Disassociated indiscernibles


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Disassociated indiscernibles
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iii, 37 leaves : ; 29 cm.
Leaning, Jeffrey Scott, 1971-
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Thesis (Ph.D.)--University of Florida, 1999.
Includes bibliographical references (leaf 36).
Statement of Responsibility:
by Jeffrey Scott Leaning.
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I would first like to thank my family. Their perpetual love and support is

behind everything good that I have ever done. I would like to thank my fellow

graduate students for their friendship and professionalism. David Cape, Scott and

Stacey Chastain, Omar de la Cruz, Michael Dowd, Robert Finn, Chawne Kimber,

Warren McGovern, and in fact all of my colleagues at the University of Florida have

contributed positively to both my mathematical and personal development. Finally,

I would like to thank my advisor Bill Mitchell. He is a consummate scientist, and has

taught me more than mathematics.


ACKNOWLEDGEMENTS ............................ ii

ABSTRACT .................................... 1


1 INTRODUCTION ... ..................... ....... 2

2 PRELIMINARIES .............. ........... ..... 4
2.1 Prerequisites ............................... 4
2.2 Ultrafilters and Ultrapowers ................. ... ..... 4
2.3 Prikry Forcing ........ ... .................. 6
2.4 Iterated Prikry Forcing ..... ......... ........ 9

3 THE FORCING .................... ............ 11
3.1 Definition .................... ........... 11
3.2 Basic Structure ........................... 12
3.3 Prikry Property ... .................. ..... 14
3.4 More Structure ........................... 16
3.5 Genericity Criteria .................... .... 18

4 GETTING TWO NORMAL MEASURES .... .......... .. 24
4.1 The Measures ................... ........ 24
4.2 Properties ................... ...... .. 27

5.1 Definition ................. .......... ...... 30
5.2 Basic Structure ............................ 31
5.3 Prikry Property ........................... 32
5.4 More Structure .................... .. ... .. 34


REFERENCES ..................... .............. 36

BIOGRAPHICAL SKETCH ..................... ...... 37

Abstract of Dissertation Presented to the Graduate School
of the University of Florida in Partial Fulfillment of the
Requirements for the Degree of Doctor of Philosophy



Jeffrey Scott Leaning

August 1999

Chairman: William J. Mitchell
Major Department: Mathematics

By 'measure', we mean a K-complete ultrafilter on some cardinal K. This cor-
responds to the sets of positive measure for a two-valued but nontranslation invariant

measure in the sense of Lebesgue. We study normal measures, that is, measures U

such that for any regressive (f(a) < a for all a) function f : K -+ K we have that f is

constant on some set X E U. We use the technique of forcing to increase the number

of normal measures on a single cardinal. We introduce a variant of Prikry forcing.

After defining a new hierarchy of large cardinals, we get many normal measures in

the presence of less consistency strength than was previously known, strictly less than

that of a measure of Mitchell order two.


Logic. The art of thinking and reasoning

in strict accordance with the limitations and incapacities

of the human misunderstanding.

-Ambrose Bierce

This paper addresses a mathematical problem in the subject of Set Theory. Set
Theory is the branch of mathematical logic concerned with infinity. Georg Cantor

discovered in 1873 that the infinite can be sensibly quantized and that there are

actually different sizes of infinity. In fact, there are infinitely many different sizes of

infinity. Although Cantor established many properties of how the infinite is arranged,

some fundamental questions about its structure eluded him. Almost a century later,

Paul Cohen of the United States discovered that the structure of infinity is plastic;
that is, there are many different possible arrangements of the infinite. This research

is an exploration of some of them.

In Chapter 2 we discuss some preliminary material from which we draw our in-

spiration and much of our technique. Cohen developed a technique he called 'forcing'

to prove his results about the infinite. We describe Karl Prikry's version of forcing

for altering the cofinality of a measurable cardinal and Magidor's iteration of Prikry's

technique which leaves some of the cardinals measurable. We state some important

properties of both, like the so-called Prikry Property, which is essential for showing

that cardinals are preserved, and Mathias' genericity criterion.

Readers familiar with the preliminary material will readily understand the

original ideas in this paper. In Chapter 3 we introduce a forcing for what is essentially

the union of the range of a function generic for Magidor's iterated Prikry forcing. We

prove the Prikry property. We give a genericity criteria. And as an application,

we show in Chapter 4 that under certain interesting conditions it forces cardinals

with exactly two normal measures. In Chapter 5 we define an iterated version of

this forcing and show in Chapter 6 that cardinals with several normal measures are

possible in the presence of less consistency strength than was previously known. The

djinn is in the detalles. Our arguments for the Prikry property and the genericity
criteria require a lot of infinitary combinatorics. Our proofs are longer than their

analogs. And the measurable cardinals we deal with contain additional subtlety.

Our motivation was the folk question, 'How many normal measures can a mea-

surable cardinal have?, perhaps due to Stanislaw Ulam. Kunen and Paris [4] forced

a measurable cardinal K to have the maximal number, 22", of normal measures. Rel-

ative to a measurable cardinal whose measure concentrates on measurable cardinals,

Mitchell [7] proved that that exactly two normal measures are possible. We confirm

this statement and similar results while reducing the hypothesis. Although generally

believed, it is still unproven that the consistency strength of a measurable cardinal

having exactly two normal measures is that of a single measurable cardinal.


2.1 Prerequisites

We assume the reader is knowledgable about the basic Set Theory associated

with large cardinals and independence proofs. In particular, he or she should be

familiar with measurable cardinals and the theory of forcing. The author highly

recommends [1], [2], and [3] as references. Since this work is an exploration of an

alteration of iterated Prikry forcing, readers that are already familiar with Prikry

forcing and its generalizations will be better prepared to proceed than those who are


2.2 Ultrafilters and Ultrapowers

Definition 2.2.1. A measure is a K-complete nonprinciple ultrafilter on a cardinal K.

If we remove the translation-invariance requirement from Lebesgue measure

and restrict to two values instead of non-negative real values, then our measures above

correspond to the class of sets of measure one.

The main reason that set theorists are interested in measures is that the cor-

responding ultrapowers (Definition 2.2.2) of absolutely well-founded models of Set

Theory are absolutely well-founded. Hence the ultrapower of the class of all sets V

is embeddable into itself by Mostowski's collapse isomorphism. This gives a homo-

morphism of V into a proper subclass of itself.

Definition 2.2.2. Let U be a measure on K. Consider the equivalence relation on

functions f, g : K -- K,

f ~ g iff {7 < c : f(7r) = g(r)} E U.

Let Ult(V, U) be the model whose universe consists of equivalence classes [f] of
functions f : K -+ K under the relation ~. Interpret 'E' in Ult(V, U) as

[f] E [g] iff {17 < f : f(77) E g(7)} E U.

Let ju : V -+ 93 Ult(V, U) be the natural map given by x [c,] where c, is

the constant function c.,(r) = x, and mt is Mostowski's transitive isomorph of the


An element of Ult(V, U) is a class of functions which are equal 'almost every-
where', to use the measure theoretic terminology, modulo U. Elements are included

in other elements if their representative functions are pointwise included almost ev-


Definition 2.2.3. A measure U on K is normal if every regressive (Va f(a) < a)
function f : K -+ K is constant on some set X E U.

The property of normality gives a clear description of which equivalence class
of functions represents which set in the ultrapower. In fact, for normal measures, the

equivalence class of the identity function f(a) = a represents the cardinal K.

Definition 2.2.4. Let (X,),<, be a collection of subsets of n. The diagonal inter-

section of these sets is the set A,<,X, = {7 < K : Vt < 7 (7 E X,)}.

Proposition 2.2.5. Let U be a measure on K. The following are equivalent.

1. U is normal.

2. VX(X n K E U E ju(X)).

3. U is closed under diagonal intersection.

Definitions 4.1.1 and 6.3 are modeled after a forcing-syntactic version of prop-
erty 2 above.

We will use the following combinatorial observation, due to Rowbottom, in-

finitely often in what follows.

Theorem 2.2.6. Let U be a normal measure on K. Let f : [n]<" -+ 7l, where rl < K.

There exists a set X E U such that f is constant on [X]" for each n < w.

The set X above is said to be homogeneous for f.

2.3 Prikrv Forcing

Prikry forces an unbounded sequence of type w through a measurable cardinal

in his dissertation [9]. We include his definition, and statements of some basic results

here, mainly for comparison with our present work.

Let K be a measurable cardinal and let U, be a normal measure on K.

Definition 2.3.1. Prikry forcing, P(U,), has as conditions pairs (s, X) such that

1. s E [Kr]<

2. X U,,.

A condition extends another, (s, X) < (s',X'), iff

1. s is an end extension of s'

2. X X'

3. s \ s' C X'.

If G is P(U,) generic, let 6 = U{s : 3X E U,(s, X) E G}. The set 6 has order

type w, and a standard density argument proves that it is unbounded in K. After

forcing with P(U,) the cardinal K has cofinality w. A crucial component in the theory

is the following.

Theorem 2.3.2. (Prikry Property) Let a be a formula in the language of P(U,,) and
let (s, X) be any condition. Then there exists a condition (s, X') < (s, X) that decides

a, that is, such that either (s,X) I- a or (s, X) It -a

We will occasionally use the notation 'p II a' for 'p decides a'.

Corollary 2.3.3. Prikry forcing preserves cardinals.

Adrian Mathias [6] proved not only that 6 all but finitely included in every

X E U,, but in fact the almost-containment of 6 exactly characterizes the measure
one sets:

Theorem 2.3.4. (Mathias' Genericity Criteria) Let 6 C K. Then 6 is P(U,) generic
if and only if for all X C K,

16 \ X < R0 -= X e U,.

We devote the rest of this section to examining some properties of the set 6.

Definition 2.3.5. Let 9t = (M, <,=) be a model and let I be a simply ordered
subclass of M. Then I consists of first-order indiscernibles over 9t1 iff for any formula

P(x1,..., x,) and any sequences y7 < < 7n and 7- < < 7- from I we have

aj [-W[i,..,7n] 1 9A h Wc[-Y: ...,./].

This next theorem is intended to explain our usage of the term 'indiscernibles'

when referring to a Prikry sequence. Kunen noticed the first part, but the second
part was noticed either by Solovay, Mathias, or Kunen the history is a bit cloudy.

Theorem 2.3.6. Let Uo be a normal ultrafilter on Ko in a model 9mo of ZFC. Let

in,n+ ? : n -+ 9n+1i Ult(9)n, Un) be the map from 3n to the transitive collapse of
the ultrapower of 9,, by Un. Denote Un+j = in,n+i(Un) and Kn+l = in,n+i(Kn). Let
9rW be the transitive collapse of the direct limit of the above system of ultrapowers,

and let n, and U, be the images of Ko and Uo respectively. Finally, denote 6 = {n, :

n < w}. Then:

1. 6 is a set of first-order indiscernibles over 9M,.

2. 6 is P(U,)-generic over 9M,.

Although Prikry sequences do not in general consist of first-order indiscernibles
in the above sense, this is nearly the case.

Theorem 2.3.7. Let 6 be P(U)-generic, where U is a measure on K. Then for each

formula p(xz,...,x,) there is a A, < K such that for any sequences y7 < *** <

y7,, 71 < ...* < 7y from 6 \ A,, we have

V[1l,...,7n] Y = -1,-.., '].

Proof: First, let U" be the ultrafilter consisting of sets X such that for some

H E U we have [H]" C X. Consider the sets

T = {(71,...,7n) : 71 < *** < tn < p and p[7l,,...,7,n]},

F = {(,... ,) : < ... < <7 < and v[71,...,7n]},

where y is the critical point of U. Since T is the complement of F, one of these sets

must be in Un. Without loss of generality, suppose T E Un. Thus, for some H E U,

we have [H]" C T. It follows that the set {(s,X) : [X]" C T} is dense in P(U). Hence

all but finitely many elements of 6 come from H. The theorem follows. *

We now show that the above theorem is the best possible. That is, we cannot

in general get a Prikry generic sequence consisting of first order indiscernibles over
the ground model.

Theorem 2.3.8. There is a model 9 of "ZFC + there is a measurable cardinal" and
a class of parameter-free formulas {1Vx(v)}JeWm such that each p, is satisfied by x and

only x.

Proof: Consider the minimal set model La [U] of "ZFC + there is a measurable

cardinal". Let 91 be the set of elements definable over La [U] by parameter-free

formulas. That is, let

9 = {x : for some formula Cp, L,[U] = V[xz}.

We claim that 931 is an elementary submodel of La,[U]. We verify that existen-

tial formulas with parameters from 9X1 holding in La[U] also hold in 91. The converse

is similar. Suppose

L,[U] 3v (p[v, a,,..., a,],

where al,...,a,, E 93. Since al,...,a, are definable without parameters, we can

assume that p has only one free variable. The model L,,[U] has a definable well-

That is, let f(v) p(v) A Vv'('(v') -+ v
in La [U] without parameters, and x satisfies p. It follows that x E 97 and moreover,

91t ) p[x]. Thus 91 = 3v Vp(v). This establishes the claim.
Since 91A -< L,[U] and since the latter is the minimal model, it follows that

91 = L,,[U]. Therefore, every element x of L,a[U] is definable in La,[U] via a parameter-
free formula Vp,(v). The class of formulas {px(v)}xeL,[u] satisfies the theorem. 0

The property of the above class of formulas precludes getting any class of

first-order indiscernibles for this model. In particular, no Prikry generic sequence can

consist of first-order indiscernibles, since for each element y of such a sequence there

is a formula 4(ij satisfied only by 7.

2.4 Iterated Prikry Forcing

Magidor [5] formulated his iterated Prikry forcing to obtain results compar-

ing compact cardinals with both measurable and supercompact cardinals. Starting

with a compact cardinal K, he forces to kill all the measurable cardinals below while

preserving the compactness of n, thereby establishing that that the first compact car-

dinal can be the first measurable cardinal. He also proves that the first supercompact

cardinal can be the first compact cardinal.

Let A be a set of measurable cardinals. For each p E A, pick a normal measure

Up on p not containing the set of measurable cardinals below p. Denote U = (U,),Ea.

Definition 2.4.1. The iterated Prikry forcing for U, denoted M(U), has as condi-

tions pairs (s, X,),,Ea such that

1. for each p E A, s, E [p]
2. for each p E A, Xt E U1

3. UtEast is finite.

A condition extends another, (sw,X,),tE < (s', X'),eA, iff for each p E A we
have (s.,X,) < (s'A, X,) in the P(U,) order.

If G is M(U) generic then for each p E A, let 6, = U{s, : 3(s,, Xt),a E G}.

Each 6, has order type w. Magidor's forcing adds a function = {G,},AA which

associates to each measurable cardinal p in the domain a Prikry sequence 6, C t.

In the work that follows we describe a forcing which adds a set of indiscernibles

which are essentially the union of the range of this function, UPEA,'. With sufficient

consistency strength some of the elements can be decoupled, so to speak, from any

particular measurable cardinal.

Theorem 2.4.2. M(U) satisfies the Prikry Property.

Theorem 2.4.3. Iterated Prikry forcing preserves cardinals.


3.1 Definition

Let A be a set of measurable cardinals. For each p E A, pick a normal measure

U, on p which gives measure zero to the set of measurable cardinals below p. Denote

U = (UPA,

K = sup(A).

These conventions will remain fixed throughout the rest of this paper. For conve-

nience, let us assume that the only normal measures in the universe are those that

appear in U. We might as well assume that our ground model is the core model [8]

for this sequence.

The following definitions are relative to U.

Definition 3.1.1. Let the filter of long measure one sets be

(U) = {X C K: X rn E Up for all p E A}.

Definition 3.1.2. The disassociated indiscernible forcing, D(U), has as conditions

pairs (s, X) such that

1. s E [Kj
2. X E Z(U).

A condition extends another, (s, X) < (s', X'), iff

1. s D s',

2. XCX'

3. s \ s' C X'.

For p = (s,X) E D(U), let the support of p, denoted by supt(p), be the set s. A
condition p is a direct extension of q, denoted p <* q, if p < q and supt(p) = supt(q).

The conditions and extension criteria of D(U) closely resemble those for iter-
ated Prikry forcing (Definition 2.4.1). The main difference is that instead of getting

a function associating cardinals with sets of indiscernibles, we get a single set of
indiscernibles: denote 6 = {rj < s : 3p E G such that 7r E supt(p)}.

We will generally adhere to the following conventions. Lower case letters
denote natural numbers (i, j, n,...), finite sets of ordinals (s, t, u,...), or conditions

(p, q, r,... ). Upper case letters (U, X, Z,...) denote measures or measure one sets.
Upper case script letters (U, X, Z,...) denote sequences of measures or long measure
one sets. Lower case Greek letters (3, 7, 7,...) usually represent ordinals.

3.2 Basic Structure

Definition 3.2.1. Relative to U, a measure U, is a vertical repeat point if for all

X E U, there exists some p < K such that X n f E U,.

The idea is that the measure one sets of U, are entangled in measure one sets

from cardinals below. We will exploit this later in order to get indiscernibles that can

not be associated with any particular measure.
The following proposition shows that a vertical repeat point has strictly less

consistency strength than a measure of Mitchell order two.

Proposition 3.2.2. Let L[U] be the minimal model for the existence of a measure

of Mitchell order two on some cardinal K. Then U(t, 0) contains the set of cardinals

pi < K such that U(pt, 0) is a vertical repeat point.

Proof: The class U is a function such that each U(p, 7) is a normal measure
on i and 7 is its Mitchell order index.
We first note that U(n, 0) itself is a vertical repeat point. In fact, consider

the ultrapower ju(,1) : L[U] -+ L[U'] P Ult(L[U],YU(K,1)). Note that in L[U'],

jU(4,x)(K) is measurable. Let X E U(n,,0). Then jU(K,i)(X) E ju(K,I)(U(K,O)). But
then X = K n ju(K,I)(X) E U'(K, 0). By elementarity, there exists some 77 < such
that X n 77 E U(/(, 0), so U(K, 0) is a vertical repeat point. Now note that U(K, 0)
remains a vertical repeat point in L[U']. By normality, K = [I]u(K,1), where I is the
identity function. The proposition then follows from Log' Theorem. *

Proposition 3.2.3. Suppose max(A) = K. That is, suppose sup A = K E A.

1. If K is a vertical repeat point then D(U) = D(U [ K).

2. If K is not a vertical repeat point then forcing with D(U) is equivalent to forcing
with D(U [ K) x P(UK), where P(U,) is Prikry forcing for U,.

By "equivalent" in the second part above, we mean that from a generic object
for one forcing, we may construct a generic object for the other.


1. We need only show that a set that is long measure one for U [ K is long measure
one for U. Let X be long measure one for U [ K. We must show that X has K
measure one, that is, that X E U,. Suppose not. Then K \ X has measure one.
Since UK is a vertical repeat point, for some 7 < K, we have that yn(K\X) E U,.
But then 7 n X U,. This contradicts that X is long measure one for U [ K.

2. Let N be a set in U. witnessing that K is not a vertical repeat point. In other
words, let NE U, be such that for all p < n, we have that pn N U,. Consider
the map
i : D(U) -+ Dy(\N(UrK) x P((U,),

given by (s, X) ((s \ N,X \ N), (s n N,X n N)), where D,\N(U r) is the
disassociated indiscernible forcing for U rK below the condition (0, K \ N), and

PC(U,) is Prikry forcing below the condition (0, N) with the extension of con-
ditions, (s, X) < (s',X'), broadened to allow s D s' instead of just s an end

extension of s'. It is a simple matter to confirm that i is a complete embedding.
That is, i preserves incompatibility and extension of conditions, and i"D(U) is
a dense subset of Da\N(UrK) x PC(U,). It follows that the domain and target
of i are equivalent as notions of forcing. The first term in the target product
is obviously equivalent to D(U ( ), so we are left to convince the reader that
PC(UK) is equivalent to Prikry forcing. Consider the map

: P(UK) PC(U")

given by (s, X) 4 (s n N, (X n N) \ max(s) + 1). This map is also a complete
embedding, and this concludes the proof of the proposition. 0

The above proposition is the first clue concerning how disassociated indis-
cernibles forced from vertical repeat points might behave.

3.3 Prikrv Property

Definition 3.3.1. The relation s < t means that s, t are ordinals or finite sets of
ordinals (they need not be the same type of object) and that sup(s) < sup(t).

Definition 3.3.2. Let (X,),E[,]<. be a sequence of sets. The diagonal intersection of
this collection is the set

AE[H<. X = {Y < Kt : Vs y(y E X,)}.

Proposition 3.3.3. Let X, E (U) for s E [Kt]<". Then A,5[,]
Proof: Let A be the diagonal intersection. Fix p E A. We will show A4 f/A E
U,. Consider the ultrapower map iu, : V --+ 93 m, Ult(V, U,). We need to show
that / E iu,(A). Denote X = (X,),[Kl p E iu,(X),. Notice that for s < we have iu.(X), = iug(X,). Since X, n p E U,
for s < p, we have that p E iu,(X,). The theorem follows. n

Theorem 3.3.4. Let a be a formula in the language of D(U) and let (s, Z) be any
condition. Then there exists a direct extension of (s, Z) which decides a.

Proof: In the interest of notational simplicity, let us assume that s = 0. The
proof can be easily modified to accommodate the more general case by carrying a
constant through in some of the arguments.
Let a and (0, Z) be given.
For each p E A and each s E [p]<", let f : p \ max(s) + 1 -+ 3 via
0 if 3X (sU {7},X) Il a,
f,(7) = 1 if 3X (s U {7},X) IF -a,
2 otherwise.
Let H, E U. be homogeneous for f,. Let H, = A,ga] and for y,7' E H, \ max(s) + 1, we have that

3X (s U {7}, X) a if and only if 3X (s U {7'}, X) 1 a,

and moreover, these conditions decide a the same way.
Let Z' = Z n UEaH,. We claim that a direct extension of (0, Z') decides a.
Suppose not, and we shall reach a contradiction. Fix (sU {S}, Z") < (0, Z') with

Is U {S}l the least possible such that (s U {6}, Z") 1 where s << It is possible
that s is empty. Without loss of generality, assume (s U {8}, Z") IF a. Fix p such
that S E H,.
Now, for each y E H,, there is some X such that (s U {y}, X) forces a. Let
X, be some such X. Let X = Ay<,X,.
Let Y = {v E H, : {7 < C: v E Xy} E U,}. Let H = {7 < p : Xny = Yn3y}.
We claim H E U,. If not, then let f : p \ H -+ p be such that f(7) = v, where v
witnesses that X, n Y n 7. Then there is some v and C E U, such that f(7) = ;
for all 7- E C. Whether or not i E Y, this is a contradiction.
Let Q = ((Y U H) n X n li) U (X \ p). We claim that (s, Q) forces a, contra-
dicting the minimality of Is U {} i. It suffices to show that any extension of (s, Q)

has an extension which forces r. Let (s', Q') < (s, Q) be arbitrary. By extending
if necessary, we may assume that s' n H # 0. Denote = min(s' n H). Notice
that (s', Q' n ) < (s U {(}, X,). In fact, consider 7 E s' such that 7 > (. Since
97 E Q C Ay<,x, we have that r E Xe. Now consider 7 E s' such that 7 < We
have that Q n C (YUH) n and s' n nH = 0 by choice of 6. Since E H,
it follows that Y n ( = X4 n f; so I7 E Xt as well. Hence s' C Xt. It follows that

(s', Q' n 4) extends both (s U {(}, Xt) and (s', Q'). But (s U {~}, 4) forces a, so
(s', Q' n .4) does as well. We have produced an extension of an arbitrary extension
of (s', Q') which forces r. Hence (s', Q') itself forces r. This establishes the claim,
which contradicts the minimality of Is U {S}|, finishing the proof. *

3.4 More Structure

Definition 3.4.1. Let p = (s, X) E D(U) and let v < t. Let

1. p [(v+1)= (s n (v + ),Xn (v+ 1))

2. D,+,(U) = D(U [ (v + 1)).

3. p 1 v = (s \ ( + 1), X \ ( + 1)).

4. i(U) = {(s, X) E D(U r ( \ (v+ 1))) : (su X) n (v + 1) = 0}.

We refer to p r (v + 1) and p [ v as 'p up through v + 1' and 'p down past v',

Proposition 3.4.2. For any v < K, we have that D(U) is isomorphic to the product
D,+1(U) x VY(U) by the map p + (p r (v + 1),p I V).

Proposition 3.4.3. Fix v < K.

1. D,+i(U) has the p+-chain condition, where p = Isup(A n (v + 1))1.

2. The <* relation is 2"-closed in IW(U).

Proof: For the first assertion, notice that incompatible elements must differ
in their support. Since there are only p choices for the support of a condition in

D,1I (U), there can be at most p incompatible conditions.
For the second assertion, note that each member of A \ (v + 1) is greater
than v and a strong limit. Fix r7 < A. The claim follows from the fact that if

Xy E Z(U [ (K \ (v + 1))) for 7 < rq, then n,<,,X, E (U r (Kc \ (V + 1))). I

Theorem 3.4.4. Fix v < n. Let a be a formula in the language of D,+1(U) x D (U).
Let (p,q) E DL+ (U) x D (U). There is a q' <* q such that for any (p',q") (p,q')
such that (p', q") decides a, we have that (p', q') decides a (the same way that (p, q")

Proof: For each a E D,+i (U), there is a q(a) <* q, q(a) E D (U) such that

(a, q(a)) decides the formulas

3q E G6 (a,q) IF- a (Ta)

3q E G" (a, q) IF (Fa)

Since, in DW(U), the direct extension relation is 2k-closed, there is some q' such that

q' < q(a) for all a E D,+1 (). We claim that this q' satisfies the theorem. Suppose

(p', q") < (p, q') such that (p', q") decides a. Without loss of generality, suppose
(p', q") II- It follows that (p', q") forces Tp. Now consider (p, q'). Since (p', q') <
(p, q(p)), we have that (p', q') decides Tp. But (p',q') is compatible with (p', q"). We
have that (p', q') decides Tp and is compatible with a condition that forces Tp,. Hence

(p', q') must also force Tp,. With this fact in hand, let us suppose for a contradiction
that (p', q') does not force a. Since (p', q') does not force a, there is a (p", q") < (p', q')
such that (p", q") It- a. Since (p", q") extends (p', q'), we have that (p", q") also forces
Tp,. This means that for some q"' compatible with q", the condition (p, q"') forces a.

But (p", q") is compatible with (p, q"'), and the former forces a while the latter
forces a. This contradiction establishes the theorem. *

Theorem 3.4.5. Disassociated indiscernible forcing preserves cardinals.

Proof: Suppose for a contradiction that A is the least cardinal in V that is
not a cardinal in V[G].
First note that A is a successor cardinal since if all cardinals below a limit
cardinal are preserved then the limit cardinal must be preserved as well. Hence the
least collapsed cardinal cannot be a limit.
Next, A < n. In fact, D(U) has the K+ chain condition and hence preserves
cardinals above K.
We now claim that A is not a cardinal in V[Gr(A + 1)]. To see this, we
demonstrate that the power set of A is the same in V[G] and V[G[(A + 1)]. Let S be
a D(U)-name for a subset of A. Suppose p II- S C A. For each v < A, apply Theorem
3.4.4 to p at A +1 for the formula rv E S' and get a condition q(v) E DA+I (U). Since
the q(v) have identical support, by Proposition 3.4.3 there is a condition q extending
all of them. Consider T, the name for the set {v < A : (p r (A + 1),q) IF v E S}. It
follows from the definition that 5G = j.Gr(A+1), which establishes the claim.

But now notice that DA+I (U) has the A chain condition. Hence A must remain
a cardinal in V[G[(A + 1)], which is a contradiction. U

3.5 Genericity Criteria

Theorem 3.5.1. (Genericity Criteria) Let 6 C K. Then 6 is D(U) generic if and
only if for all X C K,

16 \XI < o = X E (U). (3.1)

Proof: The proof of this theorem will take up the whole section. Clearly, if
G is D(U) generic, then 6 = U{s : (s,X) E G} satisfies (3.1). Suppose now that

6 satisfies (3.1). Let G[6] = {(s,X) : s 6 and 6 \s C X}. Let D be dense and
open in I(U). We shall prove that D) n G[G] # 0.

Definition 3.5.2. For any dense D, let

D* = {(s, X) : 3 E A \ max(s) {7 < p : (s U {7},X) E D} E U,}.

The following is the main technical lemma used in the proof.

Lemma 3.5.3. Let D be dense open. There is a long measure one set Q(D) such that
for anyt, if is the least possible ordinal such that t < S and (0, Q(D)) > (t U {6}, X) E D

for some X, then (t, Q(D)) E D*.

Proof: The issue here is that the long measure one part Q(D) of the condition
in the conclusion is fixed. We obtain this set by exploiting the fact that D is open; a
large part of the proof consists of getting Q(D) contained in enough sets X such that

(t, X) E D. The set D* provides the relevant concept of 'enough.' As in the proof
of the Prikry Property, we begin by labeling the relevant long measure one sets. For
each t E [x]] X. Since D is open, the goal becomes finding a set Q(D) that is contained in enough
sets Xt.
Let us first take care of getting a set E eventually contained in each Xt. Let

E = AtXt.

Then for each t E [c]<" we have

C \ max(t) + 1 C Xt. (3.2)

We will use this to ensure that enough sets Xt contain the part of Q(D) above t.
Getting enough sets Xt containing the part of Q(D) below t will prove more difficult.

We continue by defining measure one sets H each of which is homogeneous
for getting a condition in D. For each p E A, each t E []<" such that t << p, let

ft : j_ -+ 2 such that

S 0 if 3X (t {7}, X) D,
JA T 1 otherwise.

Let H, E U,, be homogeneous for ft and maximal. Let

H, = AtE[,
Thus, if 7, 7' E H, and t < 7, 7' then

3X (t U {7}, X) E D 3X (t U {7'},X) E D. (3.3)

We can now define sets J and W with the property that, roughly speaking,
W is homogeneous for indexing long measure one sets X that contain an initial part
of J see (3.4).
First we need some auxiliary notation. Define a partial function p : [Kn]< -+ A
as follows. Let p(t) be the least p such that {7 : 3X (t U {7}, X) E D} E U,. All of
the following presumes that this function is defined. Note that p(t) is the least p (if
it exists) such that ft"H = {0}.
For each t E [K]
g9(Y) = tu{,}n 7.

Let Wt E U,(t) be homogeneous for gt. Let

Wt = n,<(t)W,.

J'= U (-yn Xtu,}).

Each Jt is long measure one up to p(t). To get a long measure one set up to K, let
Jt be Jt with the interval [p(t), K) adjoined. Let

J = At- .

We have, for any t such that Wt is defined and for 7 E Wt,

Xtu{y} D3 n 7. (3.4)

This will ensure that each Xtu{, contains the part of Q(D) below 7.
Finally, let
Q(D) = En n U H,.
We claim that Q(D) satisfies the lemma. For some t, 6, and X, we have that

(0, Q(D)) 2 (t U {}, X) E D; pick t and 6 so that t < 6 and so that S is as small as
possible. We claim that this condition is in D*. In fact, we show

{7 : (t U {7}, Q(D)) E D} E U,(t).

Note that J E H, for some p with f,"H, = {0}. By minimality of S, we have that
u = p(t). Furthermore, (3.3) implies that

{7 : (t U {7}, Xu{-)) E D} E U,(t).

By (3.2) and (3.4) we have that Q(D) is contained in each Xtu{y} above. Since D is
open, we have that (t U {7}, Q(D)) E D for 7 in a measure one set. This concludes
the proof of the lemma. I
Let us define the dense sets to which we will apply Lemma 3.5.3.

Definition 3.5.4. For each s E []

D,o0 = {(s', X): (s U S',X) E D},

We will use these sets to get a condition that is in G[6] n Do,m for some m E w.
Let so = 6 \ Q(Do,o). Then (so, Q(Do,o)) E G[E]. Suppose now that i = (So,..., si)
has been defined. For notational simplicity, let us identify si with the set so U U si.

z(gi) = nl,< n,,, A,,2Q(DZ)u,,n),

si+ = \ ((s;) U s,).

Then (si+x,Z(si)) E G[6], where s,+1 = (so,...,si+i).
Claim 1: For each i < w e have si+x < si.

Proof: Denote 8 = max(si+,). By definition of si, we have that 6 8 Z(Si).
However, since Z(.i) is a diagonal intersection, if max(si) < 6, then 6 E Z(.i). This
impossibility establishes the claim. *
Since our universe is well founded, infinite descending chains of membership
are excluded. Hence for some k < w we have that sk+1 = 0.
Claim 2: For some m < w, we have (Sk+1, Z(Sk)) E Do,m.

Proof: For some {i6,... ,m} and some X, we have

(0, Z(gk)) > ({86,..., m}, X) E DJsk+,0. We show by induction on i < m that
({61,...,6m-i}, Z(Sk)) E Dik+,,,. Suppose that ({6J,...,6 -i},Z(Sk)) E D,k+,,;. Pick
_'-, the least possible such that for some X

(0, Z(k)) 2 ({I,...,6, -i,_},X) E D8,+li.

Since Q(Dik+,,j) C Z(sk), we may apply Lemma 3.5.3 and use the fact that D, k,,i is
open to see that

({81,..., S-(i+l)},Z(Sk)) E Da~k+,i+l.


(Sk+1 U {1, ., ,m-i-},Z(Sk)) E Do,i+,.

This establishes the claim n

Finally, we claim that for some {71,... ,ym} we have that

(9k+, U {71, ...,7m}, Z(,k)) E Do,o n G[6].

We proceed by induction on i < m. Suppose

( 1k+1 U {7'1, ... ,yi, Z(Sk)) E Do,m-i n G[G].

Select 'Yi+1 as follows. Since Do,m-i = D*,m-i_-, there is a measure one set of elements

- such that adding 7 to the support of the above condition gets the condition into

Do,m-i-1. Since 6 is unbounded in U, for all p E A by hypothesis, we may select

7i+i from the aforementioned measure one set intersected with 6. The augmented
condition will be in Do0,m-i-i n G[6]. This completes the induction and establishes

the claim.

We have found a condition in both V and G[6], and since D was an arbitrary

dense open set it follows that G[6] is generic. This concludes the proof of Theorem

3.5.1. *


4.1 The Measures

Assume that UK is a vertical repeat point. We show that after disassociated
indiscernible forcing not only does i remain measurable, but in fact K has exactly
two normal measures. In this section we really do need to assume that our ground
model is the minimal model having a repeat point.

Definition 4.1.1. We define Uj for i E {0, 1}. Let j : V -+ 9t = Ult(V, U,) be the
ultrapower map via the normal measure U,. A condition (s, X) in the generic filter
forces a set X to be in Uo, (ie. (s,X) IF X E Uo) iff for some X' E (j(U)) with
X' n n = X we have (s,X') IF- K E j(X). Similarly, (s,X) forces X to be in U), iff
for some X' E (j(U)) with X' n K = X we have (s U {f}, X') It- r E j(X). In either
case, the condition which forces K to be in X is said to be a witness of (s, X) forcing
X in UP.

Here is an informal discussion of why the definition works and why a vertical
repeat point is required. Note that a condition q which witnesses that p forces X to
be in Ur is compatible with j(p). In fact we require the long measure one part of p
to match the long measure one part of q up to K. Since UK is a vertical repeat point,
long measure one sets in ultrapowers by any U, D Ur are measure one for UK (even
though t cannot be measurable in any such ultrapower). Hence the long measure one
part of q is measure one for U,. This is key. Since U, is a vertical repeat point, we
have that q r (c + 1) is in the original forcing for U. In other words, the fact that UK
is a vertical repeat point ensures that the restriction to K + 1 of images of conditions

in D(U) remain in D(U). Further, if U, were not a vertical repeat point we could have
some q (compatible with the image of p) forcing the part of the set of indiscernibles
below K to be disjoint from a set in UK. But the set of indiscernibles must be identical
to its image up to K under any ultrapower with critical point K. In other words, for a
condition p to force the set of indiscernibles to be in some normal measure extending
UK, conditions compatible with the image of p must have a long measure one part
which ensures that the initial segment of the indiscernibles comes from sets in UK.
Otherwise, the part of the image of the indiscernibles below the critical point could
differ from the original set of indiscernibles.

Theorem 4.1.2. In the generic extension, U, and U, are normal K-complete non-
principle ultrafilters with 6 U, and 6 E U,.

Proof: We first need to show that the Ui are well-defined. That is, p IF X E
U' does not depend on choice of name for X. Suppose p IF X = Y and p II- X E U".
Then j(p) I- j(X) = j(Y), and if q < j(p) with q It K E X, then q IF K E Y. So the
definition is name-independent.
Next, each U, is a nonprinciple ultrafilter. We show that U/ is closed under
enlargements; the proofs of closure under intersection, nonprincipality, and maximal-
ity are similar. Suppose p IF rX E U, A X C Y'. There is a witness q < j(p) of p
forcing X in U,. So q IF rK E j(X) A j(X) C j(Y)'. It follows that q IF K E j(Y).
Hence q is also a witness of p forcing Y in UI.
Let us now show that U. is K-complete. Suppose for some q/ < K we have
p IF U,<,X, = K. We find a p' < p and a y' < 7 such that p' It X E U,,. By
elementarity, j(p) IF U,<,j(X.) = j(K). For each 7 < r, apply Theorem 3.4.4 to j(p)
at K to get a pair (r,, r(7)) E D,+l(j(U)) x D1(j(U)) such that r, = j(p) [ (K + 1),

r"(7) <* J(p) L K, and such that if (a, b) < (r,, r"(7)) with (a, b) deciding K E j(X,),
then (a,rr"()) decides K E j(X,) the same way. Since D'(j(U)) is c+-closed and

the r"(7) are compatible, there is a condition r" < r"(7) for all 7 < r7. Note that
r, = j(p) r (t + 1) for all y < 77, so that (j(p) F (K + 1),r") < (r, r"(7)) for all
y7 i. Now since (r,,r") IF U,<,j(X,) = j(K), for some q < (p', r') and some
7' < r, we have that q IF Ki E j(Xy,). Denote p' = q r (K + 1). Since U, is a vertical
repeat point, p' E D(U). We claim that (p', r) is a witness to p' forcing X, in U,,
where the i depends on whether or not n is in the support of q. It is clear from
the fact that r" <* j(p) [ K that the support of (p', r") is equal to the support of
p'. Also j((p',r")) r (r + 1) = j(p') r (t + 1). And by definition of the (r,, r") we
have that (p', r") IF E (X,,). Thus (p', r") is a witness as claimed. It follows that
p' I- Xy E U, so that Uf is K-complete.
Finally, we show that the UI are normal. Suppose p IF- rf : A V7 <
K f(7-) < 7'. We find a p' < p, a 7' < K, and a set X such that p' IF rX E
Ur A f"X = 7''. For every 7 < t, apply Theorem 3.4.4 to j(p) at K to get a pair
(r,r'(7Y)) E DI+1+(j(U)) x ID (j(U)) such that r, = j(p) r (. + 1), r"(7) <* j(p) [ t,
and such that if (a, b) < (r,,r"(-)) with (a, b) deciding j(f)(K) = 7, then (a,r"(7))
decides j(f)(K) = 7 the same way. Since ]D(j(U)) is K-closed, there is a condition
r" < r"(-) for all 7 < r7. Now, since j(p) > (r,, r") II j(f) : j(K) -- j(K), there is
some -' < t and some < (r,, r") such that q I- j(f)(K) = -'. Denote p' = q r (K+1),
which is in D(U) since U, is a vertical repeat point. We claim that (p', r") is a witness
to p' forcing {7 < K : f(-y) = 7'} in U,. It is clear from the fact that r" <* j(p) I[ K
that the support of (p', r) is equal to the support of p'. Also j((p', r")) r (K + 1) =
j(p') r (n + 1). And by definition of the (r,, r) we have that (p', r") II- j(f)(K) = 7'.
In other words, (p', r) I- Kt E {7 < j(K.) : j(f)(7) = 7'}. That (p',r") is the witness
as claimed follows by elementarity. *

Theorem 4.1.3. The only normal ultrafilters on K in the generic extension are U0
and UK.

Proof: Suppose p = (s, X) II rX E W A W is a normal measure on K1. We
find a q < p such that q IF X E U, for some i E {0, 1}.
In the generic extension there is an ultrapower map jw. The restriction of this
function to the ground model factors through the ultrapower j = ju".
We appeal to Theorem 3.4.4. For each v < K there is a q, = (s, Q,) <* p [ v
in IY(U) such that if (a, b) < (1,q,) and (a,b) E D,+i(U) x DW(U) decides v E X
then (a, q,) decides v E X the same way.
Let Q = A<,, and let p' = (s,X n Q).
Now p' IF X E W. Hence for any generic filter G with p' E G we have
V[G'] = n E j(X), where j(p') E G'. It follows that there is some q compatible with

j(p) such that q II K E j(X). Denote q' = (q [ (r + 1),j(p') [ K). Note that q' is also
compatible with j(p'). We can assume that q' r n < p'.
Then q' IF X E Ui, where

i= 0 if E supt(q r (K + 1)),
t 1 if K 0 supt(q [ (n + 1)).

By the elementarity of j and the definition of p', we have q' II- K E j(X). The claim,
and hence the theorem, follows. n

4.2 Properties

Theorem 4.2.1. In the generic extension, the following hold:

1. For each X E U there is some Y E U, such that Y \ 6 C X.

2. For each X E U, there is some Y E U, such that Yn 6 C X.

Proof: Suppose p = (s, X) I- X E U, for some i E {0,1}. The two cases are
not exclusive, but without loss of generality we can assume that exactly one holds.
By Theorem 3.4.4, for each v < K, there are q, = (s \ (v + 1), Q,) E W (U) such that
if (a, q') < (1,q,), where (a,q') E D,+i(U) x 1B(U) such that (a, q') decides v E X,

then (a, q,) decides v E X the same way that (a, q) does. In other words, q, decides
v EX above v. Let Q = ll, Since q IF X E U,, there is an r < j(q) such that r r (Kc + 1) = j(q) [ (K + 1),
r IF- t E j(X), and supt(r) \ supt(q) is either {0} or {c}, depending on whether X is
forced to be in U, or U, by p. In the former case, let r' E D,(U) be the restriction

of r to v + 1. In the latter case, let ri be the above restriction augmented by adding
v to its support.
Let Yi = {v < c : (r, q [ v) IF v E X}. Note that j(Y1) = {v < j(t) :
(j(r),,j(q) v) IF v E j(X)}. Since j(r)' = r r (K + 1) and (r r (n + 1),j(q) L[ K) IF
K E j(X), we have that K E j(Y). It follows that Y E U,.
Suppose now q' < q and either q' IF v E Y n 6 or q' IF v E Y \ 6.
Note that the above decides whether or not v E supt(q'). In either case, we have
(r, q [ v) IF v E X by definition of Y. Since r, = q r (v + 1) > q' r (v + 1), we have
(q' r (v + 1),q L v) IF v E X. Since q' I v < q L v, we have q' IF v E X. The theorem
follows. M

Compare the following with Theorem 2.3.6.

Theorem 4.2.2. For i E {0, 1}, consider the restriction of the ultrapower

ju : L[U][6] -+ L[Ul][6'] Ult(L[U][6], U)

to the ground model

ji: L[U] -, L[U'].


1. ji is an iterated ultrapower along measures from U

2. every element of G' \ G is a critical point of the iterated ultrapower.

Proof: The first point is established in [8]. Let us demonstrate the second.
Suppose for a contradiction that v E 6' \ 6 is not a critical point in the iteration.
Then for some function f : K -+ K, there is a 7 < v such that v = juv(f)(y). Thus
v is in the set {rq < jut(K) : 7 E juv(f)"'r}. In other words v E juv(Z), where
Z = {r7 < K : 77 E f"r)}. Note that for all y E A we have ZA ip U,. Denote
D = {(s,X n Z) : (s,X) E D(U)}. Now D is dense, so G n D # 0 for all generic G.
Suppose p is a condition in the intersection. We have ju((p) E juv(G). But since p
has finite support bounded below n, we have that v is not in the support of ju((p).
Neither is v in the long measure one set. Since 6' = j. (6), we cannot have v E 6'.
This contradiction establishes the theorem. n


5.1 Definition

Definition 5.1.1. We define functions fy : It -+ r for 7 < r+ by recursion. Let

fo(r7) = 0. Let f-y+1(r) = fy(r) + 1. For limit ordinals 7, let f,(r) = U,<,fy,(rq),
where (% ),<, is onto 7.

The definition of f. for limit 7 is the diagonal union. Note that [f-,]u = Y.
That is, these functions are canonical representations of the ordinals less than jK+ in

the ultrapower by U,.

The following is relative to our fixed sequence of measures U = (U,),4A.

Definition 5.1.2. We define vertical repeat point of order 7 < x+ by recursion on

7. Let ord(U,) = 0 if for some X E UK, for all p < K we have X nf pj U,. Let

ord(U,) > 7 if for every X E UK, the set {/ < K : ord(U,) = f,(p) and X n p E U,}
is stationary in K.

The definition for successor ordinals is an analogy to being Mahlo for the

predecessor order.

Theorem 5.1.3. The consistency strength of a vertical repeat point of order 7 < Ic+

is strictly less than that of a measure concentrating on measurable cardinals.

We define iterated disassociated indiscernible forcing of order up to and in-
cluding K.

For each 7 < K, let A, = {/ E A : ord(U,) > f.(p)}. For each 7 < K, let

= (U),EA,.

Definition 5.1.4. Let I be an interval of ordinals. The I-iterated disassociated in-
discernible forcing for U, denoted D(U, I), has as conditions sequences (sy, Xy)yE
such that

1. for all 7 E I, we have s, E [Kn]<

2. all but finitely many s., are empty

3. for all 7 E I, we have X, E (UY)

4. if 7 y 7' then s, n sy = 0.

A condition extends another, (s', X.)., E < (Sy, X).yE,6, if for all 7 E I we have
(s', X,) < (s, Xy) in the D(U) order.

The definition of direct extension is analogous to that for disassociated indis-
cernible forcing. We sometimes use the notation (s, X) for conditions when the index
is obvious. We'll use the 'adjoin' operator '^' to display the contents of g, regarding
the partial function as a set of pairs:

(io, 7o, )^ -(i, 7, )^ .-^ in, 7n).

Assume from now on that max(A) = I, and that U, is a vertical repeat point of order
< K.

5.2 Basic Structure

Definition 5.2.1. Fix v < K. Let p = (sy, X-)YIE E D(U, A). Let

1. p r(v + 1)= (s, n (v + 1), Xn (v + 1)),y<

2. p [v = (s, \ (v + 1),X, \ (v + 1)),
3. D,+ (, A) = D((U,)PEAn(v.+), A).

4. ID(U, A) = {(s-, X E)<, D((U,),eA\(+1), A) : V7 < A((sUX,)n(v+1) = 0)}.

We do not disallow some of the long measure one entries being empty in
p + 1).

Proposition 5.2.2. For any v < n, we have that D(U, A) is isomorphic to the prod-
uct D,+ (U, A) x IYV(U, A) by the map p (-p (p (V+ 1),p L v).

It is interesting to note that this forcing and its conditions can be factored in
two ways: 'vertically', as above, and although we will not utilize it, horizontally:

Definition 5.2.3. Let p = (s.y, X4)Y E D(U, A). Fix ( < A. Denote

1. p+ ( = (s, X-Y)-Y<

2. p+ C = (s,, X-,)eA\

Proposition 5.2.4. For any C < A, we have that D(U, A) is isomorphic to the product
D(U, () x D(U, a \ ) by the map p (p+ C, p+ C).

5.3 Prikry Property

Definition 5.3.1. For any set K, fix some bijection e : K -+ K. Let (Xi)ieK be
given. Let the e-diagonal intersection be AeX, = -{ < i : Vi < y7( E Xe(.))}.

Proposition 5.3.2. Let (Xi)iEK be given such that each Xi E (U). Let X = AeXi.

1. For each i we have that X \ (e-1(i) + 1) C Xi.

2. x e (U).

Definition 5.3.3. If s' is a set of pairs, denote supgs = max{i,j : (i,j) E s}. The
relation < means that s', fare either ordinals or finite partial functions from K to
Theorem 5.3.4. Let a be a formula in the language of D(U, A) and let (sy, X),<
be any condition, where A < K. Then there exists a direct extension of (s, Z,), that decides a.

Proof: The proof is essentially the same as the proof for the disassociated
indiscernible Prikry Property with some slight modifications to handle the iterations.
Select some bijection e : I -+ [ x c]<" such that for each cardinal ( we have that
e"C = [( x (]<. We will use e to define our diagonal intersections. We will assume
that each s, = 0 to simplify the notation.
Let a and (0, X) be given.
For each p E A, s, and i < A let f y2 : p 3 via

f 0 if 3' (S -,Y), X) IF-0,

2 otherwise.
Let H,,i E U, be homogeneous for f/;. Let H,i = A;Hfg;. For any S, any 7, E
H,,i \ (sup S'+ 1), we have that

32X (S (i, 7), ) | a if and only if 3X (' (i, 7'), X) jI a,

and these conditions decide a the same way.
For each i < A, let Hi = UAE, He,i. Let Z' = (Hi n Zi)i, direct extension of (0, Z') decides a. Suppose not, and we shall reach a contradiction.
Fix (s (j, 8), Z") < (0, Z') with I ^ (j, 6)1 the least out of all conditions deciding a,
where s'< S. Without loss of generality, assume (s(-(j, S), Z") I- a. Let p be such
that 8 E Hj.
From now on, s,1 8 and j are fixed.
For each pair (i, 7) where i < 7 < p, if there is some X such that (5 (i, 7), X)
decides a, then let X(i,") be some such X. Let X = (Ae, )(ikY'' For each k < A, let Yk = {v < t : {7 < : v E E U } and let
H* = {7 < p : 7 = Yk n 7}. Then each H; E U,. Let H* = Ak Then for any 7 < p we have Xk(j') n 7 = Yk n for all k < 7.
Let Qk = ((Yk U H*) n Xk n y) U (k \ /). We will show that (s, Q) forces
a. Let (t, R) < (s, Q). Without loss of generality, i contains a pair (j, ) such that

SE H*. Consider the least such (. We claim (t, (lk n X )k<,) < (S -^j, ), X ).
In fact, let (k,7) E f. If > then 7 E Qk \ (e-((j,)) + 1) C X'). Suppose
7- < Since C E H* and k < 7 < ( we have that Xk(1 n Y = Yk n so (k, y) E Xk).
Either way tf\ comes from X,'4). This establishes the claim. But (S' "(j, ), X(U'())
forces a. Hence (t, (Rk n Xk <)k (S, Q), forces a. It follows that (s, Q) forces a. This contradicts the minimality of
jS ^(j, 6)I, finishing the proof. n

5.4 More Structure

Theorem 5.4.1. Fix v < tc. Let a be a formula in the language of DI(U, A) x
II(U,A). Let (p,q) E D (U,A) x IY'(U,A). There is a q' <* q such that for any
(p', q") < (p, q') such that (p', q") decides a, we have that (p', q') decides r.

Proof: The proof is virtually the same as the proof of Theorem 3.4.4. U

Theorem 5.4.2. Iterated disassociated indiscernible forcing preserves cardinals.

Proof: The obvious alteration of Theorem 3.4.5 works. 0


Assume that K is a vertical repeat point of order A < K. After A-iterated disassociated
indiscernible forcing, K has exactly A normal measures.

Definition 6.3. We define U/ for 7 < A. Let j be the ultrapower map via the normal
measure U,. A condition (sy, X,), (ie. (sy, X,),< IF X E ) iff there is a (s', X,).,< compatible with j((s,, Xy)<,)
such that s' = sU U {K}, s' = s, for 7 5 rq, and X, = X' nK for all 7 < A.

For each r < A, denote 6, = {( < K : 3(sy,,X.,)',E E G such that C E s,}.

Theorem 6.4. In the generic extension, the normal measures on r are exactly the
U for < A.

Proof: The proof is almost identical to those for Theorems 4.1.2 and 4.1.3,
with Theorem 5.4.1 used in place of Theorem 3.4.4. *


[1] Thomas Jech Set Theory, Pure and Applied Mathematics v. 79, Academic Press,
[2] Akihiro Kanamori The Higher Infinite, Perspectives in Mathematical Logic,
Springer-Verlag, 1991.
[3] Kenneth Kunen Set Theory: An Introduction to Independence Proofs, Studies
in Logic and the Foundations of Mathematics v. 102, Elsevier, 1980.
[4] Kenneth Kunen and Jeffrey B. Paris Boolean extensions and measurable cardi-
nals, Ann. Math. Logic, 1971, pp.359-377.
[5] Menachem Magidor, How large is the first strongly compact cardinal? or A study
on identity crises, Ann. Math. Logic, 1976, pp.33-57.
[6] A. R. D. Mathias, On sequences generic in the sense of Prikry, J. Australian
Math. Soc., 15 1973, pp. 409-414.
[7] William J. Mitchell, Sets constructible from sequences of ultrafilters, J. Symbolic
Logic, 1974, pp.57-66.
[8] William J. Mitchell, The core model for sequences of measures. I, Math. Proc.
Camb. Phil. Soc., 1984, pp.229-260.
[9] Karl Prikry Changing measurable cardinals into accessible cardinals, Disserta-
tiones Math., 1970, pp.5-52.


Jeff Leaning was born in Luton England on September 1, 1971. He was raised

outside of Detroit in Farmington Hills Michigan. In 1989 he moved to Gainesville


I certify that I have read this study and that in my opinion it conforms to
acceptable standards of scholarly presentation and is fully adequate, in scope and
quality, as a dissertation for the degree of Doctor of Philosoby

WilliarJ". Mitchell, Chairman
Professor of Mathematics

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Douglaaenzer 0
Professor of Mathematics

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quality, as a dissertation for the degree of Doctor of Philosophy.

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Professor of English

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acceptable standards of scholarly presentation and is fully adequate, in scope and
quality, as a dissertation for the degree of Doctor of Philosophy.

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Professor of Mathematics

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acceptable standards of scholarly presentation and is fully adequate, in scope and
quality, as a dissertation for the degree of Doctor of Philosophy.

Jorie Martinez
Professor of Mathematics

This dissertation was submitted to the Graduate Faculty of the Department of
Mathematics in the College of Liberal Arts and Sciences and to the Graduate School
and was accepted as partial fulfillment of the requirements for the degree of Doctor
of Philosophy.

August 1999
Dean, Graduate School


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