THE FIRST MEASURABLE CARDINAL CAN BE THE FIRST UNCOUNTABLE REGULAR CARDINAL AT ANY SUCCESSOR HEIGHT

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1 THE FIRST MEASURABLE CARDINAL CAN BE THE FIRST UNCOUNTABLE REGULAR CARDINAL AT ANY SUCCESSOR HEIGHT ARTHUR W. APTER, IOANNA M. DIMITRÍOU, AND PETER KOEPKE Abstract. We use techniques due to Moti Gitik to construct models in which for an arbitrary ordinal ρ, ℵ ρ+1 is both the least measurable and least regular uncountable cardinal. 1. Introduction In this paper, we study cardinal patterns where the least uncountable regular cardinal is the least measurable cardinal. Jech [Jec68] and Takeuti [Tak70] independently showed that if we assume the consistency of ZFC+ There is a measurable cardinal, then the theory ZF+DC+ ω 1 is the least measurable cardinal is consistent, and this is an equiconsistency. We can also ask whether it is consistent for the least uncountable regular cardinal κ to be the least measurable cardinal and also be such that κ > ω 1. The first author has proved that this is indeed the case. In particular, it follows from the work of [Apt96] that relative to the consistency of AD, it is consistent for ℵ 2 to be both the least measurable and least regular uncountable cardinal. The methods used in the proof of this result may be extended to show that relative to the consistency of AD, it is possible to obtain models in which, e.g., ℵ ω+1 is both the least regular uncountable and least measurable cardinal, or ℵ ω+2 is both the least regular uncountable and least measurable cardinal, or there is an uncountable limit cardinal which is both the least regular and least measurable cardinal. However, these methods do not easily extend to cardinals such as ℵ 3 or ℵ ω+3. In this paper we will prove the following theorem, which allows us to handle cardinals different from those provided by AD. Theorem 1.1. Let V = ZFC + ρ > 0 is an ordinal + There is a sequence κ ξ ; ξ < ρ of strongly compact cardinals such that each limit point of the sequence κ ξ ; ξ < ρ is singular, and with a measurable cardinal κ ρ above the supremum of the sequence. There is then a partial ordering in the ground model V and a symmetric model V (G) of the theory ZF + For each 1 β ρ, ℵ β is singular + ℵ ρ+1 is a measurable cardinal carrying a normal measure. Theorem 1.1 generalises earlier work of the first author (different from that found in [Apt96]). Specifically, the first author, in [Apt85] and with Henle in [AH91] and Date: May 23, The authors were partially supported by the DFG-NWO cooperation grant KO 1353/5-1 Infinitary combinatorics without the Axiom of Choice, and the Hausdorff Center for Mathematics of the University of Bonn. The first author was also partially supported by various PSC - CUNY grants. The first author wishes to thank the members of the Bonn Logic Group, and especially his coauthors, for all of the hospitality shown to him during his visits to Bonn. The contents of this paper also appeared as part of the second author s PhD thesis (see [Dim11]). We thank the referee for the conscientious and constructive reading and very helpful comments of several versions of this paper. 1

2 2 ARTHUR W. APTER, IOANNA M. DIMITRÍOU, AND PETER KOEPKE Magidor in [AM95], showed how to symmetrically collapse a measurable cardinal κ to be the successor of a singular cardinal of cofinality ω while preserving the measurability of κ. The large cardinal hypotheses used (instances of supercompactness in [Apt85] and [AM95] and instances of strong compactness in [AH91]) are considerably stronger than the existence of one measurable cardinal. Note that the standard proof for the existence of a normal measure for a measurable cardinal requires the use of DC. Since ℵ 1 has cofinality ω in V (G), both DC and AC ω are false in this model. It is therefore especially relevant that ℵ ρ+1 carries a normal measure in V (G). The forcing mentioned in the theorem above is a modification of Gitik s forcing in [Git80]. In addition to what Gitik proved about this forcing, we prove that in our version, none of the strongly compact cardinals collapses in the symmetric model V (G), but all other regular uncountable cardinals below κ ρ do. We modify Gitik s construction in several ways. This is because Gitik s (class sized) forcing does not focus on exactly which cardinals are preserved in his symmetric model N G, but rather on preserving the power set axiom. In particular, the remarks found in [Git80, page 62, paragraph immediately following Theorem II] mention that the (well-ordered) cardinals of N G are ω, the ground model strongly compact cardinals, and the (singular) limits of the ground model strongly compact cardinals. However, other than [Git80, Lemma 3.4], which indicates that every limit ordinal has cofinality ω in N G, there is neither an explicit proof nor hint at any point in [Git80] as to how to determine the cardinal structure of N G. By carefully reworking the definition of Gitik s forcing conditions in the context of set forcing, and by providing a detailed analysis of the nature of our partial ordering, we are able to determine precisely (see Corollary 2.9) the relevant cardinal structure of our symmetric inner model V (G). To collapse the intervals between the strongly compacts, we will choose fine ultrafilters over the P κξ (α) for each α (κ ξ, κ ξ+1 ) in order to make parts of the forcing be isomorphic with strongly compact Prikry forcings. The isomorphisms will ensure the relevant collapses. To ensure that an α (ω, κ 0 ) has collapsed we will use a fine ultrafilter over P ω (α) and the same proof. If there are any limit ordinals λ < ρ (e.g., if ρ = ω + 1) then similar arguments using fine ultrafilters will collapse the cardinals in the open interval ( ξ<λ κ ξ, κ λ ). Forcing at each κ ξ will be done with a κ ξ -complete measure over κ ξ. Finally, a small forcing argument guarantees that κ ρ remains measurable in the symmetric forcing extension and carries a normal measure. 2. The Gitik construction For our construction, we assume knowledge of forcing as presented in [Kun80, Ch. VII], [Jec03, Ch. 14], and of symmetric submodels as presented in [Jec03, Ch.15]. Let ρ Ord. We start with an increasing sequence of cardinals, κ ξ ; ξ < ρ, such that for every ξ < ρ, κ ξ is strongly compact, and such that the sequence has no regular limits. Let κ ρ > ξ<ρ κ ξ be a measurable cardinal. We will construct a model with a sequence of ρ + 1-many successive singular cardinals in which κ ρ is the first regular uncountable cardinal, and it is still measurable. We will do this by modifying and adapting Gitik s construction of [Git80], whose notation and terminology we freely use.

3 THE FIRST MEASURABLE CAN BE THE FIRST UNCOUNTABLE REGULAR 3 Our construction is a finite support product of Prikry-like forcings which are interweaved in order to prove a Prikry-like lemma for that part of the forcing. Call Reg κρ the set of regular cardinals α [ω, κ ρ ) in V. For convenience call ω =: κ 1. For an α Reg κρ we define a cf α to distinguish between the following categories. (type 1) This occurs if there is a largest κ ξ α (i.e., α [κ ξ, κ ξ+1 )). We then define cf α := α. If α = κ ξ and ξ 1, then let Φ κξ be a measure for κ ξ. If α = ω then let Φ ω be any uniform ultrafilter on ω. If α > κ ξ is inaccessible, then let H α be a κ ξ -complete fine ultrafilter over P κξ (α) and let h α : P κξ (α) α be a bijection. Define Φ α := {X α ; h 1 α X H α }. This is a uniform κ ξ -complete ultrafilter over α. If α > κ ξ is not inaccessible, then let Φ α be any κ ξ -complete uniform ultrafilter over α. (type 2) This occurs if there is no largest strongly compact α. We then let β be the largest (singular) limit of strongly compacts α. Define cf α := cfβ. Let κ α ν ; ν < cf α be a fixed ascending sequence of strongly compacts cf α such that β = {κ α ν ; ν < cf α}. If α is inaccessible, then for each ν < cf α, let H α,ν be a fine ultrafilter over P κ α ν (α) and h α,ν : P κν (α) α a bijection. Define Φ α,ν := {X α ; h 1 α X H α,ν }. Again, Φ α,ν is a κ α ν -complete uniform ultrafilter over α. If α is not inaccessible, then for each ν < cf α, let Φ α,ν be any κ α ν - complete uniform ultrafilter over α. This cf α will be used when we want to organise the choice of ultrafilters for the type 2 cardinals. We use the fine ultrafilters H α and H α,ν to make sure that in the end only the strongly compacts and their singular limits remain cardinals below κ ρ. For type 1 ordinals we will do some tree-prikry like forcings to singularise in cofinality ω. Type 1 cardinals in the open intervals (κ ξ, κ ξ+1 ) will be collapsed to κ ξ because enough of these forcings will be isomorphic to strongly compact Prikry forcings (or fake strongly compact Prikry forcings in the case of ξ = 1). This is why we use fine ultrafilters for these cardinals. To singularise type 2 ordinals Gitik used a technique he credits in [Git80] to Magidor, a Prikry-type forcing that relies on the countable cofinal sequence c that we build for cf α to pick a countable sequence of ultrafilters Φ c(n) ; n ω. To show that the type 2 cardinals are collapsed, we use again the fine ultrafilters. As usual with Prikry-type forcings, one has to prove a Prikry-like lemma (see [Git80, Lemma 5.1]). For the arguments one requires the forcing conditions to grow nicely. These conditions can be viewed as trees. These trees will grow from left to right in order to ensure that a type 2 cardinal α will have the necessary information from

4 4 ARTHUR W. APTER, IOANNA M. DIMITRÍOU, AND PETER KOEPKE the Prikry sequence 1 at stage cf α. Let us take a look at the definition of the stems of the Prikry sequences to be added. Definition 2.1. For t Reg κρ ω κ ρ we define the sets dom(t) :={α Reg κρ ; m ω γ Ord((α, m, γ) t)}, and dom 2 (t) :={(α, m) Reg κρ ω ; γ Ord((α, m, γ) t)}. Let P 1 be the set of all finite subsets t of Reg κρ ω κ ρ, such that for every α dom(t), t(α) := {(m, γ) ; (α, m, γ) t} is an injective function from some finite subset of ω into α. To add a Prikry sequence to a type 2 cardinal α, we want to have some information on the Prikry sequence of the cardinal cf α. We also want to make sure that these stems are appropriately ordered for the induction in the proof of the aforementioned Prikry-like lemma. So we define the following. Definition 2.2. Let P 2 be the set of all t P 1 such that the following hold. (1) For every α dom(t), cf α dom(t) and dom(t(cf α)) dom(t(α)). (2) If {α 0,..., α n 1 } is an increasing enumeration of dom(t) \ κ 0, then there are m, j ω, such that m 1, j n 1 with the properties that for every k < j we have that dom(t(α k )) = m + 1 and for every k {j,..., n 1} we have that dom(t(α k )) = m. These m and α j are unique for t and are denoted by m(t) := m and α(t) := α j. (3) If cf α(t) < α(t) then m(t) dom(t(cf α(t))). We may think of the point (α(t), m(t)) as the point we have to fill in next, in order to extend t. As we will see in the next definition, the value of t(cf α(t))(m(t)) will decide how the condition t will grow at the point (α(t), m(t)). Let us call elements of P 2 stems. In the following image we can see roughly what a stem t with a domain {α 0, α 1, α 2 } above κ 0 looks like. In order to add Prikry sequences, we will use the ultrafilters and define the partial ordering with which we will force. Definition 2.3. Let P 3 be the set of pairs (t, T ) such that (1) t P 2, (2) T P 2, (3) t T, (4) for every t T we have t t or t t, and dom(t ) = dom(t), (5) for every t T, if t = r {(α(r), m(r), β)} then t := r T, i.e., T is tree-like, 1 For the rest of this paper, we will abuse terminology by using phrases like Prikry sequence and Prikry forcing when referring to our Prikry-like forcing notions.

5 THE FIRST MEASURABLE CAN BE THE FIRST UNCOUNTABLE REGULAR 5 (6) for every t T with t t, if α(t ) is of type 1 (i.e., cf (α(t )) = α(t )) then Suc T (t ) := {β ; t {(α(t ), m(t ), β)} T } Φ α(t ), and (7) for every t T with t t, if α(t ) is of type 2 (i.e., cf α(t ) < α(t )) and m(t ) dom(t (cf (α(t )))) then Suc T (t ) := {β ; t {(α(t ), m(t ), β)} T } Φ α(t ),t (cf α(t ))(m(t )). For a (t, T ) in P 3 and a subset x Reg κρ we write T x for {t x ; t T }. We call t the trunk of (t, T ). This P 3 is the forcing we are going to use. It is partially ordered by (t, T ) (s, S) : dom(t) dom(s) and T dom(s) S. A full generic extension via this P 3 adds too many subsets of ordinals and makes every ordinal in the interval (ω, κ ρ ) countable. By restricting to sets of ordinals which can be approximated with finite domains we ensure that the former strongly compacts are still cardinals in the symmetric model to be constructed, and the power sets stay small. To build a symmetric model we need an automorphism group of the complete Boolean algebra B = B(P 3 ) that is induced by P 3, as in [Jec03, Corollary 14.12]. We start by considering G to be the group of permutations of Reg κρ ω κ ρ whose elements a satisfy the following properties. For every α Reg κρ there is a permutation a α of α that moves only finitely many elements of α, and is such that for each n ω and each β α, a(α, n, β) = (α, n, a α (β)). The finite subset of α that a α moves, we denote by supp(a α ), which stands for support of a α. For only finitely many α Reg κρ is a α not the identity. This finite subset of Reg κρ we denote by dom(a). We extend G to P 3 as follows. For a G and (t, T ) P 3, define a(t, T ) := (a t, {a t ; t T }), where a t := {(α, n, a α (β)) ; (α, n, β) t}. Unfortunately, in general a(t, T ) is not a member of P 3 because of the branching condition at type 2 cardinals. In particular, it is possible that for some α dom(t) of type 2, and some t T with α = α(t ), we have that a cf α(t (cf α)(m(t ))) = γ t (cf α)(m(t )), and even though we had before Suc T (t ) Φ α,t (cf α)(m(t )), it is not true that Suc T (t ) Φ α,γ. To overcome this problem, for an a G, define P a P 3 as follows. (t, T ) is in P a iff the following hold: (1) dom(t) dom(a), (2) for every α dom(t) we have that dom(t(α)) = dom(t(cf α)), and (3) for every α dom(t), we have that rng(t(α)) {β supp(a α ) ; q T (β rng(q(α)))}. The equality in (2) ensures that there will be no severe mixup in the requirements for membership in ultrafilters of the form Φ α,γ. In (3) we require that the stem of each condition already contains all the ordinals that the a α could move. This will prevent any trouble with membership in the ultrafilters. One may think that this requirement should be supp(a α ) rng(t(α)) but this is not the case; note that

6 6 ARTHUR W. APTER, IOANNA M. DIMITRÍOU, AND PETER KOEPKE there might be some γ in a α which doesn t appear in the range of any q T. Now, we have that a : P a P a is an automorphism. Also, as mentioned in [Git80, page 68], for every a G the set P a is a dense subset of P 3. Therefore, a can be extended to a unique automorphism of the complete Boolean algebra B. We denote the automorphism of B with the same letter, and also by G the automorphism group of B that consists of these extended automorphisms. By [Jec03, (14.36)], every automorphism a of B induces an automorphism of the Boolean valued model V B. Proceeding to the definition of our symmetric model, for every e Reg κρ define E e := {(t, T ) P 3 ; dom(t) e}, I := {E e ; e Reg κρ is finite and closed under cf }, fixe e := {a G ; α e(a α is the identity on α)}, and let F be the normal filter (see [Jec03, (15.34)]) over G that is generated by {fixe e ; E e I}. For each ẋ in the Boolean valued model V B, define its symmetry group sym(ẋ) := {a G ; a(ẋ) = ẋ}. A name ẋ is called symmetric iff its symmetry group is in the filter F. The class of hereditarily symmetric names HS is defined by recursion on the rank of the name, i.e., HS := {ẋ V G ; ẏ tc dom (ẋ)(sym(ẏ) F)}, where tc dom (ẋ) is defined as the union of all x n, which are defined recursively by x 0 := {ẋ} and x n+1 := {dom(ẏ) ; ẏ x n }. We will say that an E e I supports a name ẋ HS if fixe e sym(ẋ). For some V -generic ultrafilter G on B we define the symmetric model V (G) := {ẋ G ; ẋ HS}. By [Jec03, Lemma 15.51], this is a model of ZF, and V V (G) V [G]. For each (t, T ) P 3 and each E e I, define (t, T ) E e = (t e, {t e ; t T }). According to [Git80, Lemma 3.3.], if φ is a formula with n free variables, ẋ 1,..., ẋ n HS, and E e I is such that sym(ẋ 1 ),..., sym(ẋ n ) fixe e then we have that for every (t, T ) P 3 (t, T ) φ(ẋ 1,..., ẋ n ) (t, T ) E e φ(ẋ 1,..., ẋ n ). This implies the following lemma, which we will refer to as the approximation lemma. Lemma 2.4. If X V (G) is a set of ordinals, then there is an E e I such that X V [G E e ], where G E e := {(t, T ) E e ; (t, T ) G}.

7 THE FIRST MEASURABLE CAN BE THE FIRST UNCOUNTABLE REGULAR 7 Proof. Because of [Git80, Lemma 3.3.] mentioned above, because of the symmetry lemma [Jec03, (15.41)], and because canonical names ˇα are not moved by automorphisms of B, we have that if Ẋ HS is a P 3 -name for X and E e I supports Ẋ then the set Ẍ := {(ˇα, (t, T ) E e ) ; (t, T ) ˇα Ẋ} is an E e -name for X. We will use the approximation lemma in all our subsequent proofs. Theorem 2.5. For every 0 ξ ρ, κ ξ is a cardinal in V (G). Consequently, their (singular) limits are also preserved. The proof proceeds by construing a finite support part of P 3 as a two-step iterated forcing E Q, where the first component is small forcing relative to κ ξ, and the second component satisfies a Prikry lemma and does not add bounded subsets to κ ξ. This corresponds to the intuition behind the definition of P 3. The details of this proof, however, are largely technical, and in order to highlight the structure of this proof, we relegate them to the appendix. Proof. Assume towards a contradiction that there is some δ < κ ξ and a bijection f : δ κ ξ in V (G). Let f be a name for f with support E e I. Note that e is a finite subset of Reg κρ that is closed under cf. By the approximation lemma (Lemma 2.4), since f may be coded by a set of ordinals, there is an E e -name for f, i.e., for this e, f V [G E e ]. We will show that this is impossible, by taking a dense subset of E e and showing that it is forcing equivalent to an iterated forcing construction, the first part of which has cardinality less than κ ξ, and the second part of which does not collapse κ ξ (by not adding bounded subsets to κ ξ, similarly to Prikry forcing). It s not hard to check that J := {(t, T ) E e ; q T α κ ξ n < ω( if (α, n) dom 2 (q) \ dom 2 (t) and cf α < α then the ultrafilter Φ α,q(cf α)(n) is κ ξ -complete)} is dense in E e (see also Lemma A.1 in the appendix and the beginning of the proof of [Git80, Theorem 5.4]). Without loss of generality assume that e κ ξ. Define the sets E :={(t, T ) E e κξ ; (t, T ) J}, and P 2 :={t (e \ κ ξ ) ; t P 2 }. For s P 2 we can define α(s) and m(s) as we did for the s P 2, in Definition 2.2(2). Let G be an arbitrary E-generic filter and note that for every α e \ κ ξ such that cf α < κ ξ, the set t(cf α)(m) ; T ((t, T ) G m dom(t(cf α))) is the Prikry sequence that is added to cf α by E. In V [G ] we define a partial ordering Q by (s, S) Q : (1) s P 2, (2) S P 2, (3) s S, (4) for all s S, dom(s ) = dom(s) = e \ κ ξ, and either s s or s s, (5) for every s S and every s P 2, if s s then s S, i.e., S is tree-like,

8 8 ARTHUR W. APTER, IOANNA M. DIMITRÍOU, AND PETER KOEPKE (6) for every s S with s s, if α(s ) is of type 1 then {β ; s {(α(s ), m(s ), β)} S} Φ α(s ), (7) for every s S with s s, if α(s ) is of type 2 and cf α(s ) κ ξ then {β ; s {(α(s ), m(s ), β)} S} Φ α(s ),s (cf α(s ))(m(s )), and (8) for every s S with s s, if α(s ) is of type 2 and cf α(s ) < κ ξ then {β ; s {(α(s ), m(s ), β)} S} Φ α(s ), G (cf α(s ))(m(s )). Q is partially ordered by (s, S) Q (r, R) iff S R. This definition means that Q is like P 3 but restricted to ordinals and ultrafilters at and above κ ξ. Because of this we can get a canonical E-name Q for Q and that J densely embeds into E Q (Lemma A.2 in the appendix). Therefore, Q can be seen as the top part of the forcing E e, cut at κ ξ. For the rest of the proof we will work with Q inside V [G ]. At this point we have that the presumed collapsing function f : δ κ ξ is in some generic extension V [G E e ], that E e is forcing equivalent to J, which in turn is forcing equivalent to E Q. We know that E is too small to add a function like f, so f must be added by Q. To derive the desired contradiction, it remains to show that Q cannot add a function like f. To show this, as is standard with Prikry-like forcing notions, we have, proved as Lemma A.3 in the appendix: The Prikry lemma for Q. In V [G ], let τ 1,..., τ k be Q-names, and φ be a formula with k free variables. Then for every forcing condition (s, S) Q there is a stronger condition (s, W ) Q with the same trunk which decides φ(τ 1,..., τ k ). The proof of this lemma resembles the proof of Gitik s Prikry-like lemma in [Git80, Lemma 5.1], but with an application of the Lévy-Solovay theorem [LS67] in the usual way. That is, using the fact that E is small forcing with respect to κ ξ, we get that all ultrafilters involved in the definition of Q can be extended to κ ξ -complete ultrafilters in V [G ]. Then every time we have to intersect conditions in Q, we get a pseudo-condition with splitting sets in the extended ultrafilters. But by the definition of these extended ultrafilters we can always find subsets in the original ultrafilters of the ground model, and thus get a stronger condition that is in Q. Using the Prikry lemma for Q, we get, using the standard Prikry-style arguments, the κ ξ -completeness of the ultrafilters in the definition of Q, and the usual application of the Lévy-Solovay theorem, that Q does not add bounded subsets to κ ξ (Lemma A.4 in the appendix). Therefore E e cannot collapse κ ξ, and so the collapsing function f : δ κ ξ cannot exist in V [G E e ]. This completes the proof of Theorem 2.5. Next we will see that we singularised the targeted ordinals. This is similar to [Git80, Lemma 3.4]. Lemma 2.6. Every cardinal in (Reg κρ ) V has cofinality ω in V (G). Thus every cardinal in the interval (ω, κ ρ ) is singular. Proof. Let α Reg κρ. For every β < α, the set D β := {(t, T ) P 3 ; n < ω(t(α)(n) β)} is dense in (P 3, ). Hence f α := {t(α) ; (t, T ) G} is a function from ω onto an unbounded subset of α. This function has a symmetric name, which is supported

9 THE FIRST MEASURABLE CAN BE THE FIRST UNCOUNTABLE REGULAR 9 by E a, where a is the smallest subset of Reg κρ that contains α and is closed under cf. Therefore f α V (G). Usually in symmetric models built from ZFC-models with large cardinals, there is some combinatorial residue from the large cardinal. Such is the case also here. First we will show that in the interval (ω, κ ρ ), only the former strongly compact cardinals and their (singular) limits remain cardinals, i.e., that all cardinals of V that are between the κ ξ and their (singular) limits have collapsed. Lemma 2.7. For every ordinal ξ [ 1, ρ) and every α (κ ξ, κ ξ+1 ), ( α = κ ξ ) V (G). Proof. Fix an ordinal ξ [ 1, ρ). Since strongly compact cardinals are limits of strongly inaccessible cardinals, it suffices to show that for every strongly inaccessible α (κ ξ, κ ξ+1 ), we have that ( α = κ ξ ) V (G). Fix α (κ ξ, κ ξ+1 ) strongly inaccessible. We have a bijection h α : P κξ (α) α (see the definition of type 1 ordinals). We will use this bijection to show that E {α} is isomorphic to the following partial ordering. Let P s α be the forcing that consists of all injective H α -trees, i.e., of sets T such that T consists of finite injective sequences of elements of P κξ (α), (T, ) is a tree, where denotes end extension, T has a trunk tr T, i.e., an element such that for every t T, either t tr T or tr T t, and for every t T such that t tr T, the set {x P κξ (α) ; t x T } of the -successors of t in T, is in the ultrafilter H α. The forcing P s α is partially ordered by S T S T. Towards the isomorphism, define a function f from the injective finite sequences of elements of P κξ (α) to P 2 by f(t) := {(α, m, β) ; m dom(t) β = h α (t(m))}. Define another function i : P s α E {α} by i(t ) := (f(tr T ), {f(t) ; t T t tr T }). This i is indeed a function from T to E {α} because h α is a bijection. In fact, this i is a bijection itself. It is easy to see that it also preserves the relation of the forcings, so P s α and E {α} are isomorphic. Now let Ĝ be P α-generic. s Because H α is fine we have that for every β < α the set D β := {T Pα s ; m ω(β tr T (m))} is dense in P 3. So α = ( Ĝ)(n). n<ω But each ( Ĝ)(n) is in P κξ (α), hence it has cardinality less than κ ξ < α. Therefore in any forcing extension of V via Pα, s α has become a countable union of sets of cardinality less than κ ξ and therefore is collapsed to κ ξ. So there is an E {α} -name for a collapsing function from κ ξ to α, which can be seen as a P 3 -name in HS for such a function, supported by E {α}. Next we show that the regular cardinals of type 2 have collapsed to the singular limit of strongly compacts below them.

10 10 ARTHUR W. APTER, IOANNA M. DIMITRÍOU, AND PETER KOEPKE Lemma 2.8. For every α of type 2, if β is the largest limit of strongly compacts below α, then ( α = β) V (G). Proof. Similarly to the proof of the previous lemma, we assume that α is inaccessible and we look at each of the bijections h α,ν : P κ α ν (α) α. Let e be the smallest finite subset of Reg κρ that contains α and is closed under cf. Look at V [G E e ]. Let γ i ; i ω be the Prikry sequence added to cf α, and let α i ; i ω be the Prikry sequence added to α. For each i ω, let A i := h 1 α,γ i (α i ). We want to show that for each δ α, there is some i ω such that δ A i. Fix δ α. For all i ω, the V -ultrafilter H α,γi is fine, so So for every i ω, the set Define the set {A P κ α γi (α) ; δ A} H α,γi. Z i := {ζ α ; δ h 1 α,γ i (ζ)} Φ α,γi. D δ := {(t, T ) E e ; i dom(t)(δ h 1 α,γ i (t(α)(i)))}. This is dense in E e and δ was arbitrary. Therefore in V [G E e ], we have that α = i ω A i is a countable union of β-sized sets, and thus there is a symmetric name for a collapse of α to β, supported by E e. We summarise our results on the cardinal structure of the interval (ω, κ ρ ). Corollary 2.9. An uncountable cardinal of V (G) that is less than or equal to κ ρ is a successor cardinal in V (G) iff it is in {κ ξ ; ξ ρ}. Thus in V (G), for every ξ ρ we have that κ ξ = ℵ ξ+1. Also, an uncountable cardinal of V (G) that is less than or equal to κ ρ is a limit cardinal in V (G) iff it is a limit in the sequence κ α ; α < ρ in V. Proof. This follows inductively, using Theorem 2.5, Lemma 2.7, and Lemma 2.8. Before we go into the combinatorial properties in V (G), let us mention that the Axiom of Choice fails really badly in this model. The following is [Git80, Theorem 6.3]. Lemma In V (G), countable unions of countable sets are not necessarily countable. In particular, every set in H κρ is a countable union of sets of smaller cardinality. Here x has a smaller cardinality than y means that x is a subset of y and there is no bijection between them. 3. Results We will now prove our main result, Theorem 1.1. We will use the approximation lemma for V (G) and the Lévy-Solovay theorem [LS67], which says that measurability is preserved under small forcing. In particular, it says that if U is a normal measure over κ ρ, then the set generated from U by taking supersets is still a normal measure after small forcing. Proof. By Corollary 2.9, we only need to show that the measurability of κ ρ and the existence of a normal measure over κ ρ is preserved to V (G). We will prove that if U is a normal measure for κ ρ in the ground model, then the following set defined in V (G), U := {Y κ ρ ; X U(X Y )},

11 THE FIRST MEASURABLE CAN BE THE FIRST UNCOUNTABLE REGULAR 11 is a normal measure over κ ρ in V (G). This U is clearly a filter in V (G), so it remains to show that it is also a κ ρ -complete normal ultrafilter. For this, we need to use the approximation lemma for V (G). To show that U is an ultrafilter, let X κ ρ, X V (G), and let Ẋ HS be a name for X, supported by E e I. By the approximation lemma, we have that X V [G E e ], so we can use the Lévy-Solovay theorem [LS67] to see that either X U or κ ρ \ X U. To show that U is κ-complete, let γ < κ ρ and X δ ; δ < γ be a sequence of sets in U. Let σ HS be a name for this sequence and let E e I be a support for this sequence. Since a sequence of sets of ordinals can be coded into a set of ordinals, we can use the approximation lemma to get that the sequence is in V [G E e ]. Again by the Lévy-Solovay theorem [LS67] we get that its intersection is in U. Therefore U is a measure for κ ρ in V (G). To show that U is normal, let f : κ ρ κ ρ, f V (G) be regressive. Since f can be coded by a set of ordinals, by the approximation lemma, f V [G E e ] for some E e I. Again, by the Lévy-Solovay theorem [LS67], we will get a set in U on which f is constant. The construction in this paper is a generalised construction. For particular results, e.g., ℵ ω+3 becoming both the first uncountable regular cardinal and a measurable cardinal, we just put ρ = ω + 2. Thus we can immediately get a theorem such as the following. Theorem 3.1. If V is a model of There is an ω + 2-sequence of strongly compact cardinals with a measurable cardinal above this sequence, then there is a symmetric model in which ℵ ω+3 is both a measurable cardinal and the first regular cardinal. We can replace measurable by some large cardinal properties that are preserved under small forcing, and which are of the form For every set of ordinals X, there is a set Y such that φ(x, Y ) holds for certain formulas φ with two free variables. This is because for such properties we can capture the arbitrary set of ordinals in an intermediate ZFC model that is included in the symmetric model and use small forcing arguments to prove that such a large cardinal property is preserved. This allows us to construct models in which the first ρ uncountable cardinals are singular and ℵ ρ+1 is, e.g., weakly compact, Erdős, Ramsey, etc. Appendix A. Details for the proof of Theorem 2.5 This appendix contains the details on the proof of Theorem 2.5. Therefore all subsequent notation is the notation in the proof of that theorem. The arguments in the proof of the next lemma are part of the beginning of the proof of [Git80, Theorem 5.4] where it is shown that the powerset axiom holds in the class version of this model. Lemma A.1. The following set is dense in E e. J := {(t, T ) E e ; q T α κ ξ n < ω( if (α, n) dom 2 (q) \ dom 2 (t) and Proof. First notice that the set cf α < α then the ultrafilter Φ α,q(cf α)(n) is κ ξ -complete)} D := {(t, T ) E e ; α dom(t)(dom(t(α)) = dom(t(cf α)))} is dense in E e. This proof is similar to the proof that for a G, P a is dense in P 3. Now we will prove that for every (t, T ) D there is a T T such that (t, T ) J. For every α dom(t) \ κ ξ such that cf α < α, let λ α be the least ordinal ν < cf α such that κ α ν κ ξ.

12 12 ARTHUR W. APTER, IOANNA M. DIMITRÍOU, AND PETER KOEPKE We have that (t, T ) J iff for all q T, if α(q) κ ξ and cf α(q) < α(q) then q(cf α(q))(m(q)) λ α(q). This equivalence is true because the right hand side of the implication above ensures that the ultrafilter Φ α,q(cf α),m(q) is κ ξ -complete. Define and for β b define b := {cf α ; α dom(t) \ κ ξ cf α < α}, c β := max{λ α ; α dom(t) \ κ ξ and cf α = β < α}. Then (t, T ) J if for all q T and (α, m) dom 2 (q) \ dom 2 (t) such that cf α b we have that q(cf α)(m) c cf α. So let T :={q T ; α b m < ω(if (α, m) dom 2 (q) \ dom 2 (t) then q(α )(m) c α )}. Clearly, this (t, T ) J. Lemma A.2. There is an E-name Q for Q and J densely embeds into E Q. Proof. An obvious name Q for Q is the following. For (t, T ) E, (σ, (t, T )) Q iff (a) there is an s P2 and an E-name σ such that s t P 2, σ = (š, σ)ˇ, and for all π dom( σ) there is a s P2 such that š = π. (b) (t, T ) š σ, (c) (t, T ) π(π σ dom(π) = dom(š) (π š π š)), (d) (t, T ) π(π σ α(π) = cf α(π) ˇX( ˇX ˇΦ α(π) ˇβ( ˇβ ˇX π {(ˇα(π), ˇm(π), ˇβ)} σ))), (e) (t, T ) π(π σ α(π) > cf α(π) cf α(π) κ ξ ˇX( ˇX ˇΦ α(π),π(cf α(π))(m(π)) ˇβ( ˇβ ˇX π {(ˇα(π), ˇm(π), ˇβ)} σ))), (f) (t, T ) π(π σ α(π) > cf α(π) cf α(π) < κ ξ ˇX( ˇX ˇΦ α(π)γ(cf α(π))(m(π)) ˇβ( ˇβ ˇX π {(ˇα(π), ˇm(π), ˇβ)} σ))), where Γ is the standard E-name for {t ; T ((t, T ) G )}, (g) (t, T ) π π (π σ π ˇP 2 π π π σ). The name for the ordering on Q is defined as ((σ, τ)ˇ, (t, T )) Q : (t, T ) σ τ. From the forcing theorem we have that t T such that t t define Q G = (t, T ) (t ) := (t, {t T ; t t or t t }), Q. For every (t, T ) P 3 and the extension of (t, T ) with trunk t. If t t then we conventionally take (t, T ) (t ) := (t, T ). Define a map i : J E Q. For (r, R) J, we take i((r, R)) = ((r 1, R 1 ), ρ) : (i) (r 1, R 1 ) := (r, R) E e κξ, (ii) ρ = (ř 2, ρ)ˇ, where r 2 := r (e \ κ ξ ) and for all π dom( ρ), there is an r R such that (π = (r (e \ κ ξ )), (iii) For all r R with r r we have that ((r (e \ κ ξ )), (r 1, R 1 )) ρ, (iv) For all r R with r r we have that ((r (e \ κ ξ )), (r 1, R 1 ) (r (e κ ξ ))) ρ. (v) No other elements are in ρ.

13 THE FIRST MEASURABLE CAN BE THE FIRST UNCOUNTABLE REGULAR 13 Claim 1. For all (r, R) J, i((r, R)) = ((r 1, R 1 ), ρ) E Q. Proof of claim. That (r 1, R 1 ) E is immediate. So we must show that (r 1, R 1 ) ρ Q. Requirement (a) clearly holds with r 2 := r (e \ κ ξ ). For (b) we want that (r 1, R 1 ) ř 2 ρ which holds because (ř 2, (r 1, R 1 )) ρ. For (c) we want that or equivalently that (r 1, R 1 ) π(π ρ dom(π) = dom(ř 2 ) (π ř 2 π ř 2 )) π V E (b, B) (r 1, R 1 ) (b, B ) (b, B) ((b, B ) π ρ or (b, B ) (dom(π) = dom(ř 2 ) (π ř 2 π ř 2 ))). Let π V E and (b, B) (r 1, R 1 ) be arbitrary and let (b, B ) (b, B) decide the formula π ρ. Assume that (b, B ) π ρ. Then we have that (b, B ) π ρ. By the definition of ρ there is some r R such that (b, B ) π = (r (e \ κ ξ )). Since (r, R) is a condition in P 3 we get that (b, B ) dom(π) = dom(ř 2 ) (π ř 2 π ř 2 ). For (d) we want to show that for every π V E, (r 1, R 1 ) (π ρ α(π) = cf α(π) ˇX( ˇX ˇΦ α(π) ˇβ( ˇβ ˇX π {(ˇα(π), ˇm(π), ˇβ)} ρ))). As before, let π V E and (b, B) (r 1, R 1 ) be arbitrary and let (b, B ) (b, B) decide the formula π ρ α(π) = cf α(π). Assume that (b, B ) (π ρ α(π) = cf α(π)). Then (b, B ) (π ρ α(π) = cf α(π)). Let r R be such that (b, B ) (r 1, R 1 ) (r (e κ ξ )) and (b, B ) π = (r (e \ κ ξ )). Call r 1 := r (e κ ξ ) and r 2 := r (e \ κ ξ ). We want to show that (b, B ) ˇX( ˇX ˇΦ α(π) ˇβ( ˇβ ˇX π {(ˇα(π), ˇm(π), ˇβ)} σ)). Case 1, if α(r ) = α(r 2) κ ξ. Then let and note that X Φ α(r 2 ) and X := Suc R (r ) = {β ; r {(α(r ), m(r ), β)} R}, r {(α(r ), m(r ), β)} R ((r 2 {(α(r 2), m(r 2), β)}), (r 1, R 1 ) (r 1)) ρ β X. Let ˇβ V E be arbitrary. We want that (b, B ) ( ˇβ ˇX π {(ˇα(π), ˇm(π), ˇβ)} ρ) ( ˇβ ˇX π {(ˇα(π), ˇm(π), ˇβ)} ρ),

14 14 ARTHUR W. APTER, IOANNA M. DIMITRÍOU, AND PETER KOEPKE or equivalently that (c, C) (b, B ) (c, C ) (c, C) ((c, C ) ( ˇβ ˇX π {(ˇα(π), ˇm(π), ˇβ)} ρ) or (c, C ) ( ˇβ ˇX π {(ˇα(π), ˇm(π), ˇβ)} ρ)). So let (c, C) (b, B ) be arbitrary and let (c, C ) (c, C) be stronger than (r 1, R 1 ) (r 1) and decide π {(ˇα(π), ˇm(π), ˇβ)} ρ. Clearly this (c, C ) satisfies and we re done with this case. (c, C ) ( ˇβ ˇX π {(ˇα(π), ˇm(π), ˇβ)} ρ) or (c, C ) ( ˇβ ˇX π {(ˇα(π), ˇm(π), ˇβ)} ρ) Case 2, if α(r ) < κ ξ, then let r r be such that r (e \ κ ξ ) = r 2 and α(r ) = α(r 2) κ ξ. The rest follows as in Case 1. For (e), (f), and (g) we proceed similarly. Claim 1 Claim 2. The map i is a dense embedding. Proof of claim. Let ((t, T ), σ) E Q be arbitrary. We want to define an (r, R) J such that i((r, R)) ((t, T ), σ). Define r := t s and let σ = (š, σ). By (a) of the definition of Q, r P 2. If r P 2 is such that r r then let r R. For r r we define R recursively as follows. Let r P 2 be such that r r and r R. If α(r ) < κ ξ then r {(α(r ), m(r ), β)} R : (r {(α(r ), m(r ), β)}) (e κ ξ ) T. If α(r ) κ ξ then r {(α(r ), m(r ), β)} R : (t, T ) (r (e κ ξ )) ((r (e \ κ ξ )) {(α(r ), m(r ), β)}) σ. Subclaim 1. For every r R, call r 1 := r (e κ ξ ) and r 2 := r (e \ κ ξ ). Then (t, T ) (r 1) ř 2 σ. Proof of subclaim. Since by the definition of R and Q this holds for all r r, we ll use induction with base case r = r. For r = r it holds with (t, T ) = (t, T ) due to (b) of the definition of Q. So assume it holds for r and let r {(α(r ), m(r ), β)} R be arbitrary. If α(r ) < κ ξ then α(r ) = α(r 1) and by the definition of R we have that (r 1 {(α(r ), m(r ), β)}) T. By the induction hypothesis we get that (t, T ) (r 1 {(α(r ), m(r ), β)}) (t, T ) (r 1) ř 2 σ. If α(r ) κ ξ then α(r ) = α(r 2) and by the definition of R we have that (t, T ) (r 1) (r 2 {(α(r 2), m(r 2), β)}) σ. Subclaim 1 Subclaim 2. (r, R) J Proof of subclaim. To show that (r, R) P 3 we only need to verify (6) and (7) of the definition of P 3.

15 THE FIRST MEASURABLE CAN BE THE FIRST UNCOUNTABLE REGULAR 15 For (6), let r R with r r and α(r ) of type 1. Call r 1 := r (e κ ξ ) and r 2 := r (e \ κ ξ ). If α(r ) < κ ξ then r {(α(r ), m(r ), β)} R iff r 1 {(α(r 1), m(r 1), β)} T. Since (t, T ) E we get that Suc R (r ) = {β ; r 1 {(α(r 1), m(r 1), β)} T } Φ α(r ). So assume that α(r ) κ ξ. We have that (t, T ) π(π σ π š α(π) = cf α(π) ˇX( ˇX ˇΦ α(π) ˇβ( ˇβ ˇX π {(α(π), m(π), β)} σ))). By Subclaim 1 we have that (t, T ) (r 1) ř 2 σ ř 2 š α(ř 2) = cf α(ř 2), thus (t, T ) (r 1) ˇX( ˇX ˇΦ α(ř 2 ) ˇβ( ˇβ ˇX ř 2 {(α(ř 2), m(ř 2), β)} σ)). So for some (t, T ) (t, T ) (r 1) there is some X (t,t ) Φ α(r 2 ) such that for every ˇβ V E we have that (1) (t, T ) ˇβ ˇX (t,t ) ř 2 {(α(ř 2), m(ř 2), β)} σ. Define the set X := {X (t,t ) ; (t, T ) (t, T ) (r 1) and (1) holds}. Since κ ξ is inaccessible, E < κ ξ, and Φ α(r 2 ) is κ ξ -complete, we have that X Φ α(r 2 ) and (1) holds for (t, T ) (r 1) and X. Let β X. Then (t, T ) (r 1) ř 2 {(α(ř 2), m(ř 2), ˇβ)} σ which by the definition of R means that r {(α(r ), m(r ), β)} R. So X Suc R (r ) Φ α(r ). For (7), let r R with r r and α(r ) of type 2. Again, call r 1 := r (e κ ξ ) and r 2 := r (e \ κ ξ ). If α(r ) κ ξ and cf α(r ) < κ ξ then we have that for every π V E, (t, T ) (π σ π š α(π) > cf α(π) < κ ξ ˇX( ˇX ˇΦ α(π),γ(cf α(π))(m(π)) ˇβ( ˇβ ˇX π {(α(π), m(π), β)} σ))). By Subclaim 1 we have that (t, T ) (r 1) ř 2 σ ř 2 š α(ř 2) > cf α(ř 2) < κ ξ. With the same arguments as for (6), there is some X V such that (t, T ) (r 1) ˇX ˇΦ α(ř 2 ),Γ(cf α(ř 2 ))(m(ř 2 )) and for every ˇβ V E we have that (t, T ) (r 1) ˇβ ˇX ř 2 {(α(ř 2), m(ř 2), ˇβ)} σ. But since r 1 r 2 = r P 2, we have that (t, T ) (r 1) decides the value of Γ(cf α(ř 2))(m(ř 2)) to be γ := r 1(cf α(r 2))(m(r 2)). So (t, T ) (r 1) ˇX ˇΦ cf α(ř 2 ),γ. Since X V and Φ cf α(ř 2 ),γ V we have that X Φ cf α(ř 2 ),γ. So take an arbitrary β X. Then (t, T ) (r 1) (r 2 {(α(r 2), m(r 2), β)}) σ which by the definition of R means that r {(α(r ), m(r ), β)} R and consequently X Suc R (r ) Φ α(r ),r (cf α(r ))(m(r )). Similarly for the other cases where α(r ) > κ ξ and cf α(r ) κ ξ, and α(r ) < κ ξ. To conclude Subclaim 2 we want to show that the last condition for membership in J is fulfilled, i.e., if r R, α κ ξ, and n < ω are such that (α, n) dom 2 (r ) \ dom 2 (t) and cf α < α, then Φ α,r (cf α)(n) is κ ξ -complete. Let q r be such that for some q R, q = q {(α, n, β)}, α(q ) = α, and m(q ) = n. Note that r (cf α)(n) = q (cf α)(n).

16 16 ARTHUR W. APTER, IOANNA M. DIMITRÍOU, AND PETER KOEPKE If cf α κ ξ then clearly Φ α,q (cf α)(n) is κ ξ -complete. If cf α < κ ξ then note that q (e κ ξ ) T and (t, T ) E, i.e., for some (s, S) J, (s, S) E e κξ = (t, T ). So q (cf α)(n) must be high enough for the ultrafilter Φ α,q (cf α)(n) to be κ ξ -complete. Subclaim 2 Lastly, we want to show that i((r, R)) E Q ((t, T ), σ). Let i((r, R))) = ((r 1, R 1 ), ρ), and ρ = (ř 2, ρ)ˇ. By the definition of R and of i we immediately get that (r 1, R 1 ) E (t, T ). So it remains to show that (r 1, R 1 ) ρ σ, i.e., or equivalently that (r 1, R 1 ) π(π ρ π σ), π V E (b, B) (r 1, R 1 ) (b, B ) (b, B)((b, B ) π ρ or (b, B ) π σ). So let π V E and (b, B) (r 1, R 1 ) be arbitrary. There is some (b, B ) (b, B) that decides π ρ. Assume that (b, B ) π ρ. Then (b, B ) π ρ. By the definition of ρ, there must be some r R such that (b, B ) π = (r (e \ κ ξ )) and (b, B ) is compatible with (r 1, R 1 ) (r (e κ ξ )). Call r 1 := r (e κ ξ ), r 2 := r (e\κ ξ ), and let (b, B ) be stronger than both (b, B ) and (r 1, R 1 ) (r 1). By Subclaim 1, (t, T ) (r 1) ř 2 σ. Then (r 1, R 1 ) (r 1) ř 2 σ so (b, B ) ř 2 σ. Claim 2 So we have shown that Q can indeed be seen as the top part of E e, cut in κ ξ. Lemma A.3. (The Prikry lemma for Q) In V [G ], let τ 1,..., τ k be Q-names, and φ be a formula with k free variables. Then for every forcing condition (s, S) Q there is a stronger condition (s, W ) Q which decides φ(τ 1,..., τ n ). This proof is almost identical to Gitik s Prikry style lemma [Git80, Lemma 5.1]. Proof. Work in V [G ]. Let (s, S) Q. Let r S. If α(r) is of type 1 then call Φ r := Φ α(r). If α(r) is of type 2 and cf α(r) κ ξ then call Φ r := Φ α(r),r(cf α(r))(m(r)). If α(r) is of type 2 and cf α(r) < κ ξ then let γ cf α be such that G (cf α(r))(m(r)) = γ, and call Φ r := Φ α(r),γ. For all r P 2 (e \ κ ξ ) ω κ ρ define Φ r := {X α(r) ; Y Φ r (Y X)}. For each r S, the ultrafilter Φ r is at least κ ξ complete. Since E < κ ξ, we can use arguments from the Lévy-Solovay theorem [LS67] to get that in V [G ], each Φ r is at least κ ξ -complete as well. Also define the following sets. S 0 := {r S ; r s and R S((r, R) Q and (r, R) Q φ(τ 1,..., τ k ))} S 1 := {r S ; r s and R S((r, R) Q and (r, R) Q φ(τ 1,..., τ k ))} S 2 := {r S ; r s and R S(if (r, R) Q then (r, R) does not decide φ(τ 1,..., τ k ))}.

17 THE FIRST MEASURABLE CAN BE THE FIRST UNCOUNTABLE REGULAR 17 Clearly, S = S 0 S 1 S 2. Let e \ κ ξ := {α 0,..., α n 1 }. We will now enumerate the set ((e \ κ ξ ) ω) \ dom 2 (s) from left to right and upwards, by a function x that is recursively defined as follows. First, x(0) := (α(s), m(s)). Now let x(i) = (α j, m) for some α j e \ κ ξ and m ω. If j < n 1 then let and if j = n 1 then For every i ω define x(i + 1) := (α j+1, m), x(i + 1) := (α 0, m + 1). F i := {r S ; dom 2 (r) \ dom 2 (s) = x (i + 1)}. Now for i j we will define recursively on j i a set of functions F i,j : Fi 3. For l < 3 and r F i let F i,i (r) = l : r S l. For i < j let F i,j (r) = l : the set {β ; r {(α(r), m(r), β)} S and F i+1,j (r {(α(r), m(r), β)}) = l} is in the ultrafilter Φ r. Define recursively on i < ω a subset F i F i. Using the definition and the ω-completeness of the ultrafilter Φ s, we find (in V ), a set F 0 F 0 which is homogeneous for all functions in the set {F 0,j ; j < ω} and which is such that the set {β ; s {(α(s), m(s), β)} F 0} is in Φ s. By homogeneous here we mean that for all t 1, t 2 F 0 and for all 0 j < ω, F 0,j (t 1 ) = F 0,j (t 2 ). For i > 0 we take F i := {r F i ; r F i 1 and i j(f i,j (r) = F i 1,j (r ))}, where r is defined as in Definition 2.3(5). By the induction hypothesis, it follows that F i is homogeneous for all functions in the set {F i,j ; i j < ω}. The definition of the functions F i,j implies that for every r F i 1, (2) {β ; r {(α(r), m(r), β)} F i } Φ r. Define the set F := {s} { F i ; i < ω}. Claim. If s 1, s 2 F, (s 1, A 1 ), (s 2, A 2 ) Q, and (s 1, A 1 ), (s 2, A 2 ) Q (s, S), then it is impossible to have that (s 1, A 1 ) Q φ(τ 1,..., τ k ) and (s 2, A 2 ) Q φ(τ 1,..., τ k ). Proof of claim. We have that for some i 1, i 2 < ω and for every j = 1, 2, dom 2 (s j ) = dom 2 (s) {x(l) ; l < i j }. Without loss of generality we may assume that i 1 i 2. If i 1 < i 2 then we can increase the dom 2 (s 1 ), one step at a time until we get i 1 = i 2. We have that the set E := {β < α(s 1 ) ; s 1 {(α(s 1 ), m(s 1 ), β)} F i 1 } is in Φ s1 and since (s 1, A 1 ) is a condition in Q we have that also the set E := {β < α(s 1 ) ; s 1 {(α(s 1 ), m(s 1 ), β)} A 1 } is in Φ s1.

18 18 ARTHUR W. APTER, IOANNA M. DIMITRÍOU, AND PETER KOEPKE Let β E E, let s 1 := s 1 {(α(s 1 ), m(s 1 ), β)}, and let Ā1 := {t A 1 ; t s 1 }. Then we have that s 1 F i 1 F and that ( s 1, Ā1) Q (s 1, A 1 ). Therefore, ( s 1, Ā1) Q φ(τ 1,..., τ n ), and dom 2 ( s 1 ) = dom 2 (s) {x(k) ; k < i 1 + 1}. This way we keep increasing i 1 until we get i 1 = i 2. Denote i 2 i 1 by i. If i = 0 then s 1 = s 2 = r, therefore (s 1, A 1 ) and (s 2, A 2 ) are compatible which is a contradiction. If i > 0 then s 1, s 2 F i 1. Because (s 1, A 1 ) Q (s, S) and (s 1, A 1 ) Q φ(τ 1,..., τ n ), we have that s 1 S 0, thus F i 1,i 1 (s 1 ) = 0. Similarly we get that F i 1,i 1 (s 2 ) = 1 which contradicts the homogeneity of F i 1 for F i 1,i 1. Claim So to finish the proof of this lemma, we first show that (s, F ) is indeed a condition in Q. It suffices to show that for every s F, the set Suc F (s ) of successors of s in F is in the ultrafilter Φ s. We have that Suc F (s ) ={β < α(s ) ; s {(α(s ), m(s ), β)} F } ={β < α(s ) ; s {(α(s ), m(s ), β)} F i+1, } where i ω is such that s F i. By (2) we get that Suc F (s ) Φ s. Finally, let (s 1, A 1 ) Q be any condition that decides φ(τ 1,..., τ n ) and extends (s, F ). Without loss of generality assume that (s 1, A 1 ) Q φ(τ 1,..., τ n ). By the definition of Q we can assume that dom(s 1 ) = e \ κ ξ, and hence s 1 F and A 1 F. Suppose that (s, F ) does not decide φ(τ 1,..., τ n ), then there must be some (s 2, A 2 ) Q (s, F ) such that (s 2, A 2 ) Q φ(τ 1,..., τ n ). But this contradicts the Claim. Therefore (s, F ) decides φ(τ 1,..., τ n ). Lemma A.4. In V [G ], the partial order (Q, Q ) does not add bounded subsets to κ ξ. Proof. We work in V [G ]. Consider the relation Q Q, defined as (s, S) Q (r, R) iff s = r and S R. Claim. (Q, Q ) is κ ξ-closed. Proof of claim. Let γ < κ ξ and let (s ζ, S ζ ) ; ζ < γ be a Q-descending sequence of elements in Q. Since E is small forcing with respect to κ ξ, we will use the Lévy- Solovay theorem [LS67] to get that all ultrafilters involved in the definition of Q can be extended to κ ξ -complete ultrafilters in V [G ]. Let s = s ζ for all ζ < γ and define a set S ζ<γ S ζ inductively (above s) as follows. If s s then s S. Assume that s s S. Since for every ζ < γ, s S ζ, there is a set Φ, that is a κ ξ -complete ultrafilter over α(s ) in V, and such that for every ζ < γ, Suc Sζ (s ) := {β ; s {(α(s ), m(s ), β)} S ζ } Φ. In particular, we choose these Φ to fit with the definition of P 3, i.e., if α(s ) is of type 1 then Φ = Φ α(s ), if α(s ) is of type 2 and cf α(s ) κ ξ then Φ = Φ α(s ),s (cf α(s ))(m(s )), and if α(s ) is of type 2 and cf α(s ) < κ ξ then Φ = Φ α(s ), G (cf α(s ))(m(s )). By the Lévy-Solovay theorem [LS67], Φ := {X α(s ) ; Y X(Y Φ)} is a κ ξ -complete ultrafilter over α(s ). So ζ<γ Suc S ζ (s ) Φ and consequently there is some Y s Suc Sζ (s ) such that Y s Φ. We define then s {(α(s ), m(s ), β)} S : β Y s.

19 THE FIRST MEASURABLE CAN BE THE FIRST UNCOUNTABLE REGULAR 19 Clearly (s, S) Q and for every ζ < γ, (s, S) Q (s ζ, S ζ ). Claim Now to show that (Q, Q ) does not add new bounded subsets to κ ξ, we proceed as usual with Prikry type forcings. Let (s, S) Q, let τ be a Q-name, γ < κ ξ, and assume that (s, S) Q τ ˇγ. Using the Prikry Lemma for Q (by Lemma A.3) and that (Q, Q ) is κ ξ-closed (by the claim), we get a Q-decreasing sequence (s, S ζ ) ; ζ < γ of conditions in Q such that for each ζ < γ, (s, S ζ ) decides the statement ˇζ τ. Again by using the claim we get a condition (s, S ) stronger than all (s, S ζ ), therefore one such that (s, S ) τ = ˇx, where x = {ζ γ ; (s, S ) ˇζ τ}. References [AH91] Arthur W. Apter and James M. Henle. Relative consistency results via strong compactness. Fundamenta Mathematicae, 139: , [AM95] Arthur W. Apter and Menachem Magidor. Instances of dependent choice and the measurability of ℵ ω+1. Annals of Pure and Applied Logic, 74: , [Apt85] Arthur W. Apter. Successors of singular cardinals and measurability. Advances in Mathematics, 55: , [Apt96] Arthur W. Apter. AD and patterns of singular cardinals below Θ. Journal of Symbolic Logic, 61: , [Dim11] Ioanna M. Dimitríou. Symmetric Models, Singular Cardinal Patterns, and Indiscernibles. PhD thesis, Rheinische Friedrich-Wilhelms-Universität Bonn, [Git80] Moti Gitik. All uncountable cardinals can be singular. Israel Journal of Mathematics, 35:61 88, [Jec68] Thomas J. Jech. ω 1 can be measurable. Israel Journal of Mathematics, 6: , [Jec03] Thomas J. Jech. Set theory. The third millenium edition, revised and expanded. Springer, [Kun80] Kenneth Kunen. Set theory: an introduction to independence proofs. Elsevier, [LS67] Azriel Lévy and Robert M. Solovay. Measurable cardinals and the continuum hypothesis. Israel Journal of Mathematics, 5: , [Tak70] Gaisi Takeuti. A relativization of strong axioms of infinity to ω 1. Annals of the Japan Association for Philosophy of Science, 3: , A. W. Apter, The Graduate Center of The City University of New York, Mathematics, 365 Fifth Avenue, New York, NY 10016, USA & Department of Mathematics, Baruch College, One Bernard Baruch Way, New York, NY 10010, USA address: awapter@alum.mit.edu URL: I. M. Dimitríou, Mathematisches Institut der Universität Bonn, Endenicher Allee 60, Bonn, Germany address: ioanna.m.dimitriou@gmail.com URL: P. Koepke, Mathematisches Institut der Universität Bonn, Endenicher Allee 60, Bonn, Germany address: koepke@math.uni-bonn.de URL: