DISASSOCIATED INDISCERNIBLES
By
JEFFREY SCOTT LEANING
A DISSERTATION PRESENTED TO THE GRADUATE SCHOOL OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY
UNIVERSITY OF FLORIDA
1999 ACKNOWLEDGEMENTS
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.
11 TABLE OF CONTENTS
ACKNOWLEDGEMENTS ii
ABSTRACT 1
CHAPTERS
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 ITERATED DISASSOCIATED INDISCERNIBLES 30 5.1 Definition 30 5.2 Basic Structure 31 5.3 Prikry Property 32 5.4 More Structure 34
6 GETTING MANY NORMAL MEASURES 35
REFERENCES 36
BIOGRAPHICAL SKETCH 37
iii 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
DISASSOCIATED INDISCERNIBLES
By
Jeffrey Scott Leaning
August 1999
Chairman: William J. Mitchell Major Department: Mathematics
By ‘measure’, we mean a «-complete ultrafilter on some cardinal «. 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 (/(a) < a for all a) function /:«->« we have that / is constant on some set X 6 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.
1 CHAPTER 1 INTRODUCTION
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.
2 3
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. CHAPTER 2 PRELIMINARIES
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 not.
2.2 Ultrafilters and Ultrapowers
Definition 2.2.1. A measure is a «-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. a Let U be measure on «. Consider the equivalence relation on functions f,g:K—tn,
f ~ g iff {r/ < k : f(g) = g(g)} e U.
4 5
Let Ult(V, U) be the model whose universe consists of equivalence classes [/] of
functions f : k k under the relation ~. Interpret *€’ in Ult(V, U) as
[/] € [fir] iff {t7 < /« : f(r¡) € g{r¡)} € U.
Let ju : V —> 9JÍ w Ult(V, U) be the natural map given by x i-> [cx] where cx is
the constant function cx(r]) = x, and VJl is Mostowski’s transitive isomorph of the
ultrapower.
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¬
erywhere.
Definition 2.2.3. A measure U on k is normal if every regressive (Va f(a) < a)
function f : k —y k is constant on some set X € 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 /(a) = a represents the cardinal k.
Definition 2.2.4. Let (Xv)r, section of these sets is the set A^^X^ = {7 < k : Vr; < 7 (7 E AT,,)}. Proposition 2.2.5. Let U be a measure on k. The following are equivalent. 1. U is normal. 2. VX(X n k£U <=> k € ju(X)). 3. U is closed under diagonal intersection. Definitions 4.1.1 and 6.3 are a modeled after forcing-syntactic version of prop¬ erty 2 above. 6 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 : [/c] There exists a set X G U such that f is constant on [X]n for each n < u. The set X above is said to be homogeneous for f. 2.3 Prikry Forcing Prikry forces an unbounded sequence of type ui 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 UK be a normal measure on k. Definition 2.3.1. Prikry forcing, F(UK), has as conditions pairs (s,X) such that 1. s e [k] 2. X € UK. A condition extends another, (s,X) < (s',X'), iff 1. s is an end extension of s' 2. X C X' 3. s \ s' C X'. If G is P(£/k) generic, let © = U{.s : 3X € UK{s,X) € G}. The set © has order type u>, and a standard density argument proves that it is unbounded in k. After forcing with P([/K) the cardinal k has cofinality u. A crucial component in the theory is the following. 7 Theorem 2.3.2. (Prikry Property) Let a be a formula in the language ofF(UK) and let (s, X) be any condition. Then there exists a condition (s, X') < (s, X) f/iai decides a, that is, such that either (s,X) lh a or (s,X) II— We will occasionally use the notation lp || cr’ for lp decides a’. Corollary 2.3.3. Prikry forcing preserves cardinals. Adrian Mathias [6] proved not only that © all but finitely included in every X G UK, but in fact the almost-containment of © exactly characterizes the measure one sets: Theorem 2.3.4. (Mathias’ Genericity Criteria) Let © C «. Then © is P(UK) generic if and only iffor all X C k, I© \ X\ < No X € UK. We devote the rest of this section to examining some properties of the set 6. Definition 2.3.5. Let VJl = (M, <,=) be a model and let I be a simply ordered subclass of M. Then / consists of first-order indiscernibles over ÜJI iff for any formula 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 Uq be a normal ultrafilter on Ko in a model QJlo of ZFC. Let in,n+1 '• 97in ~^ 97tn+i ~ Ult(9Jtn, Un) be the map from 97tn to the transitive collapse of the ultrapower of by Un. Denote Un+i = in,n+i(Un) and /cn+1 = ¿n,n+i(«n)- Let dJlu be the transitive collapse of the direct limit of the above system of ultrapowers, 8 and let nw and Uu be the images of kq and Uq respectively. Finally, denote 6 = {/c„ : n < uj}. Then: 1. & is a set of first-order indiscernibles over VJlw. 2. © is F(UU)-generic over DJIw. 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 © be IP(U)-generic, where U is a measure on k. Then for each formula ip(xi,...,x„) there is a < k such that for any sequences 71 < ••• < 7nj 7Í < • •' < 7Ó from 6 \ \v, we have Proof: First, let Un be the ultrafilter consisting of sets X such that for some H E U we have [//]" C X. Consider the sets T = {(71,- ..,7») : 7i < ••• < 7n < M and <^[71,... ,7„]}, F = {(71 > • • •»7«) : 7i < ••• < 7n < H and ^[71» • • •»7«]}, where p. 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 € Un. Thus, for some H € U, we have [H]n C T. It follows that the set {(s,X) : [X]n C T} is dense in P(i7). 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. 2.3.8. a Theorem There is model DJI of “ZFC + there is a measurable cardinal” and a class of parameter-free formulas {y>x(u)}xg£ut such that each ipx is satisfied by x and only x. 9 Proof: Consider the minimal set model La[U] of “ZFC + there is a measurable cardinal”. Let 9Jt be the set of elements definable over La[U] by parameter-free formulas. That is, let 9JÍ = {x : for some formula ip, La[U] \= <¿>[x]}. We claim that is an elementary submodel of La[U]. We verify that existen¬ tial formulas with parameters from 9JÍ holding in La[U] also hold in 9JL The converse is similar. Suppose La[U] \= 3v v?[u,aj,...,an], where ai,...,an € ÜJI. Since ai,...,an are definable without parameters, we can assume that p has only one free variable. The model La[U] has a definable well¬ ordering That is, let ip(v) = p(v) AVv'( 9JÍ |= p[x\. Thus 971 (= 3wy>(u). This establishes the claim. Since 9JÍ -< La[U] and since the latter is the minimal model, it follows that VJX = La[U]. Therefore, every element x of La[U] is definable in La[U] via a parameter- free formula px(v). The class of formulas {«^(^^xeLag/] satisfies the theorem. ■ 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 7 of such a sequence there is a formula (p{7} 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 10 preserving the compactness of k, 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 € A, pick a normal measure U, on p not containing the set of measurable cardinals below p. Denote U — ({7m)m£a- Definition 2.4.1. The iterated Prikry forcing forU, denoted has as condi¬ tions pairs (s^, A/1)/i€a such that 1. for each p € A, G [p] 2. for each p € A, € V^ 3. UMeAis finite. A condition extends another, (s^, a < (.s^, X^eA, iff for each /i € Awe have < (sj,, X'¿) in the P{U¡f) order. If G is M(W) generic then for each p € A, let 0^ = U{sM : 3(sm,Xm)mGa eG}. —# Each 6M has order type u. Magidor’s forcing adds a function 6 = {6m}m£a which associates to each measurable cardinal p in the domain a Prikry sequence C p. 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, U^aS^. With sufficient consistency strength some of the elements can be decoupled, so to speak, from any particular measurable cardinal. Theorem 2.4.2. M(¿/) satisfies the Prikry Property. Theorem 2.4.3. Iterated Prikry forcing preserves cardinals. CHAPTER 3 THE FORCING 3.1 Definition Let A be a set of measurable cardinals. For each /x € A, pick a normal measure Uft on /x which gives measure zero to the set of measurable cardinals below fi. Denote K — {U 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 £(£/) = {A* C k : X H /x € for all /x € A}. Definition 3.1.2. The disassociated indiscernible forcing, D(W), has as conditions pairs (s, X) such that 1. s € [k] 2. X e £(W). A condition extends another, (s,X) < (s', A"), iff 1. sD s', 2. A C A" 3. s \ s' C A". 11 12 For p = (s,X) € D(¿f), 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(qi). The conditions and extension criteria of D(£f) 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 © = {77 < k : Bp (E G such that q E supt(p)}. We will generally adhere to the following conventions. Lower case letters denote natural numbers (:, 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 (¿, 7, q,...) usually represent ordinals. 3.2 Basic Structure Definition 3.2.1. Relative to U, a measure UK is a vertical repeat point if for all X € UK there exists some p < k such that X fl p € Ull. The idea is that the measure one sets of UK 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[¿/] be the minimal model for the existence of a measure of Mitchell order two on some cardinal k. Then U(k, 0) contains the set of cardinals p < k such that U(p,0) is a vertical repeat point. Proof: The class U is a function such that each ¿/(p, 7) is a normal measure on p and 7 is its Mitchell order index. We first note that U(k, 0) itself is a vertical repeat point. In fact, consider the ultrapower ju(K,i) '• L[M] —► L\U'] « Ult(L[¿Y],¿/(«, 1)). Note that in L[W], 13 ju(K,1)(«) is measurable. Let X G ¿/(/c,0). Then ju(K,i)(X) € 0)). But then X = k H j¿v(/4il)(.Y) G U'(k,0). By elementarity, there exists some r¡ < k such that X fl r] G U(r],0), so £/(/c,0) is a vertical repeat point. Now note that U(k,0) remains a vertical repeat point in L[U']. By normality, k = [7]w(*,i), where / is the identity function. The proposition then follows from Los’ Theorem. ■ Proposition 3.2.3. Suppose max(A) = k. That is, suppose sup A = k G A. 1. If k is a vertical repeat point then = D(¿Y \ «). 2. If k is not a vertical repeat point then forcing with B{U) is equivalent to forcing with f k) x IP(UK), where P(UK) is Prikry forcing for UK. 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. Proof: 1. We need only show that a set that is long measure one for U f 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 G UK. Suppose not. Then k \ X has measure one. Since UK is a vertical repeat point, for some 7 < n, we have that 7fl(/c\X) G U.r But then 7ÍIT ^ Uy. This contradicts that X is long measure one for U f k. 2. Let N be a set in UK witnessing that k is not a vertical repeat point. In other words, let NG Uk be such that for all p < k, we have that pC\N £ U^. Consider the map given by (s,X) 1-4 ((s \ iV,X \ N), (s fl iV,X fl A^)), where is the disassociated indiscernible forcing for U\k below the condition (0,/c \ N), and Ffc(UK) 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 14 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 *”D{U) is a dense subset of DKyv(£/f/c) x P^(i/K). 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^f/c), so we are left to convince the reader that is equivalent to Prikry forcing. Consider the map i:P(Uk)^F%{Uk) given by (s, X) A (s D N, (X D ./V) \ max(s) + 1). This map is also a complete embedding, and this concludes the proof of the proposition. ■ The above proposition is the first clue concerning how disassociated indis¬ cernibles forced from vertical repeat points might behave. 3.3 Prikry Property Definition 3.3.1. The relation s <3C 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 (A’*)se[K]<« be a sequence of sets. The diagonal intersection of this collection is the set A,e[«]<«;r4 = {7 < k : Vs < 7(7 e xs)}. Proposition 3.3.3. Let X, € £(U) for s € [k]<ü\ Then A4e[*j<*,TJ € £(£/). Proof: Let A be the diagonal intersection. Fix n € A. We will show £ Consider the ultrapower map : V -4 271 « Ult(V, U^). We need to show that n € iupi-A). Denote X = ( —* —* H € iuII(X)s. Notice that for s Theorem 3.3.4. Let a be a formula in the language of ID(ZV) 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 (7 and (0,Z) be given. For each fi G A and each s G [//]<", let /* : fi \ max(s) + 1 —> 3 via '0 if 3* (s U {7}, X) lb cr, W) « 1 if 3X (iU {7},*) lb-a, 2 otherwise. Let f/® G Un be homogeneous for /*. Let = As6[M]<«//®. So, for any s G [fi] 3X (s U {7}, X) || cr if and only if 3X (s U {77}, X) || cr, and moreover, these conditions decide a the same way. Let Z' = Z fl We claim that a direct extension of (0,Z') decides a. Suppose not, and we shall reach a contradiction. Fix (sU{¿},2") < (0,Z') with |s U {¿}| the least possible such that (s U {<£}, Z") || cr, where s 8. It is possible that s is empty. Without loss of generality, assume (s U lb cr. Fix p such that 5 G Hf,. Now, for each 7 G Hthere is some X such that (s U {7},^) forces cr. Let X\ be some such X. Let X = A7 Let Y = {is G : {7 < /x : is G We claim H G U^. If not, then let / : fi \ H —> // be such that /(7) = is, where is witnesses that Let Q = ((F U H) fl X fl /i) U (X \ p). We claim that (s, Q) forces cr, contra¬ dicting the minimality of |s U {¿}|. It suffices to show that any extension of (s, Q) 16 has an extension which forces cr. Let (s', Q') < (s, Q) be arbitrary. By extending if necessary, we may assume that s' fl H ^ 0. Denote £ = min(s' D H). Notice that (s\Q' fl Xf) < (s U fact, consider rj G s' such that rj > £. Since r) € Q C A-1 (s', Q' fl X¿) extends both (s U and (s'i Q')- But (s U forces cr, so (s', Q' fl X¿) does as well. We have produced an extension of an arbitrary extension of (s', Q') which forces 3.4 More Structure Definition 3.4.1. Let p = (s,X) 6 WJA) and let u < k. Let 1. pf(i/ + i) = (sn(i/+i),A’n(i/ +1)) 2. a+i(w) = D(w rO'+i)). 3. p[u= (s\(i/ + l),*\(i/ + l)). 4. IIy{U) = {{s,X) eBiU r(/c\(i/+l))):(su^)n(i/+l)=0}. We refer to p f {u -f 1) and p [ v as ‘p up through v + 1’ and lp down past i/\ respectively. Proposition 3.4.2. For any v < k, we have that D(U) is isomorphic to the product Dv+i{U) x D"(U) by the map p i-> (p [ [y + l),p [ u). Proposition 3.4.3. Fix u < k. 1. has the p+ -chain condition, where p = | sup(A fl (u + 1))|. 2. The <* relation is 2"-closed in W(U). 17 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 0„+i(W), 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 r¡ < A. The claim follows from the fact that if X1 E £(U f (k \ (y + 1))) for 7 < 77, then C\1 Theorem 3.4.4. Fix v < k. Let a be a formula in the language ofDv+l(U)xW(U). Let (p,q) E D„+i(£/) x Iy[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 cr (the same way that (p',q") does.) Proof: For each a E D^+i (U), there is a q(a) <* q, q(a) € W(U) such that (a,q(a)) decides the formulas 3q e G1' (a,q) lh a (Ta) 3«? G Gu (a, q) lh -> Since, in IT (¿7), the direct extension relation is 2l/-closed, there is some q' such that q' ^ ?(a) f°r a £ 1IX+i(¿0- We claim that this q' satisfies the theorem. Suppose < (P)9/) such that (p',q") decides cr. Without loss of generality, suppose (p\q") II- a- If 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 cr. Since (p', q') does not force Tp/. This means that for some q'" compatible with 9", the condition (p, q'") forces a. 18 But (p",q") is compatible with (p, q'"), and the former forces —’ cr while the latter forces cr. 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 < k. In fact, has the k+ chain condition and hence preserves cardinals above k. We now claim that A is not a cardinal in V[G|"(A + 1)]. To see this, we demonstrate that the power set of A is the same in V[G] and V[GT(A + 1)]. Let S be a D(¿/)-name for a subset of A. Suppose p lb S' C A. For each u < A, apply Theorem 3.4.4 to p at A + 1 for the formula rv G S'"1 and get a condition q(u) € Da+i (U). Since the q(u) 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 [ (A + 1 ),q) lb v G S’}. It follows from the definition that SG = j,Gf(A+1), which establishes the claim. But now notice that Da+i (W) has the A chain condition. Hence A must remain a cardinal in V[Gf(A + 1)], which is a contradiction. ■ 3.5 Genericitv Criteria Theorem 3.5.1. (Genericity Criteria) Let 6 C k. Then © is D)(¿/) generic if and only iffor all X C k, |6 \ X\ < N0 <í=^ bf G £{U). (3.1) Proof: The proof of this theorem will take up the whole section. Clearly, if G is D(¿/) generic, then © = U{s : (s, X) G G} satisfies (3.1). Suppose now that 19 6 satisfies (3.1). Let G[6] = {(s, X) : s C & and 6 \ s C X}. Let V be dense and open in D(¿/). We shall prove that V D (?[©] ^ 0. Definition 3.5.2. For any dense D, let D* = {(s, X) : 3fi G A \ max(s) {7 < p. : (s U {7},^) G D} G {/„}. 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 any t, if S is the least possible ordinal such that t 8 and (0, Q(D)) > (t U {¿}, X) e D for some X, then (t, Q(D)) € 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) € 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 G [k] 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 8 eventually contained in each Xt. Let £ = ^t<&K.Xt. Then for each t G [«] 8 \ 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. 20 We continue by defining measure one sets H each of which is homogeneous for getting a condition in D. For each /x G A, each t G [«] 0 if 3X (*11(7},*) e D, 1 otherwise. Let H* G Up be homogeneous for /* and maximal. Let Hp — AteM Thus, if 7,7' € Hp and <<7,7' then 3* (t U {7}, X) € D *=> 3X (t U {7'}, <¥) € 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 /x : [/c] For each t G [k] 4(t0 = ^u{7} n 77. Let W* G Up(t) be homogeneous for g*v. Let w* = \ Let j‘= U (-?n ^i)- -yew* 21 Each Jl is long measure one up to n(t). To get a long measure one set up to k, let Jl be Jx with the interval [/z(f),/c) adjoined. Let J = At We have, for any t such that W* is defined and for 7 € Wl, ,Ttu{7} D 3 H 7- (3-4) This will ensure that each Q(D) = £njn (J Hfi. liea We claim that Q(D) satisfies the lemma. For some í, 8, and X, we have that (0, Q(D)) > (t U {J}, A’) € D; pick t and 5 so that t <£ 5 and so that 8 is as small as possible. We claim that this condition is in D*. In fact, we show {7 : <* U {7}, Q(D)) € D] € UMt). Note that 5 G for some /i with = {0}. By minimality of 8, we have that H = n(t). Furthermore, (3.3) implies that {7 : (í U € D} 6 U^t)- By (3.2) and (3.4) we have that Q(D) is contained in each Xtu{i} above. Since D is open, we have that (t U {7}, Q(D)) € D for 7 in a measure one set. This concludes the proof of the lemma. ■ Let us define the dense sets to which we will apply Lemma 3.5.3. Definition 3.5.4. For each s G [«] Ds,o = {(s',X):(SUs',X)eV}, Ds,n+1 = Ds,n- 22 We will use these sets to get a condition that is in (?[©] D L>o,m for some m G u>. Let s0 = 6 \ Q(Do}q). Then (s0, Q{DQfl)) G G[6]. Suppose now that s,- = (s0,... , s,) has been For defined. notational simplicity, let us identify s,- with the set soU- • -Us,. Let Z>(Si) = r)j S,+i = 6 \ (2(Si) U Si). Then (si+1,Z(s,)) € G[6], where s,+i = (s0,...,s,+i). Claim 1: For each i < u> we have s,+i -C s¡. Proof: Denote S = max(st+i). By definition of s,, we have that S £ Z(s{). However, since Z(si) is a diagonal intersection, if max(s,) < S, then S G Z{sí). This impossibility establishes the claim. ■ Since our universe is well founded, infinite descending chains of membership are excluded. Hence for some k < u> we have that = 0. Claim 2: For some m < u>, we have (sk+i, Z(sk)) G A),m- Proof: For some {¿j,..., and some X, we have (0,Z(s*)) > ({¿i,...,¿m}, ({¿i,...,Jm_,},Z(sfc)) € Djk+Ui. Suppose that ({Í1,...,ám_í},2(5*)) G Dik+ui. Pick S'm_i the least possible such that for some X <0,Z(*.)) > ({S, e D,w,. Since Q(Dik+ui) C Z{sk), we may apply Lemma 3.5.3 and use the fact that is open to see that ({^l> • • • 1 ^m-(«'+l)}) 2{$k)) £ DSk+lti+l. Hence ($k+l U , • • • , &m—i— 1}) Z(sk)) G D0 This establishes the claim ■ 23 Finally, we claim that for some {71,... ,7m} we have that {$k+i U {71,... ,7m}, Z{sk)) € Do,o H G[©]. We proceed by induction on i < m. Suppose {sk+i U{7i,...,7i},Z(sk)) e 50,m-¡nG[6]. Select 7,+1 as follows. Since A),m-t = T*o,m-¿-i> there is a measure one set of elements 7 such that adding 7 to the support of the above condition gets the condition into Since © is unbounded in U^ for all n € A by hypothesis, we may select 7,+i from the aforementioned measure one set intersected with ©. The augmented condition will be in Do,m-i-i fl G[©]. This completes the induction and establishes the claim. We have found a condition in both V and G[©], and since V was an arbitrary dense open set it follows that G[©] is generic. This concludes the proof of Theorem 3.5.1. ■ CHAPTER 4 GETTING TWO NORMAL MEASURES 4.1 The Measures Assume that UK is a vertical repeat point. We show that after disassociated indiscernible forcing not only does k 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 ¿P for i € {0,1}. Let j : V -* 9JÍ = Ult(V, UK) be the ultrapower map via the normal measure UK. A condition (s,X) in the generic filter forces a set X to be in f/°, (ie. (s,X) Ib X G t/°) iff for some X' € £(j(l/)) with X' fl k = X we have (s, X') lb k € j(X). Similarly, (s, X) forces X to be in £/* iff for some X' € £(j{U)) with X' D k = X we have (s U {/c}, X') lb k € 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 U'K. 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 U'K 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 UK are measure one for UK (even though k cannot be measurable in any such ultrapower). Hence the long measure one part of q is measure one for UK. This is key. Since UK is a vertical repeat point, we have that q [ (k + 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 -b 1 of images of conditions 24 25 in D(£0 remain in D(¿/). Further, if UK 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 n-complete non¬ principle ultrafilters with & £ U® and 6 E £/|. Proof: We first need to show that the U\ are well-defined. That is, p Ib X E U\ does not depend on choice of name for X. Suppose p Ib X = Y and p Ib X E U'K. Then j(p) lb j(X) = j(Y), and if q < j(p) with q lb k E X, then q lb k E Y. So the definition is name-independent. Next, each U'K is a nonprinciple ultrafilter. We show that U\ is closed under enlargements; the proofs of closure under intersection, nonprincipality, and maximal¬ ly are similar. Suppose p Ib rX E U'K A X C Y~*. There is a witness q < j(p) of p forcing X in U'K. So q lb r« E j(X) A j(X) C j(K)n. It follows that q lb k E j{Y). Hence q is also a witness of p forcing V in U'K. Let us now show that U'K is /c-complete. Suppose for some q < k we have p lb U-/ the r“(7) are compatible, there is a condition r* < r*(7) for all 7 < r). Note that rK = j(p) r (* + 1) f°r 7 < »?, so that (j(p) f (/c + 1), rK) < (rK,r*(7)) for all 7 < 77. Now since (rK,rK) lh Uy 7' < 7, we have that q lb « € j(Xy>). Denote p' = <7 f (« + 1). Since UK is a vertical repeat point, p' € We claim that (p',r") is a witness to p' forcing Xy in U'K, where the i depends on whether or not « is in the support of q. It is clear from the fact that rK <* j(p) [ « that the support of (p',r*) is equal to the support of p'. Also j((p/,rK)) \ (« + 1) = j(p') \ (« + 1). And by definition of the (rK,rK) we have that (p', rK) lh « 6 (Ay). Thus (p',rK) is a witness as claimed. It follows that p' lh Xy € UlK so that C/‘ is «-complete. Finally, we show that the UlK are normal. Suppose p lh rf : « -» « A V7 < « f(~l) < 7n- We find a p' < p, a 7' < «, and a set X such that p' lh rX € U'K A f”X = 7,n. For every 7 < «, apply Theorem 3.4.4 to j(p) at « to get a pair (rK, r**(7)) € lD>«+1(j(ZY)) x D~(i(W)) such that rK = j(p) f (« + 1), r"(7) <* j(p) ( k, and such that if (a, 6) < (^,^(7)) with (a, b) deciding j(/)(«) = 7, then (a,r*(7)) decides j(/)(«) = 7 the same way. Since D* (,)(£/)) is «-closed, there is a condition r* < r"(7) for all 7 <7. Now, since j(p) > (r^r*) lh j(f) : j(«) —► j(«), there is some 7' < « and some 9 < (rK,rK) such that q lh j(/)(«) = 7'. Denote p' = q \ (« + 1), which is in ID>(W) since UK is a vertical repeat point. We claim that (p', r") is a witness to p1 forcing {7 < « : /(7) = 7'} in U'K. It is clear from the fact that r* <* j(p) [ « that the support of (p', r") is equal to the support of p'. Also j((p',rK)) f (« + 1) = j(p') f (« + 1). And by definition of the (rK,r*) we have that (p', rK) lh j(/)(«) = 7'. In other words, (p', r*) lh « 6 {7 < j(«) : j{f)(7) = 7'}. That (p',rK) is the witness as claimed follows by elementarity. ■ Theorem 4.1.3. The only normal ultrafilters on « in the generic extension are U° and U\. 27 Proof: Suppose p = (s, X) II- rX £ VP A VP is a normal measure on /c”1. We find a q < p such that q lb X € U'K for some i £ {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 = juK- We appeal to Theorem 3.4.4. For each u < k there is a q„ = (s, Qu) <* p [ v in W{U) such that if (a, b) < and (a, 6) £ H>v+i{U) x W(U) decides v £ X then (a,q„) decides v G X the same way. Let Q = A„ Now p' II- X G VP. Hence for any generic filter G with p' E G we have V[G/] |= k G j{X), where j(p') 6 G'. It follows that there is some q compatible with j(p) such that q II- k (E j(X). Denote q' = (q f (k + l),i(p#) [ «). Note that q' is also compatible with j(p'). We can assume that q' f k < p'. Then q' Ih X € Í7’, where . _ k 1 f 0 if G supt(^ f (k + 1)), \ 1 if K ^ supt(<7 f (k + 1)). By the elementarity of j and the definition of p', we have q1 Ih k € j(X). The claim, and hence the theorem, follows. ■ 4.2 Properties Theorem 4.2.1. In the generic extension, the following hold: 1. For each X € U® there is some Y € UK such that Y \ © C X. 2. For each X 6 U\ there is some Y (z UK such that Y fl 6 C X. Proof: Suppose p = (s,X) Ib X € 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 u < k, there are qv = (s \ [u + 1), Qu) £ ly (U) such that if (a,q') < (1 ,<7„), where (a,q') £ Ou+i(U) x W(U) such that (a, q') decides v £ X, 28 then (a, q„) decides v G X the same way that (a, q') does. In other words, q„ decides v G X above v. Let Q = AU Since q Ib X € U\, there is an r < j(q) such that r f (/c + 1) = j(q) \ (k + 1), r lh k G j(X), and supt(r) \ supt(^) is either {0} or {«}, depending on whether X is forced to be in U% or U\ by p. In the former case, let r° G D„(W) be the restriction of r to i/ + 1. In the latter case, let be the above restriction augmented by adding v to its support. Let Y¡ = {v < k : (r],,q v) Ib v G X}. Note that j(Y¡) = {u < j(/c) : (j(r)l„,j{q) [ v) lb v G Since j(rYK = r \ (k + 1) and (rf(/c + 1), j{q) [ k) lb k G j{X), we have that k G j(Y). It follows that Y G UK. Suppose now q' < q and either q' lb v G Y fl © or q' lb u G Y \ 0. Note that the above decides whether or not v G supt(g'). In either case, we have (rj,, q [ u) lb v G X by definition of Y. Since rj, = q \ (v + 1) > q' \ (u + 1), we have (q1 P {y + 1)) Q L u) lb v € X. Since q' [ v < q [ v, we have q' Ib u G X. The theorem follows. ■ Compare the following with Theorem 2.3.6. Theorem 4.2.2. Fori G {0,1}, consider the restriction of the ultrapower jui : L[U}[&) -4 L[U'][&) « Ult(L[^][6],C/‘) to the ground model ji : L[U] -> L[U'\. Then 1. ji is an iterated ultrapower along measures from U 2. every element of & \ © is a critical point of the iterated ultrapower. 29 Proof: The first point is established in [8]. Let us demonstrate the second. Suppose for a contradiction that v G ©' \ © is not a critical point in the iteration. Then for some function / : k —> /c, there is a 7 < v such that v — Thus v is in the set {77 < : f] £ ji/iC/)”7?}- In other words v G ju>K{Z), where Z = {rj < k : r¡ G Note that for all /x G A we have Z fl /x ^ Ull. Denote D = {(5,X fl Z) : (s,X) G D(¿/)}. Now D is dense, so G fl D ^ 0 for all generic G. Suppose p is a condition in the intersection. We have ju\{p) € ju^(G). But since p has finite support bounded below k, we have that u is not in the support of jt/;(p)- Neither is v in the long measure one set. Since ©' = juiK(&), we cannot have v G This contradiction establishes the theorem. ■ CHAPTER 5 ITERATED DISASSOCIATED INDISCERNIBLES 5.1 Definition Definition 5.1.1. We define functions /7 : k -» k for 7 < k+ by recursion. Let ^0(^7) — 0. Let — f-ii7!) "t 1* For limit ordinals let f^^Tl') Oi/<7)/y1/(»7)) where (7„)„<,c is onto 7. The definition of /7 for limit 7 is the diagonal union. Note that [f~,]uK = 7- That is, these functions are canonical representations of the ordinals less than k+ in the ultrapower by UK. The following is relative to our fixed sequence of measures U = (C/i4)M£A- Definition 5.1.2. We define vertical repeat point of order 7 < /c+ by recursion on 7. Let ord(UK) = 0 if for some X G UK, for all p < k we have X D p U^. Let ord(UK) > 7 if for every X € UK, the set {p < k : ord(Un) = /7(/i) and X D p € 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 < k+ 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 A7 = {p G A : ordlfU^) > f-,(p)}. For each 7 < /c, let W = (Ey„eA,. 30 31 Definition 5.1.4. Let I be an interval of ordinals. The I-iterated disassociated in¬ discernible forcing forU, denoted D(W, I), has as conditions sequences (s7, A7)7e/ such that 1. for all 7 6 /, we have s7 £ [«] 2. all but finitely many s7 are empty 3. for all 7 G /, we have X7 € £,(W) 4. if 7 / 7' then s7 fl sy = 0. A condition extends another, (s^, X!f)^i < (s7, A^)7e/, if for all 7 6 / we have (s^Xlf) < (s7,A’7) in the D(^7) order. The definition of direct extension is analogous to that for disassociated indis- cernible forcing. We sometimes use the notation (s, A) for conditions when the index is obvious. We’ll use the ‘adjoin’ operator to display the contents of s, regarding the partial function as a set of pairs: (*o,7o>)~(ti,7i>)~---~ (*n>7n). Assume from now on that max(A) = k, and that UK is a vertical repeat point of order A < K. 5.2 Basic Structure Definition v 5.2.1. Fix < k. Let p = (s7, X^)^\ € 0(¿/,A). Let 1. p \ (i/ + l) = (s7n(i^-|-l),A'7n(i/ + l))7 2. p [ v = (s7 \ (y + 1), X^ \ {v + 1))7 3. Di/+i(¿/,A) = A). 4. W(U, A) = {(s7,A'7)7'-|-l) = 0)}. 32 We do not disallow some of the long measure one entries being empty in P \ (*>+ !)• Proposition 5.2.2. For any v < k, we have that 0>(ZY, A) is isomorphic to the prod¬ uct D„+i (U, A) x IT (U, A) by the map pi-4 (p f (v + l),p [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 = (s7,A^)7€a G A). Fix ( < A. Denote 1. p-f-C = (s~n 2* C = {sm <^y)'y6A\C* Proposition 5.2.4. For any £ < \, we have that BfU, A) is isomorphic to the product B{U,C) x tyU,a \ C) by the map p (p-f- (,p+ 0- 5.3 Prikry Property Definition 5.3.1. For any set K, fix some bijection e : k K. Let (X,),e/<- be given. Let the e-diagonal intersection be AeX{ = {7 < k : Vi/ < 7(7 € Xe(„))}. Proposition 5.3.2. Let be given such that each € £(£/). Let X = AeX¿. 1. For each i we have that X \ (e_1(i) + 1) C Xi. 2. X G £(W)- Definition 5.3.3. If s is a set of pairs, denote sups = max{í,j : (i,j) G s}. The relation means that s', t are either ordinals or finite partial functions from k to [«]<" and that sup(s) < sup(t). Theorem 5.3.4. Let a be a formula in the language ofD(U,\) and let (s1,X^)v 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 : k —> [/c x k] e”( = [C x C] that each sv = 0 to simplify the notation. Let <7 and (0, X) be given. For each fx 6 A, s, and i < A let /* ,■ : ^ > 3 via 0 if 3X (s X) lb cr, /m,¿(7) = < 1 if 3X (s~(z,7),X) lb -><7, k 2 otherwise. Let f/*, € U^ be homogeneous for Let = AgH^¿. For any s, any 7,7' € Hpj \ (sup s + 1), we have that 3X 7), X) || a if and only if 3X (s ~(t, -y'),X) || a, and these conditions decide <7 the same way. For each i < A, let //, = U^a,#/*,.'• Let Z' = (Hi fl We claim that a direct extension of (0, Z') decides <7. Suppose not, and we shall reach a contradiction. Fix (s ~(j, 6), Z") < (0, Z') with \s ~(j, i)| the least out of all conditions deciding <7, where s<(S. Without loss of generality, assume (s ~ (j,S), Z") lb <7. Let fx be such that S € H^j. From now on, s,/x,S and j are fixed. For each pair (t, 7) where i < 7 < fx, if there is some X such that (s ~(i, 7), X) decides <7, then let X^ be some such X. Let X = (Ae^^X^'^)k<\. For each k < A, let Yk = {u < \x : {7 < /¿ : v £ Xjf’^} G £/,,} and let HI = {7 < n : Xjf'^ n 7 = Yk D 7}. Then each H£ G £/M. Let i/* = Ak<\H¿ G £/„. Then for any 7 < /z we have xjf’^ fl 7 = L* fl 7 for all fc < 7. Let Q* = ((Fit U H*) U («Tjt \ /¿). We will show that (s, Q) forces <7. Let (t,7l) < (s, Q). Without loss of generality, t contains a pair (j, £) such that 34 £ € H*. Consider the least such £. We claim (t, (TZk fl X¡f’^)k In fact, let (k, 7) € t. If 7 > £ then 7 E Qk\ (e-1((j, £)) + 1) C A’*7’^. Suppose 7 < £. Since £ E H“ and k < *y <( we have that A^’^ D £ = V* fl , so (k, 7) € Xjf’®. Either way t \ s comes from X^Á). This establishes the claim. But (s forces a. Hence (t,(7lk fl Xk'^)k<\), itself an extension of an arbitrary extension of (s, Q), forces cr. It follows that (s, Q) forces a. This contradicts the minimality of l-s ~(j, i)|, finishing the proof. ■ 5.4 More Structure Theorem 5.4.1. Fix u < k. Let a be a formula in the language of X) x W(U, A). Let (p,q) E A) x D>"(£/, A). There is a q' <* q such that for any (p'iQ") < ÍPiQ') such that (p',q") decides cr, we have that (p',q') decides a. Proof: The proof is virtually the same as the proof of Theorem 3.4.4. ■ Theorem 5.4.2. Iterated disassociated indiscernible forcing preserves cardinals. Proof: The obvious alteration of Theorem 3.4.5 works. ■ CHAPTER 6 GETTING MANY NORMAL MEASURES Assume that « 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 rj < A. Let j be the ultrapower map via the normal measure UK. A condition (s7,A’7)7