension A_X (or, equivalently , XX is "smaller" than XX) if and only if the superextension XX dominates the superextension XX. In the following theorem we sum up the preceding propositions obtaining two situations in which AX is actually the smallest lattice superextension of X. The proof is immediate from Propositions 7.12 and 7.13.

PAGE 91

84 THEOREM 7 . 14 . If. T is_ weakly-normal or if^ r = T u T , then XX is_ the smallest lattice superextension of X ; specifically , if Q is any admissible subbase , then the superextension XX dominates the superextension XX, In view of the relationship between T, r and any admissible subbase Z, namely, T £ E £ F, it is natural to ask if XX is always the smallest or if XX is always the largest lattice superextension of X. In general the answer is negative to both questions. As noted previously there is a topological lattice X for which the unique dominating function f: XX ^ XX is not continuous; i.e., XX is not smaller than XX and XX is not larger than XX (8.7). Even under the restriction that r = T u T , X X is not necessarily the largest. In Example (8.6, Part II) there is a topological lattice which satisfies this restriction, but for which there is an admissible subbase E such that there is no continuous dominating function f: XX ^ XX; i.e., XX is not the largest superextension of X. The theorem below follows immediately from Proposition 7.12 (ii) and Theorem 7.14. It is stated here for completeness. THEOREM 7.15. If_ all admissible subbases for X are weakly-normal , then XX is_ the largest lattice superextension of X and XX is_ the smallest lattice superextension of X.

PAGE 92

85 DEFINITION 7.16. (Kaufman [12]) Y is called an orderable compact ificat ion of a totally ordered topological lattice X if Y is a topological compactification of X and the order on X can be extended to Y so as to yield the given topology on Y as the order (interval) topology. COROLLARY 7.17. (Kaufman) If_ X is totally , then X has a largest and a_ smallest orderable compactification . PROOF. Suppose that X is totally ordered. Thus each admissible subbase for X is weakly-normal (4.12). By Theorem 7.15, XX is the largest and AX is the smallest lattice superextension of X. It is sho^^m in Appendix I that Kaufman's orderable compactif ications of X are precisely the lattice superextensions of X; therefore, AX is the largest orderable compactification of X and AX is the smallest orderable compactification of X. In view of Kaufman's method for obtaining orderable compactif ications, we may characterize the lattice superextensions of a totally ordered space X as the topological closures of h [X] in I where I is ill the closed interval [0,1], E is any subset of {f: f is a continuous increasing map of X into I such that a f[X] = and v f[X] =1} which separates the points of X, and h : X -> I is defined by it (h (x)) = f(x) Li X Li for each f e E and each x e X.

PAGE 93

8 . EXAMPLES EXAMPLE 8.1. Let X = (0,^) U (y,l) with the usual ordering. Then the completion of X by cuts, [0,1] with its usual ordering, is iseomorphic with XX (5.13), and the largest superextension XX is 1 3 iseomorphic with [0,—] U Ct;?!] with its usual ordering. (The maximal linked system of T defined by {(0,—)} u { (0 ,a] : — < a < 1} U {[a,l): < a < — } corresponds to the point — in XX, and the one defined by {(-,1)} u {(0,a]: y < a < 1} \j {[a,l): < a < -} corresponds to 3 . — m X^X. ) EXAMPLE 8.2. Let X = (0,1) with its usual ordering. Then T = r and consequently every lattice superextension of X is iseomorphic with [0,1] with its usual ordering. EXAMPLE 8.3.. Let n be a positive integer. For each j e {1, •••, n}, let I. = (0,1) with its usual ordering. Thus each I . is a topological lattice. Let Y be the topological product of {I.: j = 1, ••*, n}, and for each j = 1, •••, n, let Z. = T. (Note that for each I., T = r.) 85

PAGE 94

87 Thus E = {TrT"^[T]: T e E . , j e {1, •••, n}} is a T^-subbase for Y (3.1), and the superextension (de Groot compactification) of the product Y with respect to Z, AY (6-.Y), is homeomorphic with the product nf the superextensions {A I.: j = 1, •••, n} (3.3) (de Groot compact if icat ions J ^ {&y I.: j = 1, •••, n} (3.4)). Since each I. = (0,1), it follows from Example 8.2 that AY J " is homeomorphic with a closed n-cell in Euclidean n-space ; therefore , a de Groot compactification of an open n-cell in Euclidean n-space is the n-cell with boundary attached. Note also that since the results in Chapter 3 (Products of Superextensions) were given for arbitrary products of T -spaces , we can generalize this example to the case that Y is the product of an arbitrary collection of copies of (0,1). In particular, in view of Anderson's result [3], a de Groot compactification of Hilbert space is the Hilbert cube. EXAMPLE 8.4. Let I = [0,1] with its usual ordering, and let X = {x e I: X is rational}. Then under the usual ordering X is a topological lattice, and since I is the completion of X by cuts, AX is iseomorphic with I (5.13). For each irrational y e I, let y and y be distinct points not in X which correspond to y. Let Y = X u{y|,:yel and y is irrational} (j {y : y e I and y is irrational}.

PAGE 95

88 We extend the ordering on X to one on Y in the following way : (a) If y c I and y is irrational, then for every x e X and x' e X such that x <^ y
PAGE 96

89 lattice inclusion. The equalities below follow readily from the definitions and some properties of the real numbers: (a) If y e I is irrational, then i~"'"[Y^(y,)] = i"^[Y^(y )] = r\ {X(x) : x € X and y <^ x} and D i Du D 'I i"^[Yj(yj^)] = i"^CYj(y^)] = n{X^(x): x e X and x <^ y}. (b) If X e X, then i~^[Yj^(i(x) )] = X^(x) and i'-'^CY^dCx) )] = X^Cx), It follows from (a) and (b) that i is continuous; i.e., i is a continuous lattice inclusion. (c) i[X] = i[X], i[X] ^ = {y : y e I and y is irrational} uiCJ<], and i[X] = {y : y e I and y is irrational} Ui[X]. It follows from (c) that Y = iCX]-"-^ UiCX]-"-^ £ i[X]^ c Y, i.e., Y = i[X]^. Thus Y is a topological lattice completion of X with respect to i (5.10). To see that X is lattice dense in Y, suppose that v, w e Y with V e i[X] i[X] , w e i[X] i[X] , and w < v. It follows from (c) that there exist y, z e I such that y and z are irrational with w = y and v = z.. Since y <„ z„, it must be the case that y <^ z u . £ u Y )l I and so there is some rational number x e X such that y < x <-r 2; i.e., w = y <,, i(x) <„ z. = v. Thus X is lattice dense in Y (6.18). -'u Y Y i It now follows that Y is iseomorphic with XX where E = {i [S]: S e Z } (6.20). To see that Y is iseomorphic with X„X, we need only

PAGE 97

90 show that Z = r. Recall that E^^-* = {Y} u (Yj^CKx)): X e X} u {Y^CKx)): x e X} y {Y (y,,): y e I snd y is irrational} \j {Y^(y ) : y el and y is irrational} (6.15). If T is a closed ideal of X, then either T = X (x) = i~ [Y (i(x))] for some xeXorT= {xeX:x
PAGE 98

91 Z A W = W A Z w A p (z) , if w e A and z e B wAz,ifw,zeA,w, zeB, orw,zeC w, if w e A u B and z e C. (i) The lattice described above can be visualized as follows; (0,0) A (1,0) (2,0) (3,0) C (4,0) (ii) AX can be visualized as follows ; point in XX which is not in e^[X]. e^EC]

PAGE 99

92 (iii) If = t" u t" then = T U (A U B , C} and XX can be visualized as follows : points in AX which are not in e [X] (iv) r = T U{A, AUB, C} and AX can be visualized as follows: points in AX which are not in e [X] e^[C] From (ii), (iii), and (iv) we see that three quite natural subbases yield different lattice superextensions. EXAMPLE 8.5. Suppose that L and M are lattices. We define the lattice direct product, X, of L and M to be the set L x M with the

PAGE 100

93 ordering defined by (a,b) < (c,d) if and only if a <, c and b < d. A L M It is easily seen that the lattice direct product is indeed a lattice, Part I Let X be the lattice direct product (0,1) x (0,1) where (0,1) is the open interval with the usual ordering. Let Y be the lattice direct product [0,1] x [0,1] , where[0,1] is the closed interval with the usual ordering. X may be visualized as follows: (0,1) (1,1) (0,0) (1,0) (i) X is_ a sublattice of Y and_ Y = X^ = X^^ X^^ , but_ the inclusion map i: X ->Y is_ not continuous . Thus Y is_ not a_ topological lattice completion of X with respect to i . PROOF. It is easily seen from the definition of lattice direct product that X is a sublattice of Y , and it follows readily that Yj_ c x^^u X^ c x^ c Y^ so that Y^ = X^ (= X^^Hx^^).

PAGE 101

94 To see that i is not continuous, let W = i C^ ( (1,—) )]. We will show that W is not a closed subset of X. Suppose that r e X such that r ^ W, and let V be any basic open (in the topology on X generated by {X U: U is in the usual subbase for X}) set containing r. It follows from the definitions that V = AOB where A = X U{Xj^( (a^,b^) ): i = 1, 2, •••, n} and B = X U{Xt.( (c.,d.) ): j = 1, 2, •••, m} for some finite set {(a^,b.), •••, (a ,b ), (c.,d-), ••, (c ,d )} c X. Let 11 nn 11 mm — a = v{a.:i = l, 2, •••,n} and d = A {d, : j = 1, 2 , • • • , m}. By the finiteness of n and m and the properties of (0,1), it is possible to find (p,q) e X where a < p < 1 and < q < d a — . It follows that (p,q) e WHaHB = WOV. Since V was an arbitrary basic open set about r, it follows that r e W W; consequently, W is not closed in X. Let Y^ = X U{(0,0), (1,1)}. Extend the order of X to Y by letting (0,0) = a X and (1,1) = v v. Y may be visualized as follows; (1,1) (0,0) *• h

PAGE 102

95 (ii) The topological lattice Y is_ iseomorphic with AX. PROOF. It is easily seen that X is a sublattice of Y and Y = X nx = X . Thus Y is the completion of X by cuts and so Y is iseomorphic with XX (5.13 and 6.13). ( iii ) In this example F = T; consequently , Y is_ the unique lattice superextension of X. PROOF. To see that F = T, let C be a closed ideal of X; i.e.jCeF. IfC = X, then C € T . If C ?< X, then since C is a closed subset of X, it follows as in the proof of (i) above that C cannot be of the form {(x,y) e X: x < a for some a e (0,1)} or of the form {(x,y) e X: y < b for some b e (0,1)}. Thus there must be some (a,b) € X such that C £ X ( (a,b) ). Now by the completeness of the real numbers, it is possible to find real numbers c and d such that c = V {x: (x,y) e C} and d = v {y: (x,y) e C}. Since C ?f and C £ X ( (a,b) ), it follows that < c < a < 1 and < d < b < 1. Hence (c,d) e X. 'It is clear from the choice of c and d that C £ X ( (c,d) ) Since C is closed in the interval topology on X, and consequently in the order topology on X [ 9 ; Theorem 12], it follows that (c,d) e C. Hence C = X ( (c,d) ) 6 T . A dual argument shows that F £ T . Hence F = T, and since if E is any admissible subbase for X, we may assume without loss of generality that T c_ i (7.10), it follows that Z = T and so A^X = A^X is iseomorphic with Y .

PAGE 103

96 Part II In this part, let X be the lattice direct product [(0,l)u(2,3)] X [(0,1)U (2,3)] where (0,l)u(2,3) has the usual order. X may be visualized as follows: (0,3) .....v:v.-.-...-.-.v.-. ,,,,.v.......v.v..:. x:::i (0,2) i:i (0,1) (0,0) -^m^: Siiiiiii (3,0) In this example it is easily seen that T = T ij T . (i_) The lattice superextension A X naturally fills in the deleted cross so that AX is_ iseomorphic with the topological lattice Y^ ( Part I). Let Y be the lattice direct product [ (0,1] u [2,3) ] X [ (0,1] u [2,3) ], and let Y^ = Y^ u {(0,0), (3,3)}. Extend the ordering of Y to Y by letting (0,0) = a Y and (3,3) = v Y .

PAGE 104

97 Thus the map i: X -> Y defined by (x,y) n(x,y) is a continuous lattice inclusion and so X is a sublattice of Y . Y can be visualized as follows : (0,0) (1,2) •:::::::::::v:.:.:.:.:, (3,3) ''•'''•'•'-' 'I'l'' (2,1) 1 . . I i'^ I I I I I f V (ii) Y i_s a_ topological lattice completion of X with respect to i. PROOF. Since i is a continuous lattice inclusion, it follows from the equalities below that Y is a topological lattice completion of X with respect to i (6.10): i[X]^ = X u{(0,0), (3,3)}, iCX]^^ i[X]^ = {(x,y): x = 1 or y = 1} {(1,2), (2,1)}, iCX]^^ i[X]^ = {(x,y): x = 2 or y = 2} {(1,2), (2,1)}, and i[X]^ (i[X]^^ Ui[X]^^) = {(1,2), (2,1)}. Thus Y^ = i[X]'

PAGE 105

98 (iii) AX is iseomorphic with the topological lattice Y PROOF. Since each proper non-principal closed ideal of X is of the form i CY ( (a,b) )] where a = 1 or b = 1 and each proper non-principal closed dual ideal of X is of the form i tY ( (a,b) )] where a = 2 or b = 2 , it follows that r = {i~ [S] : S e E } . It is easily seen (using the equalities in (ii)) that X is lattice dense in Y (6.18). Thus Y is iseomorphic with AX (6.20). Let E = T U{R,Q} where R = i """[Y ( (1,1) )] and .-Ir Q = i CY^^( (2,2) )]. Since i is a continuous lattice inclusion, it follows that Z is an admissible subbase for X (4.3). The lattice superextension AX can be visualized as follows: V R h"" (iv) There is no continuous algebraic dominating function f: A^X -> A^X,

PAGE 106

99 PROOF. In order to simplify the notation in this proof, we recall that XX is iseomorphic with Y and show that there is no continuous dominating function from Y onto AX. To do this, let f be any algebraic dominating function from Y onto AX. Thus we have the following commutative diagram: Let p be the point (1,2) in Y . We now establish a few needed lemmas. (1) Either f(p) = v R or f(p) = a Q : For each (x,y) e R, i((x,y)) < (1,1) < (1,2) = p and since f is order preserving, it follows that f(i((x,y))) = e^((x,y)) < f(p). Thus v r"*" = v e^[R] < f(p). A dual argument shows that f(p) < a Q . Now, from the definition of Z it follows that there is no_i^ e AX such that v R < <^ < a Q ; consequently V r"*" = f(p) or A q"^ = f(p). Y, Y (2) p £ {(l,y): 2 < y < 3} n{(x,2): < x < 1} : It follows from the definition of the order on Y that v {(x,2): < x < 1} = (1,2) = p = A {(l,y): 2 < y < 3}. Hence by [9; Theorem 12], 4 Y Y p e {(x,2): < X < 1} and p e {(l,y): 2 < y < 3} (3) {(l,y): 2 < y < 3} c f~^[Q^] and {(x,2): < x < 1} £ f "^[r"^] : Let 2 < y < 3. It follows from the definition of E and the properties on f that f(l,y) = A {e_((a,b)): (l,y) < i((a,b))}. Also, if (a,b) e X A^X I Y^

PAGE 107

100 such that (l,y) <„ i((a,b)), then 1 < a and 2 < y < b. Thus (2,2) < i((a,b)) and so (a,b) e Q. (Recall that there is no element (a,b) e X with 1 < a < 2.) Hence for each (a,b) e X with (l,y) < i((a,b)), it follows that a Q < e ((a,b)); therefore aq"^ S a {e_((a,b)): (l,y) < i((a,b))} = f(l,y). Thus Q e f(l,y) (4.10 (vi)) and so (l,y) e f'^^CQ"*"]. It follows that {(l,y):2
PAGE 108

101 I Let X , X € X. n m (x V x') , in case n = m n I I X vx =x vx = WxVx'),in case n ^^ m and n = n m m n / m 1, in all, other cases I I X Ax = X Ax n m m n (x A x') , in case n = m or x' = 1 n (x A x'), in case n ^ m. X can be visualized as in the following diagram where the arrows denote lattice isomorphisms from I„ onto each I : On (i_) For each n e N {0}, I is a closed dual ideal of X. PROOF. Suppose v 6 X and x < y for some x e I . Then m n m n n y = X V y and since n ^ it follows from the definition of suprema ^m n "^m "^ in X that either n = m or y =1. In either case y e I and I is m m n n therefore increasing. It also follows from the definition of infima in X that I contains x Ay when x , y el. To see that I is a n n m n m n n

PAGE 109

102 closed subset of X, suppose that y ^ I . Thus n j^ m and y ?^ 1 . If m = 0, then we may pick x e (0,1) such that < x < y. It is easily seen that I c X^(x ) u Xy^^ ) ^'^'^ Vm ^ ^n^^r.^ ^ ^T^^n^' If "> ^ 0' n — un in m u ii x n then V is not comparable with any member of I {1} so let x e (0,1). Again it is easily seen that I^ L Xq(><^) ^ ^j^^n^ ^"^ ^m '^ ^D^^n^ ^ ^I^^n^' (ii) If D is a_ closed ideal of X, then D is_ a principal ideal . PROOF. If D = X, then D = X (1). If D ?! X, then it follows from the definition of suprema in X and the fact that D contains d^ v d^ whenever d, , d„ e D that D c I^ u I for some n e N. Since D O I ?* , 12 —Onn D n I is a non-void closed ideal of I , and since 1 <^ D H I , it n n n follows from the completeness of the reals that there is some x e (0,1) such that X = v^ (D fl I ). Since D Pi I is a closed subset of X, n I n n n it follows that x £ D H I [9; Theorem 12]. Thus X (x ) c D. If n n D n — d £ D, then either d £ I , in which case d < x , or d £ I. , in which n n u case there is some x' e (0,1] such that d = x . Thus t t d = x„ < x„ V X = (x' V x) £ D n I , and so d < (x' v x) < x ; n n n n n therefore, D c X^(x ) and so D = X^(x ). — D n D n ( iii ) I£ D is_ a proper closed dual ideal of X , then D is_ either a_ principal dual ideal or D = I for some n£N-{0}. ,^ PROOF. Suppose that D is a closed dual ideal of X which is not principal and not equal to I for any n £ N {0}. Thus D ^ I^^ ^°^ any n £ N {0}, and so D meets at least two of the I 's for n £ N {0}; ' n therefore, D H I ?^ 0. Since D is not principal, it follows that I £ D. Since D is increasing, X c_ D; i.e., D = X so that D is not proper.

PAGE 110

103 In view of (i), (ii), and (iii), r = Tu{l :neN-{0}}. n Let i£ = ^. ,,I for each n e N. Since for each pair n XX n *^ m,n€N-{0},I ^ 1 and I (z! I , it follows that no two members n — m m — n of {»f : n e N {0}} are comparable. Also, since "^p. = ^ e [X] = a XX, it follows that /„ < «f for each n. X„X may be visualized as follows: On r -^ (iv) For each ^e X^X, either if e e^CX] or Jf = a l"*" for s — I 1 — n ome n e N, Turning now to XX we note that for each n e N, the onlymember of T^ which contains I is X. Thus it follows from Theorem 4. I n

PAGE 111

lOU that for each n, a e^[I^] = {X} \j T^ and hence a e^[I^] = a X^X, The lattice AX may be visualized as follows: (v) If ^ e XX, then (i^ Pi T) is contained in a unique maximal linked system of T. PROOF. Let ^ e AX. If j^ e e^[X], then (^ Pi T) is a maximal linked system of T. If S^»^ e [X], then by (iv), ^ " ^ ^^ ~ "^n for some n e N. Thus, as was noted above, the only member of T containing I is X, and it follows that ii^ O T) = {X} U {X^(x ): ^ n n D n X e I } c {X} U T^ = a ax. If ^e AX such that a AX < W, then n n — D T 1 1 there is a principal dual ideal T e ()^ H T ) (a AX) (U.V). Clearly I 52! T, so there is some r el T. Hence X^(r ) e (^ H T); theren — n n D n n fore (^ n T) cannot be contained in ^. It follows that for each n ^€ AX, iX n T) is contained in a unique maximal linked system of T.

PAGE 112

105 In view of (v), XX algebraically dominates XX via a unique dominating function f : X X ^ XX which may be defined by f(^) = {U € T: U n T) \j{U) is linked} for eachJt e X X (7.5 (i)), (vi) The dominating function f : X X -> XX is_ not continuous . PROOF. Let t £ (0,1). Then t^ e ^q -^' ^"^ ^D^^O^ ^ '''d * We will show that f [(X (t ))"] is not a closed subset of X X : To see this, let r e (t,l). It is clear that e (r ) i f~ [(X (t ))-] and we will show that there is no finite subset of r whose union covers f [(X (t ))"] and misses e (r ) . This follows from the following statements : (1) For each n e N , ^^.^ £ f""^[ (X (tQ ))'''] : If n £ N, then as was shown in the proof of (v) above, f(^ ) = a XX < v (X (t ))". Thus X_(t^) £ f(^ ) (4.10 (vi)). DO n (2) If n, m £ N {0}and n ^ m, then iC y £ = e„(l): Clearly . -^ fj, p t y £ < e(l) = V AX. If S £ i? v ^ n r , then by (ii) above nmr r nmD S = X (x, ) for some x^ £ X , and it must be the case that SHI ^ and S n I ^ 0. Hence x, £ I H I ; i.e., x, = 1 and S = X. Thus m k n m k S 6 e„(l) and so e (1) < / v ^ . Hence e^(l) = / v / . r r m n r n m (3) If X„(x, ) £ ;£ and n ?! 0, then k = n or x = 1 : If X„(x, ) e H , D— k n D T< n then X (x^) Pi I ^ and x^ £ I . Thus k = n or x = 1.

PAGE 113

106 (4) If S £ r and f"'''[(X (t ))"] c s"^, then S = X : Suppose that 5 e r^ and f~"^[ (X^^Ct^) )"] c_ s"^. By (1) , j^^^ e s"^ for each n e N, Thus by (2) and since S is an ideal of AX (4-. 10 (i)), e (1) e S ; therefore, S = \^^'^ = ^• ( 5 ) If S e r {X}, then S contains at most two members of {;f : n e N}: If S e T^ {X}, then by (ii) above S = X (x ) for some X, e X. Since S ?^ X , x ?^ 1 , and so it follows from (3) that K K S+ r){£^: n e N) = {^q.J^}. (6.) If S e r {X}, then S contains at most one member of {/ : n e N}: Suppose that S e r^ {X}. By (iii) either S = X^(x, ) — n I Ik for some x, e X or S = I, for some keN-{0}. In the first case k k S"*" n{<;;f : n e N} = and in the second case S H Lsf : n e N} = {^ } . n n k We have shown in (5) and (5) that each member of (l {X}) contains at most two members of {^^ : n e N}. Since {/ : n e N} is an infinite subset of f [(X (t ))"], it follows that no finite subcollection of r"^ can cover f~ [(X (t ))-] and miss e (r ). Thus f~ [(X (t ))-] is not closed, and so f is not continuous. EXAMPLE 8.8. Here we exhibit a topological lattice which has a topological lattice completion in which the given lattice is not lattice dense (6.18). Let each of A, B, C, D, £, and F be the

PAGE 114

107 interval (0,1) with its usual ordering. Let X be made up of these six pieces in the manner shown in the following diagram where the p.'s, i = 1,2,3,4, are lattice isomorphisms between the various parts: All elements of B are less than all elements of D \j E U F. All elements of A are less than all elements of F U E. All elements of C are less than all elements of E. An element b e B is less than p (b) e A, and an element a e A is less than p (a) e C. Dually, an J. w element d ^ D is less than p (d) e F, and an element f e F is less than p (f) € E. Finally, note that a c = and v n = .

PAGE 115

108 Let Y contain X as in the following diagram: We extend the order on X to Y by letting 0=ax=aY,I=vx=vY, and G Y is a continuous lattice inclusion. Thus Y is a topological lattice completion of X with respect to i (6.10). It is clear from the diagram that H e .[X]"""^ .[X]^, J e .[X]"^^ .[X]''", 1 11 1 and H < J. However, {x e X: H < i(x) < J} = 0; i.e., X is not lattice dense in Y (6.18). EXAMPLE 8.9. In this example we exhibit a topological lattice X for which A. X algebraically dominates AX (and even dominates it as a superextension), but for which the dominating function of AX over A is not a lattice homomorphism (2.2 (i)).

PAGE 116

109 Let A, B, and C be copies of [0,1) with the usual ordering, and let X be made up of A , B , and C as in the following diagram where p and p^ are lattice isomorphisms between the indicated parts: An element a e A is less than p (a) e B, and an element c e C is less than p (c) e B. (i_) AX can be visualized as follows : A X (ii) r = T U{A,C} and AX can be visualized as follows : A X = G It is easily seen that T is a weakly-normal admissible subbase for X. Thus the superextension AX dominates the superextension A X via a unique dominating function f: A X -> AX (7.5 and 7.12).

PAGE 117

110 ( iii ) The dominating function f : XX -> AX is_ not a_ lattice homomorphism. PROOF. The points D and F in XX are not in e [X], and by Theorem 4.8, Daf=G=aX. Since clearly f(D) = f(F) = H in XX, it follows that f(D) a f(F) = H; therefore, f(D ^^ X ^-^ ^ ^^°^ ""x X ^^^^• Hence f is not a lattice homomorphism (cf . , 2.2 (i)).

PAGE 118

APPENDIX I Kaufman's Orderable Compact ificat ions The purpose of this Appendix is to show that for totally ordered topological lattices, Kaufman's [12] orderable compactifications (7.16) are precisely the lattice superextensions. Throughout this Appendix X will denote a totally ordered topological lattice. Frink [ 9 ; Theorem 3] has shown that the order and interval topologies coincide for totally ordered spaces. Kaufman [12; Theorem 2] has shown that the orderable compactif ications of X are also totally ordered spaces; hence the order and the interval topologies in an orderable compactif ication of X coincide. PROPOSITION 1(1). Every lattice superextension of X is an orderable compact ificat ion of X. PROOF. Let E be any admissible subbase for X. Then ej,[X] is iseomorphic with X (2.16 (v) and 4-. 9) so that we may identify X with e [X]. Since XX = 3 X, X is topologically dense in XX, and since Lt Li L, Li e is a lattice inclusion (4.9), it follows that the ordering on XX is an extension of that on X. Finally, we recall that the topology on XX generated by I is the interval topology (4.11). Hence XX is a Kaufman orderable compactification (cf . , 7.16). Ill

PAGE 119

112 PROPOSITION 1(2). Every orderable compact if icat ion of X is_ a_ lattice superextension of X. PROOF. Let Y be an orderable compact if icat ion of X. We will show that Y must be a topological lattice completion of X in which X is lattice dense. The desired conclusion will then follow from Theorem 6.20. Y is a topological lattice completion of X : Since Y is compact in the interval topology, it is complete (2.9). Since X is contained in Y as both a subspace and a sublattice, it follows that the identity 2 map i: X -* Y is a continuous lattice inclusion. To see that Y = X , let y £ Y. It is easily seen that z=v{x6X: x
PAGE 120

APPENDIX II Lattice Closures The purpose of this Appendix is to show how Definition 5.1 can be extended to each ordinal in order to define the "lattice closure" of sets in a complete lattice and to show that this "lattice closure" operator is a Moore closure operator, or completion-operator as in Definition 5.1. Throughout this Appendix, Y will denote a complete lattice and Z will denote a subset of Y. Note that if a is an ordinal, then a is an element of exactly one of the classes N, L, and M where N = {1,2,''«}, L is the class of limit ordinals, and M= {a+n: ae L and n e N}. Keeping this in mind we state the following inductive definition: NOTATION IKI). Suppose that Z 7! . (i) For each ordinal a, =0 =0. (ii) Let Z,Z^,Z ,Z,Z^,Z , and Z be as defined in 6.1. (iii) If n € N is even and we have defined Z and Z , then we let let If n e N is odd and we have defined Z and Z , then we ^(ntDa __ (^na^lb ^^^ ^^n+Db __ (^^b^la^ 113

PAGE 121

114 (iv) If a e L and we have defined Z and Z for each ordinal g < a, then let Z"^ = U{Z^^: 6 < a) and z"^ = U{Z^'': g < ex). (v) If a € L and we have defined Z and Z , then let ^(a+Da ^ (2«a^la ^^^ ^^a+Db ^ (^a^^lb^ (vi) If a + n € M, n is odd, and we have defined Z and „(a+n)b ., ^ . Z , then let (a+n+l)a _ . (a+n)a.lb (a+n+l)lb _ . (a+n)b.la If a + n e M, n is even, and we have defined Z and „(a+n)b ., T . Z , then let (vii) For each ordinal a, let Z = Z Pi Z PROPOSITION 11(2). For each ordinal g. z" c z"^ u z"^ c z"'*""''. By an argument similar to the one in the proof of Proposition 6.2 (iv) where it is shown that Z ij Z £ Z , the analogous conclusion is valid for each ordinal ct. COROLLARY 11(3). For each ordinal a, z" £ z" for each ordinal a such that a < a. The proof is immediate from Proposition 11(2) and Definition 11(1) (iv) and (vii). PROPOSITION 11(4). If Z £ W c Y, then z"^ c w"^, z"^ c w"^ , and Z c W for each ordinal a.

PAGE 122

115 PROOF. If a is a non-limit ordinal, then the conclusion follows from an argument analogous to that in the proof of Proposition 6.2. If a is any limit ordinal and we have shown that Z c_ W and Z c_ W for each ordinal 3 < a, then Z"^ = uiZ^^: 3 < a} c uiW^^: 3 < a} = w"^ and Z"^ = U{Z^^ 3 < a} c u{W^^ 3 < a} = w'^^ therefore, z" = z'^^ H z"^ c w"^ H w"^ = W^ PROPOSITION 11(5). There exists a least ordinal a such that Z" = Z^"^^\ PROOF. Since Y is a set, it has some cardinality. Hence the process of adding new elements of Y to Z must eventually stop; i.e., there Ct Ct'i'l is some ordinal a for which Z = Z . Since the class of ordinal numbers is well-ordered, there is some least ordinal a for which Z'^ = Z^^\ In view of Proposition 11(5), we may define the lattice closure of Z in Y unambiguously as follows : DEFINITION 11(5). If Z is a subset of Y, then the lattice ex ot ct+1 closure of Z in_ Y is Z where a is the least ordinal such that Z = Z If Z is complete, then it follows that the lattice closure of Z • „ • r,0 m Y IS Z = Z.

PAGE 123

116 Y Y PROPOSITION 11(7). The map C: 2-^2 which takes each subset of Y onto its lattice closure in Y is_ a_ completion operator on Y . PROOF. CI. Let A c Y. Then since A = A , it follows from Corollary 11(3) that A c C(A). C2 . Let A c^ Y. Then there is some least ordinal a such that ^a ^ ^a+1 ^ Q(^)^ Itfollows from 11(1) that C(A)^ = (a")"'= (A^^^)^ c A*^^^ = a"* = C(A) = C(A)° and hence C(C(A)) = C(A)° = C(A). C3 . Suppose that A £ B c_ Y. It follows from Proposition 11(4) that C(A) c c(B). Thus the map C is a completion operator on Y (cf. Definition 5.1) <

PAGE 124

BIBLIOGRAPHY [1] Alo, R. A., and Frink, 0. "Topologies of Lattice Products," Canadian Journal of Mathematics , 18(1966), 1004-1014. [2] Alo, R. A., and Shapiro, H. L. "Normal Base Compactifications ," Mathematische Annalen , 175(1968), 337-340. [3] Anderson, R. D. "Topological Properties of the Hilbert Cube and the Infinite Product of Open Intervals," Transactions of the American Mathematical Society , 126(1967), 200-216. [4] Banaschewski , B. "On Homeomorphisms between Extension Spaces," Canadian Journal of Mathematics , 12(1960), 252-262. [5] Birkhoff, G. Lattice Theory . Colloquim Publications Vol. XXV. Providence, R. I.: American Mathematical Society, 1967. [6] de Groot , J., and Aarts, J. M. "Complete Regularity as a Separation Axiom," Canadian Journal of Mathematics , 21(1969), 96-105. [7] de Groot, J., Jensen, G. A., and Verbeek , A. Superextensions . ZW 1968-017. Amsterdam: Mathematisch Centrum, 1968. [8] Frink, 0. "Compactifications and Semi -normal Spaces," American Journal of Mathematics , 86(1964), 602-607. [9] . "Topology in Lattices," Transactions of the American Mathematical Society , 51(1942), 569-582. [10] Gillman , L., and Jerison, M. Rings of Continuous Functions . Princeton, N. J.: Van Nostrand, 1960. [11] Glicksburg, I. "Stone-Cech Compactif ication of Products," Transactions of the American Mathematical Society , 90(1959), 369-382. [12] Kaufman, R. "Ordered Sets and Compact Spaces," Colloquium Mathematicum, 17(1967), 35-39. 117

PAGE 125

118 [13] Kelley, J. L. General Topology . Princeton, N. J.: Van Nostrand, 1955. [14] Njastad, 0. "On Wallman-type Compactif ications ," Hathematische Zeitschrift , 91(1956), 267-276. [15] Ward, M. "The Closure Operators of a Lattice," Annals of Mathematics, 43(1942), 191-196.

PAGE 126

BIOGRAPHICAL SKETCH Linda Joe Wahl Smith was born May 8 , 1944 , in Jacksonville , Florida. In June, 1962, she graduated from Quincy High School, Quincy, Florida. In December, 1965, she received the degree of Bachelor of Science from Florida State University. In May, 1966, she entered the Graduate School of the University of Florida. She worked as a graduate assistant at the University Computing Center during the Spring Trimester 1965-66, and from September, 1956 through August, 1968 she was a National Science Foundation Trainee. In December, 1967, she received the degree of Master of Science from the University of Florida. From that time until the present she has worked toward the degree of Doctor of Philosophy. In September, 1968, she moved to Tallahassee, Florida, where she has worked as a part-time instructor of mathematics at Florida State University and Tallahassee Junior College . Linda Smith is the wife of Nevins Carson Smith, Jr. and the daughter of Mr. and Mrs. J. Kermit Wahl. She is a member of Phi Beta Kappa, Phi Kappa Phi, Pi Mu Epsilon, Kappa Delta Pi, Alpha Lambda Delta, and the Florida Council of Teachers of Mathematics. 119

PAGE 127

This dissertation was prepared .under the direction of the chairman of the candidate's supervisory committee and has been approved by all members of that committee. It was submitted to the Dean of the College of Arts and Sciences and to the Graduate Council, and was approved as partial fulfillment of the requirements for the degree of Doctor of Philosophy. December, 1969 Dean, Colleg nd Scxfences Dean , Graduate School Supervi^spry Committee (aui^ ; \.:.iP^

PAGE 128

r\ '?'26aO-J^