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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. 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 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 Kcomplete ultrafilter on some cardinal K. This cor responds to the sets of positive measure for a twovalued 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. 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 socalled 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. 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 Kcomplete nonprinciple ultrafilter on a cardinal K. If we remove the translationinvariance requirement from Lebesgue measure and restrict to two values instead of nonnegative 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 wellfounded models of Set Theory are absolutely wellfounded. 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 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 + 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 forcingsyntactic 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 almostcontainment 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 firstorder 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 firstorder indiscernibles over 9M,. 2. 6 is P(U,)generic over 9M,. Although Prikry sequences do not in general consist of firstorder 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 parameterfree 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 parameterfree 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 ordering 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 firstorder indiscernibles for this model. In particular, no Prikry generic sequence can consist of firstorder 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. CHAPTER 3 THE FORCING 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. Proof: 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 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] 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', respectively. 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") does.) 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 2kclosed, 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]] 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,. Let J'= U (yn Xtu,}). yEWt 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,. I6EA 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. Let 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,...,6mi}, 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. Hence (Sk+1 U {1, ., ,mi},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,mi n G[G]. Select 'Yi+1 as follows. Since Do,mi = D*,mi_, there is a measure one set of elements  such that adding 7 to the support of the above condition gets the condition into Do,mi1. 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,mii 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. * 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 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 Kcomplete non principle ultrafilters with 6 U, and 6 E U,. Proof: We first need to show that the Ui are welldefined. 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 nameindependent. 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 Kcomplete. 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 Kcomplete. 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 Kclosed, 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, 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']. Then 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 CHAPTER 5 ITERATED DISASSOCIATED INDISCERNIBLES 5.1 Definition Definition 5.1.1. We define functions fy : It + r for 7 < r+ by recursion. Let fo(r7) = 0. Let fy+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 Iiterated 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, XY)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 ediagonal 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 \ (e1(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 [c] 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) IF0, 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, 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 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 CHAPTER 6 GETTING MANY NORMAL MEASURES Assume that K is a vertical repeat point of order A < K. After Aiterated 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. * REFERENCES [1] Thomas Jech Set Theory, Pure and Applied Mathematics v. 79, Academic Press, 1978. [2] Akihiro Kanamori The Higher Infinite, Perspectives in Mathematical Logic, SpringerVerlag, 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.359377. [5] Menachem Magidor, How large is the first strongly compact cardinal? or A study on identity crises, Ann. Math. Logic, 1976, pp.3357. [6] A. R. D. Mathias, On sequences generic in the sense of Prikry, J. Australian Math. Soc., 15 1973, pp. 409414. [7] William J. Mitchell, Sets constructible from sequences of ultrafilters, J. Symbolic Logic, 1974, pp.5766. [8] William J. Mitchell, The core model for sequences of measures. I, Math. Proc. Camb. Phil. Soc., 1984, pp.229260. [9] Karl Prikry Changing measurable cardinals into accessible cardinals, Disserta tiones Math., 1970, pp.552. BIOGRAPHICAL SKETCH 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 Florida. 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 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 Philosophy. Douglaaenzer 0 Professor of Mathematics 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 Philosophy. Brandon Kershner Professor of English 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 Philosophy. Jean Larson Professor of Mathematics 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 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 .L437 UNIVERSITY OF FLORIDA 1 3 1262 08555 111 29II8ll I 11 3 1262 08555 1298 