The First Measurable Cardinal Can Be the First Uncountable Regular Cardinal at Any Successor Height

Home , Limit cardinal, Measurable cardinal, Regular cardinal

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

ARTHUR W. APTER, IOANNA M. DIMITR´IOU, AND PETER KOEPKE

Abstract. We use techniques due to Moti Gitik to construct models in which for an arbitrary ordinal ρ, ℵρ+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] indepen- dently showed that if we assume the consistency of ZFC+ “There is a measurable cardinal”, then the theory ZF+DC+“ω1 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 > ω1. 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 ℵ2 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., ℵω+1 is both the least regular uncountable and least measurable cardinal, or ℵω+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 ℵ3 or ℵω+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 hκξ ; ξ < ρi of strongly compact cardinals such that each limit point of the sequence hκξ ; ξ < ρi is singular, and with a measurable cardinal κρ 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, ℵβ is singular” + “ℵρ+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, 2014. The authors were partially supported by the DFG-NWO cooperation grant KO 1353/5-1 “In- finitary 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 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 supercom- pactness 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 ℵ1 has cofinality ω in V (G), both DC and ACω are false in this model. It is therefore especially relevant that ℵρ+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 κρ 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 NG, 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 NG 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 NG, there is neither an explicit proof nor hint at any point in [Git80] as to how to determine the cardinal structure of NG. 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 Pκξ (α) for each α ∈ (κξ, κξ+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, κ0) has collapsed we will use a fine ultrafilter over Pω(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 S collapse the cardinals in the open interval ( ξ<λ κξ, κλ). Forcing at each κξ will be done with a κξ-complete measure over κξ. Finally, a small forcing argument guarantees that κρ 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,

hκξ ; ξ < ρi, such that for every ξ < ρ, κξ is strongly compact, and such that the sequence has S no regular limits. Let κρ > ξ<ρ κξ be a measurable cardinal. We will construct a model with a sequence of ρ + 1-many successive singular cardinals in which κρ 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. 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.

κρ Call Reg the set of regular cardinals α ∈ [ω, κρ) in V . For convenience call κρ 0 ω =: κ−1. For an α ∈ Reg we deﬁne a cf α to distinguish between the following categories.

(type 1) This occurs if there is a largest κξ ≤ α (i.e., α ∈ [κξ, κξ+1)). We then deﬁne cf0α := α.

If α = κξ and ξ 6= −1, then let Φκξ be a measure for κξ. If α = ω then let Φω be any uniform ultrafilter on ω. If α > κξ is inaccessible, then let Hα be a κξ-complete fine ultrafilter

over Pκξ (α) and let hα : Pκξ (α) → α be a bijection. Deﬁne

−1 Φα := {X ⊆ α ; hα “X ∈ Hα}.

This is a uniform κξ-complete ultrafilter over α. If α > κξ is not inaccessible, then let Φα be any κξ-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 cf0α := cfβ. Let α 0 hκν ; ν < cf αi be a fixed ascending sequence of strongly compacts ≥ cf0α such that β = S α 0 {κν ; ν < cf α}. 0 If α is inaccessible, then for each ν < cf α, let Hα,ν be a fine ultrafilter over P α (α) and h : P (α) → α a bijection. Define κν α,ν κν

−1 Φα,ν := {X ⊆ α ; hα “X ∈ Hα,ν }.

α Again, Φα,ν is a κν -complete uniform ultrafilter over α. 0 α If α is not inaccessible, then for each ν < cf α, let Φα,ν be any κν - complete uniform ultrafilter over α. This cf0α will be used when we want to organise the choice of ultrafilters for the type 2 cardinals.

We use the fine ultrafilters Hα and Hα,ν to make sure that in the end only the strongly compacts and their singular limits remain cardinals below κρ. For type 1 ordinals we will do some tree-Prikry like forcings to singularise in cofinality ω. Type 1 cardinals in the open intervals (κξ, κξ+1) will be collapsed to κξ 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 0 we build for cf α to pick a countable sequence of ultrafilters hΦ~c(n) ; n ∈ ωi. 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 ARTHUR W. APTER, IOANNA M. DIMITRÍOU, AND PETER KOEPKE the Prikry sequence1 at stage cf0α. Let us take a look at the definition of the stems of the Prikry sequences to be added.

κρ Deﬁnition 2.1. For t ⊆ Reg × ω × κρ we deﬁne the sets dom(t) :={α ∈ Regκρ ; ∃m ∈ ω∃γ ∈ Ord((α, m, γ) ∈ t)}, and dom2(t) :={(α, m) ∈ Regκρ × ω ; ∃γ ∈ Ord((α, m, γ) ∈ t)}.

κρ Let P1 be the set of all finite subsets t of Reg × ω × κρ, 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 cf0α. 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 P2 be the set of all t ∈ P1 such that the following hold. (1) For every α ∈ dom(t), cf0α ∈ dom(t) and dom(t(cf0α)) ⊇ dom(t(α)). (2) If {α0, . . . , αn−1} is an increasing enumeration of dom(t) \ κ0, 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(αk)) = m + 1 and • for every k ∈ {j, . . . , n − 1} we have that dom(t(αk)) = m. These m and αj are unique for t and are denoted by m(t) := m and α(t) := αj. (3) If cf0α(t) < α(t) then m(t) ∈ dom(t(cf0α(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(cf0α(t))(m(t)) will decide how the condition t will grow at the point (α(t), m(t)).

Let us call elements of P2 stems. In the following image we can see roughly what a stem t with a domain {α0, α1, α2} above κ0 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 P3 be the set of pairs (t, T ) such that

(1) t ∈ P2, (2) T ⊆ P2, (3) t ∈ T , (4) for every t0 ∈ T we have t0 ⊇ t or t0 ⊆ t, and dom(t0) = dom(t), (5) for every t0 ∈ T , if t0 = r ∪ {(α(r), m(r), β)} then t0− := r ∈ T , i.e., T is tree-like,

1For 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. THE FIRST MEASURABLE CAN BE THE FIRST UNCOUNTABLE REGULAR 5

(6) for every t0 ∈ T with t0 ⊇ t, if α(t0) is of type 1 (i.e., cf0(α(t0)) = α(t0)) then 0 0 0 0 SucT (t ) := {β ; t ∪ {(α(t ), m(t ), β)} ∈ T } ∈ Φα(t0), and (7) for every t0 ∈ T with t0 ⊇ t, if α(t0) is of type 2 (i.e., cf0α(t0) < α(t0)) and m(t0) ∈ dom(t0(cf0(α(t0)))) then 0 0 0 0 SucT (t ) := {β ; t ∪ {(α(t ), m(t ), β)} ∈ T } ∈ Φα(t0),t0(cf0α(t0))(m(t0)).

κρ 0 0 For a (t, T ) in P3 and a subset x ⊆ Reg we write T x for {t x ; t ∈ T }. We call t the trunk of (t, T ).

This P3 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 P3 adds too many subsets of ordinals and makes every ordinal in the interval (ω, κρ) 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(P3) that is induced by P3, as in [Jec03, Corollary 14.12]. κρ We start by considering G to be the group of permutations of Reg × ω × κρ whose elements a satisfy the following properties.

κρ • For every α ∈ Reg there is a permutation aα 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, aα(β)).

The finite subset of α that aα moves, we denote by supp(aα), which stands for “support of aα”. κρ • For only finitely many α ∈ Reg is aα not the identity. This finite subset of Regκρ we denote by dom(a).

We extend G to P3 as follows. For a ∈ G and (t, T ) ∈ P3, define a(t, T ) := (a“t, {a“t0 ; t0 ∈ T }), where a“t := {(α, n, aα(β)) ; (α, n, β) ∈ t}. Unfortunately, in general a(t, T ) is not a member of P3 because of the branching condition at type 2 cardinals. In particular, it is possible that for some α ∈ dom(t) 0 0 0 0 0 of type 2, and some t ∈ T with α = α(t ), we have that acf0α(t (cf α)(m(t ))) = 0 0 0 0 γ 6= t (cf α)(m(t )), and even though we had before SucT (t ) ∈ Φα,t0(cf0α)(m(t0)), it 0 is not true that SucT (t ) ∈ Φα,γ . a To overcome this problem, for an a ∈ G, define P ⊆ P3 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(cf0α)), and (3) for every α ∈ dom(t), we have that

rng(t(α)) ⊇ {β ∈ supp(aα); ∃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 Φα,γ . In (3) we require that the stem of each condition already contains all the ordinals that the aα could move. This will prevent any trouble with membership in the ultrafilters. One may think that this requirement should be supp(aα) ⊆ rng(t(α)) but this is not the case; note that 6 ARTHUR W. APTER, IOANNA M. DIMITRÍOU, AND PETER KOEPKE there might be some γ in aα 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 a [Git80, page 68], for every a ∈ G the set P is a dense subset of P3. 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 ⊆ Regκρ deﬁne

Ee := {(t, T ) ∈ P3 ; dom(t) ⊆ e},

κρ 0 I := {Ee ; e ⊂ Reg is ﬁnite and closed under cf },

ﬁxEe := {a ∈ G ; ∀α ∈ e(aα is the identity on α)}, and let F be the normal ﬁlter (see [Jec03, (15.34)]) over G that is generated by

{fixEe ; Ee ∈ I}. For eachx ˙ in the Boolean valued model V B, define its symmetry group sym(x ˙) := {a ∈ G ; a(x ˙) =x ˙}. A namex ˙ 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., G HS := {x˙ ∈ V ; ∀y˙ ∈ tcdom(x ˙)(sym(y ˙) ∈ F)}, where tcdom(x ˙) is defined as the union of all xn, which are defined recursively by S x0 := {x˙} and xn+1 := {dom(y ˙) ;y ˙ ∈ xn}. We will say that an Ee ∈ I supports a namex ˙ ∈ HS if fixEe ⊆ sym(x ˙). For some V -generic ultrafilter G on B we define the symmetric model V (G) := {x˙ G ;x ˙ ∈ HS}. By [Jec03, Lemma 15.51], this is a model of ZF, and V ⊆ V (G) ⊆ V [G]. For each (t, T ) ∈ P3 and each Ee ∈ I, define ∗ 0 0 (t, T ) Ee = (t e, {t e ; t ∈ T }).

According to [Git80, Lemma 3.3.], if φ is a formula with n free variables,x ˙ 1,..., x˙ n ∈ HS, and Ee ∈ I is such that sym(x ˙ 1),..., sym(x ˙ n) ⊇ ﬁxEe then we have that for every (t, T ) ∈ P3 ∗ (t, T ) φ(x ˙ 1,..., x˙ n) ⇐⇒ (t, T ) Ee φ(x ˙ 1,..., x˙ 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 Ee ∈ I such that ∗ ∗ ∗ X ∈ V [G Ee], where G Ee := {(t, T ) Ee ;(t, T ) ∈ G}. 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 namesα ˇ are not moved by automorphisms of B, we have that if X˙ ∈ HS is a P3-name for X and Ee ∈ I supports X˙ then the set ¨ ∗ ˙ X := {(ˇα, (t, T ) Ee);(t, T ) αˇ ∈ X} is an Ee-name for X. qed We will use the approximation lemma in all our subsequent proofs.

Theorem 2.5. For every 0 ≤ ξ ≤ ρ, κξ is a cardinal in V (G). Consequently, their (singular) limits are also preserved.

The proof proceeds by construing a finite support part of P3 as a two-step iterated ˙ forcing E ∗ Q, where the first component is small forcing relative to κξ, and the second component satisfies a Prikry lemma and does not add bounded subsets to κξ. This corresponds to the intuition behind the definition of P3. 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 < κξ and a bijection ˙ f : δ → κξ in V (G). Let f be a name for f with support Ee ∈ I. Note that e is a ﬁnite subset of Regκρ that is closed under cf0. By the approximation lemma (Lemma 2.4), since f may be coded by a set of ordinals, there is an Ee-name for f, i.e., for this e, ∗ f ∈ V [G Ee].

We will show that this is impossible, by taking a dense subset of Ee and showing that it is forcing equivalent to an iterated forcing construction, the ﬁrst part of which has cardinality less than κξ, and the second part of which does not collapse κξ (by not adding bounded subsets to κξ, similarly to Prikry forcing).

It’s not hard to check that 2 2 J := {(t, T ) ∈ Ee ; ∀q ∈ T ∀α ≥ κξ∀n < ω( if (α, n) ∈ dom (q) \ dom (t) and 0 cf α < α then the ultrafilter Φα,q(cf0α)(n) is κξ-complete)} is dense in Ee (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 ∩ κξ 6= ∅. Define the sets ∗ E :={(t, T ) Ee∩κξ ;(t, T ) ∈ J}, and ∗ P2 :={t (e \ κξ); t ∈ P2}. ∗ For s ∈ P2 we can define α(s) and m(s) as we did for the s ∈ P2, in Definition 2.2(2). ∗ Let G be an arbitrary E-generic filter and note that for every α ∈ e \ κξ such 0 0 ∗ 0 that cf α < κξ, the set ht(cf α)(m); ∃T ((t, T ) ∈ G ∧ m ∈ dom(t(cf α)))i is the 0 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 ∈ P2 , ∗ (2) S ⊆ P2 , (3) s ∈ S, 0 0 0 0 (4) for all s ∈ S, dom(s ) = dom(s) = e \ κξ, and either s ⊇ s or s ⊆ s, 0 00 ∗ 00 0 00 (5) for every s ∈ S and every s ∈ P2 , if s ⊆ s then s ∈ S, i.e., S is tree-like, 8 ARTHUR W. APTER, IOANNA M. DIMITR´IOU, AND PETER KOEPKE

(6) for every s0 ∈ S with s0 ⊇ s, if α(s0) is of type 1 then 0 0 0 {β ; s ∪ {(α(s ), m(s ), β)} ∈ S} ∈ Φα(s0), 0 0 0 0 0 (7) for every s ∈ S with s ⊇ s, if α(s ) is of type 2 and cf α(s ) ≥ κξ then 0 0 0 {β ; s ∪ {(α(s ), m(s ), β)} ∈ S} ∈ Φα(s0),s0(cf0α(s0))(m(s0)), and 0 0 0 0 0 (8) for every s ∈ S with s ⊇ s, if α(s ) is of type 2 and cf α(s ) < κξ then 0 0 0 {β ; s ∪ {(α(s ), m(s ), β)} ∈ S} ∈ Φα(s0),S G∗(cf0α(s0))(m(s0)).

Q is partially ordered by (s, S) ≤Q (r, R) iﬀ S ⊆ R.

This deﬁnition means that Q is like P3 but restricted to ordinals and ultraﬁlters ˙ at and above κξ. 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 Ee, cut at κξ. 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 → κξ is in ∗ some generic extension V [G Ee], that Ee 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 τ1, . . . , τk 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(τ1, . . . , τk). 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 κξ, we get that all ultrafilters involved in the definition of Q can be extended to ∗ κξ-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 κξ-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 κξ (Lemma A.4 in the appendix). Therefore Ee cannot collapse κξ, and so the ∗ collapsing function f : δ → κξ cannot exist in V [G Ee]. This completes the proof of Theorem 2.5. qed Next we will see that we singularised the targeted ordinals. This is similar to [Git80, Lemma 3.4]. Lemma 2.6. Every cardinal in (Regκρ )V has cofinality ω in V (G). Thus every cardinal in the interval (ω, κρ) is singular. Proof. Let α ∈ Regκρ . For every β < α, the set

Dβ := {(t, T ) ∈ P3 ; ∃n < ω(t(α)(n) ≥ β)} S is dense in (P3, ≤). Hence fα := {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 THE FIRST MEASURABLE CAN BE THE FIRST UNCOUNTABLE REGULAR 9

κρ by Ea, where a is the smallest subset of Reg that contains α and is closed under 0 cf . Therefore fα ∈ V (G). qed 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, κρ), only the former strongly compact cardinals and their (singular) limits remain cardinals, i.e., that all cardinals of V that are between the κξ and their (singular) limits have collapsed.

Lemma 2.7. For every ordinal ξ ∈ [−1, ρ) and every α ∈ (κξ, κξ+1), (|α| = V (G) κξ) . 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 inaccessi- V (G) ble α ∈ (κξ, κξ+1), we have that (|α| = κξ) .

Fix α ∈ (κξ, κξ+1) strongly inaccessible. We have a bijection hα : Pκξ (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. s Let Pα be the forcing that consists of all injective Hα-trees, i.e., of sets T such that

• T consists of finite injective sequences of elements of Pκξ (α), • (T, E) is a tree, where E denotes end extension, • T has a trunk trT , i.e., an element such that for every t ∈ T , either t E trT or trT E t, and _ • for every t ∈ T such that t D trT , the set {x ∈ Pκξ (α); t hxi ∈ T } of the E-successors of t in T , is in the ultrafilter Hα. s The forcing Pα is partially ordered by S ≤ T ⇐⇒ S ⊆ T. Towards the isomorphism, define a function f from the injective finite sequences of elements of Pκξ (α) to P2 by

f(t) := {(α, m, β); m ∈ dom(t) ∧ β = hα(t(m))}. s Deﬁne another function i : Pα → E{α} by i(T ) := (f(trT ), {f(t); t ∈ T ∧ t D trT }).

This i is indeed a function from T to E{α} because hα is a bijection. In fact, this i is a bijection itself. It is easy to see that it also preserves the ≤ relation of the s forcings, so Pα and E{α} are isomorphic.

ˆ s Now let G be Pα-generic. Because Hα is ﬁne we have that for every β < α the set s Dβ := {T ∈ Pα ; ∃m ∈ ω(β ∈ trT (m))} is dense in P3. So [ [ α = ( Gˆ)(n). n<ω S ˆ But each ( G)(n) is in Pκξ (α), hence it has cardinality less than κξ < α. Therefore s in any forcing extension of V via Pα, α has become a countable union of sets of cardinality less than κξ and therefore is collapsed to κξ. So there is an E{α}-name for a collapsing function from κξ to α, which can be seen as a P3-name in HS for such a function, supported by E{α}. qed Next we show that the regular cardinals of type 2 have collapsed to the singular limit of strongly compacts below them. 10 ARTHUR W. APTER, IOANNA M. DIMITR´IOU, 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 : P α (α) → α. Let e be the smallest α,ν κν κρ 0 ∗ ﬁnite subset of Reg that contains α and is closed under cf . Look at V [G Ee]. 0 Let hγi ; i ∈ ωi be the Prikry sequence added to cf α, and let hαi ; i ∈ ωi be the Prikry sequence added to α. For each i ∈ ω, let A := h−1 (α ). i α,γi i

We want to show that for each δ ∈ α, there is some i ∈ ω such that δ ∈ Ai. Fix

δ ∈ α. For all i ∈ ω, the V -ultraﬁlter Hα,γi is ﬁne, so

α {A ∈ Pκ (α); δ ∈ A} ∈ Hα,γi . γi So for every i ∈ ω, the set Z := {ζ ∈ α ; δ ∈ h−1 (ζ)} ∈ Φ . i α,γi α,γi Deﬁne the set D := {(t, T ) ∈ E ; ∃i ∈ dom(t)(δ ∈ h−1 (t(α)(i)))}. δ e α,γi ∗ This is dense in Ee and δ was arbitrary. Therefore in V [G Ee], we have that S α = i∈ω Ai is a countable union of ≤ β-sized sets, and thus there is a symmetric name for a collapse of α to β, supported by Ee. qed

We summarise our results on the cardinal structure of the interval (ω, κρ).

Corollary 2.9. An uncountable cardinal of V (G) that is less than or equal to κρ is a successor cardinal in V (G) iﬀ it is in {κξ ; ξ ≤ ρ}. Thus in V (G), for every ξ ≤ ρ we have that κξ = ℵξ+1. Also, an uncountable cardinal of V (G) that is less than or equal to κρ is a limit cardinal in V (G) iﬀ it is a limit in the sequence hκα ; α < ρi in V . Proof. This follows inductively, using Theorem 2.5, Lemma 2.7, and Lemma 2.8. qed 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 2.10. In V (G), countable unions of countable sets are not necessarily countable. In particular, every set in Hκρ 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 L´evy-Solovay theorem [LS67], which says that measurability is preserved under small forcing. In particular, it says that if U is a normal measure over κρ, 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 κρ and the existence of a normal measure over κρ is preserved to V (G). We will prove that if U is a normal measure for κρ in the ground model, then the following set deﬁned in V (G), 0 U := {Y ⊆ κρ ; ∃X ∈ U(X ⊆ Y )}, THE FIRST MEASURABLE CAN BE THE FIRST UNCOUNTABLE REGULAR 11

0 is a normal measure over κρ in V (G). This U is clearly a filter in V (G), so it remains to show that it is also a κρ-complete normal ultrafilter. For this, we need to use the approximation lemma for V (G). 0 To show that U is an ultrafilter, let X ⊆ κρ, X ∈ V (G), and let X˙ ∈ HS be a name for X, supported by Ee ∈ I. By the approximation lemma, we have that ∗ X ∈ V [G Ee], so we can use the Lévy-Solovay theorem [LS67] to see that either 0 0 X ∈ U or κρ \ X ∈ U . 0 To show that U is κ-complete, let γ < κρ and hXδ ; δ < γi be a sequence of sets 0 in U . Let σ ∈ HS be a name for this sequence and let Ee0 ∈ 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 Ee0 ]. Again by the Lévy-Solovay theorem [LS67] we get that its intersection is in U 0. Therefore 0 U is a measure for κρ in V (G). 0 To show that U is normal, let f : κρ → κρ, f ∈ V (G) be regressive. Since f can ∗ be coded by a set of ordinals, by the approximation lemma, f ∈ V [G Ee00 ] for 0 some Ee00 ∈ I. Again, by the Lévy-Solovay theorem [LS67], we will get a set in U on which f is constant. qed The construction in this paper is a generalised construction. For particular results, e.g., ℵω+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 ℵω+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 ℵρ+1 is, e.g., weakly compact, Erd˝os,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 Ee. 2 2 J := {(t, T ) ∈ Ee ; ∀q ∈ T ∀α ≥ κξ∀n < ω( if (α, n) ∈ dom (q) \ dom (t) and 0 cf α < α then the ultraﬁlter Φα,q(cf0α)(n) is κξ-complete)} Proof. First notice that the set 0 D := {(t, T ) ∈ Ee ; ∀α ∈ dom(t)(dom(t(α)) = dom(t(cf α)))} a is dense in Ee. This proof is similar to the proof that for a ∈ G, P is dense in P3. Now we will prove that for every (t, T ) ∈ D there is a T 0 ⊆ T such that (t, T 0) ∈ J. 0 0 For every α ∈ dom(t) \ κξ such that cf α < α, let λα be the least ordinal ν < cf α α such that κν ≥ κξ. 12 ARTHUR W. APTER, IOANNA M. DIMITR´IOU, AND PETER KOEPKE

0 0 0 We have that (t, T ) ∈ J iff for all q ∈ T , if α(q) ≥ κξ and cf α(q) < α(q) 0 then q(cf α(q))(m(q)) ≥ λα(q). This equivalence is true because the right hand side of the implication above ensures that the ultrafilter Φα,q(cf0α),m(q) is κξ-complete. Define 0 0 b := {cf α ; α ∈ dom(t) \ κξ ∧ cf α < α}, and for β ∈ b define 0 cβ := max{λα ; α ∈ dom(t) \ κξ and cf α = β < α}. Then (t, T 0) ∈ J if for all q ∈ T 0 and (α, m) ∈ dom2(q) \ dom2(t) such that cf0α ∈ b 0 we have that q(cf α)(m) ≥ ccf0α. So let T 0 :={q ∈ T ; ∀α0 ∈ b∀m < ω(if (α0, m) ∈ dom2(q) \ dom2(t) 0 then q(α )(m) ≥ cα0 )}. Clearly, this (t, T 0) ∈ J. qed

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-nameσ ¯ such that s ∪ t ∈ P2, σ = (ˇs, σ¯)ˇ, and 0 ∗ 0 for all π ∈ dom(¯σ) there is a s ∈ P2 such thats ˇ = π. (b)( t, T ) sˇ ∈ σ¯, (c)( t, T ) ∀π(π ∈ σ¯ → dom(π) = dom(ˇs) ∧ (π ⊆ sˇ ∨ π ⊇ sˇ)), 0 ˇ ˇ ˇ ˇ ˇ ˇ (d)( t, T ) ∀π(π ∈ σ¯ ∧ α(π) = cf α(π) → ∃X(X ∈ Φα(π) ∧ ∀β(β ∈ X ↔ π ∪ {(ˇα(π), mˇ (π), βˇ)} ∈ σ¯))), 0 0 (e)( t, T ) ∀π(π ∈ σ¯ ∧ α(π) > cf α(π) ∧ cf α(π) ≥ κξ → ˇ ˇ ˇ ˇ ˇ ˇ ˇ ∃X(X ∈ Φα(π),π(cf0α(π))(m(π)) ∧ ∀β(β ∈ X ↔ π ∪ {(ˇα(π), mˇ (π), β)} ∈ σ¯))), 0 0 (f)( t, T ) ∀π(π ∈ σ¯ ∧ α(π) > cf α(π) ∧ cf α(π) < κξ → ˇ ˇ ˇ ˇ ˇ ˇ ˇ ∃X(X ∈ Φα(π)Γ(cf0α(π))(m(π)) ∧ ∀β(β ∈ X ↔ π ∪ {(ˇα(π), mˇ (π), β)} ∈ σ¯))), S ∗ where Γ is the standard E-name for {t ; ∃T ((t, T ) ∈ G )}, 0 0 ˇ∗ 0 0 (g)( t, T ) ∀π∀π (π ∈ σ¯ ∧ π ∈ P2 ∧ π ⊆ π → π ∈ σ¯). The name for the ordering on Q˙ is deﬁned as

((σ, τ)ˇ, (t, T )) ∈ ≤ ˙ : ⇐⇒ (t, T ) σ ⊆ τ. Q ˙ G∗ From the forcing theorem we have that Q = Q. For every (t, T ) ∈ P3 and t0 ∈ T such that t0 ⊇ t deﬁne (t, T ) ↑ (t0) := (t0, {t00 ∈ T ; t00 ⊆ t0 or t ⊆ t00}), the extension of (t, T ) with trunk t0. If t0 ⊆ t then we conventionally take (t, T ) ↑ (t0) := (t, T ).

˙ Deﬁne a map i : J → E∗Q. For (r, R) ∈ J, we take i((r, R)) = ((r1,R1), ρ): ⇐⇒ ∗ (i)( r1,R1) := (r, R) Ee∩κξ , (ii) ρ = (ˇr2, ρ¯)ˇ, where r2 := r (e \ κξ) and for all π ∈ dom(¯ρ), there is an 0 0 ∨ r ∈ R such that (π = (r (e \ κξ)) , 0 0 0 ∨ (iii) For all r ∈ R with r ( r we have that ((r (e \ κξ)) , (r1,R1)) ∈ ρ¯, (iv) For all r0 ∈ R with r0 ⊇ r we have that

0 ∨ 0 ((r (e \ κξ)) , (r1,R1) ↑ (r (e ∩ κξ))) ∈ ρ.¯ (v) No other elements are inρ ¯. THE FIRST MEASURABLE CAN BE THE FIRST UNCOUNTABLE REGULAR 13

˙ Claim 1. For all (r, R) ∈ J, i((r, R)) = ((r1,R1), ρ) ∈ E ∗ Q.

Proof of claim. That (r1,R1) ∈ E is immediate. So we must show that ˙ (r1,R1) ρ ∈ Q. Requirement (a) clearly holds with r2 := r (e \ κξ).

For (b) we want that (r1,R1) rˇ2 ∈ ρ¯ which holds because (ˇr2, (r1,R1)) ∈ ρ¯.

For (c) we want that

(r1,R1) ∀π(π ∈ ρ → dom(π) = dom(ˇr2) ∧ (π ⊆ rˇ2 ∨ π ⊇ rˇ2)) or equivalently that 0 0 ∀π ∈ V E∀(b, B) ≤ (r1,R1)∃(b ,B ) ≤ (b, B) 0 0 0 0 ((b ,B ) ¬π ∈ ρ¯ or (b ,B ) (dom(π) = dom(ˇr2) ∧ (π ⊆ rˇ2 ∨ π ⊇ rˇ2))).

0 0 Let π ∈ V E and (b, B) ≤ (r1,R1) be arbitrary and let (b ,B ) ≤ (b, B) decide the 0 0 0 0 formula π ∈ ρ¯. Assume that (b ,B ) 6 ¬π ∈ ρ¯. Then we have that (b ,B ) π ∈ ρ¯. By the deﬁnition of ρ there is some r0 ∈ R such that 0 0 0 ∨ (b ,B ) π = (r (e \ κξ)) .

Since (r, R) is a condition in P3 we get that 0 0 (b ,B ) dom(π) = dom(ˇr2) ∧ (π ⊆ rˇ2 ∨ π ⊇ rˇ2).

For (d) we want to show that for every π ∈ V E, 0 (r1,R1) (π ∈ ρ¯ ∧ α(π) = cf α(π) → ˇ ˇ ˇ ˇ ˇ ˇ ˇ ∃X(X ∈ Φα(π) ∧ ∀β(β ∈ X ↔ π ∪ {(ˇα(π), mˇ (π), β)} ∈ ρ¯))). 0 0 As before, let π ∈ V E and (b, B) ≤ (r1,R1) be arbitrary and let (b ,B ) ≤ (b, B) decide the formula π ∈ ρ¯ ∧ α(π) = cf0α(π). Assume that 0 0 0 (b ,B ) 6 ¬(π ∈ ρ¯ ∧ α(π) = cf α(π)). 0 0 0 Then (b ,B ) (π ∈ ρ¯ ∧ α(π) = cf α(π)). 0 0 0 0 Let r ∈ R be such that (b ,B ) ≤ (r1,R1) ↑ (r (e ∩ κξ)) and 0 0 0 ∨ (b ,B ) π = (r (e \ κξ)) . 0 0 0 0 Call r1 := r (e ∩ κξ) and r2 := r (e \ κξ).

We want to show that 0 0 ˇ ˇ ˇ ˇ ˇ ˇ ˇ (b ,B ) ∃X(X ∈ Φα(π) ∧ ∀β(β ∈ X ↔ π ∪ {(ˇα(π), mˇ (π), β)} ∈ σ¯)). 0 0 Case 1, if α(r ) = α(r2) ≥ κξ. Then let 0 0 0 0 X := SucR(r ) = {β ; r ∪ {(α(r ), m(r ), β)} ∈ R}, and note that X ∈ Φ 0 and α(r2) 0 0 0 0 0 0 ∨ 0 r ∪ {(α(r ), m(r ), β)} ∈ R ⇐⇒ ((r2 ∪ {(α(r2), m(r2), β)}) , (r1,R1) ↑ (r1)) ∈ ρ¯ ⇐⇒ β ∈ X.

Let βˇ ∈ V E be arbitrary. We want that 0 0 ˇ ˇ ˇ ˇ ˇ ˇ (b ,B ) (β ∈ X ∧π ∪{(ˇα(π), mˇ (π), β)} ∈ ρ¯)∨(β 6∈ X ∧π ∪{(ˇα(π), mˇ (π), β)} 6∈ ρ¯), 14 ARTHUR W. APTER, IOANNA M. DIMITR´IOU, AND PETER KOEPKE

or equivalently that ∀(c, C) ≤ (b0,B0)∃(c0,C0) ≤ (c, C) 0 0 ˇ ˇ ˇ ((c ,C ) (β ∈ X ∧ π ∪ {(ˇα(π), mˇ (π), β)} ∈ ρ¯) or 0 0 ˇ ˇ ˇ (c ,C ) (β 6∈ X ∧ π ∪ {(ˇα(π), mˇ (π), β)} 6∈ ρ¯)). So let (c, C) ≤ (b0,B0) be arbitrary and let (c0,C0) ≤ (c, C) be stronger than 0 ˇ (r1,R1) ↑ (r1) and decide π ∪ {(ˇα(π), mˇ (π), β)} ∈ ρ¯. Clearly this (c0,C0) satisﬁes 0 0 ˇ ˇ ˇ (c ,C ) (β ∈ X ∧ π ∪ {(ˇα(π), mˇ (π), β)} ∈ ρ¯) or 0 0 ˇ ˇ ˇ (c ,C ) (β 6∈ X ∧ π ∪ {(ˇα(π), mˇ (π), β)} 6∈ ρ¯) and we’re done with this case.

0 00 0 00 0 Case 2, if α(r ) < κξ, then let r ⊇ r be such that r (e \ κξ) = r2 and 00 0 α(r ) = α(r2) ≥ κξ. The rest follows as in Case 1.

For (e), (f), and (g) we proceed similarly. qed 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 = (ˇs, σ¯). By ˙ (a) of the definition of Q, r ∈ P2. 0 0 0 0 If r ∈ P2 is such that r ⊆ r then let r ∈ R. For r ⊇ r we define R recursively 0 0 0 as follows. Let r ∈ P2 be such that r ⊇ r and r ∈ R. 0 0 0 0 • If α(r ) < κξ then r ∪ {(α(r ), m(r ), β)} ∈ R : ⇐⇒ 0 0 0 (r ∪ {(α(r ), m(r ), β)}) (e ∩ κξ) ∈ T. 0 0 0 0 • If α(r ) ≥ κξ then r ∪ {(α(r ), m(r ), β)} ∈ R : ⇐⇒ 0 0 0 0 ∨ (t, T ) ↑ (r (e ∩ κξ)) ((r (e \ κξ)) ∪ {(α(r ), m(r ), β)}) ∈ σ.¯

0 0 0 0 0 Subclaim 1. For every r ∈ R, call r1 := r (e ∩ κξ) and r2 := r (e \ κξ). 0 0 Then (t, T ) ↑ (r1) rˇ2 ∈ σ¯.

Proof of subclaim. Since by the definition of R and Q˙ this holds for all r0 ⊆ r, we’ll use induction with base case r0 = r. For r0 = r it holds with (t0,T 0) = (t, T ) due to (b) of the definition of Q˙ . So assume it holds for r0 and let r0 ∪ {(α(r0), m(r0), β)} ∈ R be arbitrary. 0 0 0 0 If α(r ) < κξ then α(r ) = α(r1) and by the definition of R we have that (r1 ∪ {(α(r0), m(r0), β)}) ∈ T . By the induction hypothesis we get that 0 0 0 0 0 (t, T ) ↑ (r1 ∪ {(α(r ), m(r ), β)}) ≤ (t, T ) ↑ (r1) rˇ2 ∈ σ.¯ 0 0 0 If α(r ) ≥ κξ then α(r ) = α(r2) and by the definition of R we have that 0 0 0 0 ∨ (t, T ) ↑ (r1) (r2 ∪ {(α(r2), m(r2), β)}) ∈ σ¯. qed Subclaim 1

Subclaim 2. (r, R) ∈ J

Proof of subclaim. To show that (r, R) ∈ P3 we only need to verify (6) and (7) of the deﬁnition of P3. THE FIRST MEASURABLE CAN BE THE FIRST UNCOUNTABLE REGULAR 15

0 0 0 0 0 For (6), let r ∈ R with r ⊇ r and α(r ) of type 1. Call r1 := r (e ∩ κξ) and 0 0 r2 := r (e \ κξ).

0 0 0 0 0 0 0 If α(r ) < κξ then r ∪ {(α(r ), m(r ), β)} ∈ R iff r1 ∪ {(α(r1), m(r1), β)} ∈ T . Since (t, T ) ∈ E we get that 0 0 0 0 SucR(r ) = {β ; r1 ∪ {(α(r1), m(r1), β)} ∈ T } ∈ Φα(r0). 0 So assume that α(r ) ≥ κξ. We have that 0 (t, T ) ∀π(π ∈ σ¯ ∧ π ⊇ sˇ ∧ α(π) = cf α(π) → ˇ ˇ ˇ ˇ ˇ ˇ ∃X(X ∈ Φα(π) ∧ ∀β(β ∈ X ↔ π ∪ {(α(π), m(π), β)} ∈ σ¯))). By Subclaim 1 we have that 0 0 0 0 0 0 (t, T ) ↑ (r1) rˇ2 ∈ σ¯ ∧ rˇ2 ⊇ sˇ ∧ α(ˇr2) = cf α(ˇr2), 0 0 0 0 thus (t, T ) ↑ (r ) ∃Xˇ(Xˇ ∈ Φˇ 0 ∧ ∀βˇ(βˇ ∈ Xˇ ↔ rˇ ∪ {(α(ˇr ), m(ˇr ), β)} ∈ σ¯)). 1 α(ˇr2) 2 2 2 0 0 0 So for some (t ,T ) ≤ (t, T ) ↑ (r ) there is some X 0 0 ∈ Φ 0 such that for 1 (t ,T ) α(r2) every βˇ ∈ V E we have that 0 0 ˇ ˇ 0 0 0 (1) (t ,T ) β ∈ X(t0,T 0) ↔ rˇ2 ∪ {(α(ˇr2), m(ˇr2), β)} ∈ σ.¯ T 0 0 0 Define the set X := {X(t0,T 0) ;(t ,T ) ≤ (t, T ) ↑ (r1) and (1) holds}. Since κ is inaccessible, | | < κ , and Φ 0 is κ -complete, we have that X ∈ Φ 0 ξ E ξ α(r2) ξ α(r2) 0 0 0 and (1) holds for (t, T ) ↑ (r1) and X. Let β ∈ X. Then (t, T ) ↑ (r1) rˇ2 ∪ 0 0 ˇ 0 0 0 {(α(ˇr2), m(ˇr2), β)} ∈ σ¯ which by the definition of R means that r ∪{(α(r ), m(r ), β)} ∈ 0 R. So X ⊆ SucR(r ) ∈ Φα(r0).

0 0 0 0 0 For (7), let r ∈ R with r ⊇ r and α(r ) of type 2. Again, call r1 := r (e ∩ κξ) 0 0 and r2 := r (e \ κξ). 0 0 0 If α(r ) ≥ κξ and cf α(r ) < κξ then we have that for every π ∈ V E, 0 (t, T ) (π ∈ σ¯ ∧ π ⊇ sˇ ∧ α(π) > cf α(π) < κξ → ˇ ˇ ˇ ˇ ˇ ˇ ∃X(X ∈ Φα(π),Γ(cf0α(π))(m(π)) ∧ ∀β(β ∈ X ↔ π ∪ {(α(π), m(π), β)} ∈ σ¯))). By Subclaim 1 we have that 0 0 0 0 0 0 (t, T ) ↑ (r1) rˇ2 ∈ σ¯ ∧ rˇ2 ⊇ sˇ ∧ α(ˇr2) > cf α(ˇr2) < κξ. 0 With the same arguments as for (6), there is some X ∈ V such that (t, T ) ↑ (r1) Xˇ ∈ Φˇ 0 0 0 0 and for every βˇ ∈ V E we have that α(ˇr2),Γ(cf α(ˇr2))(m(ˇr2)) 0 ˇ ˇ 0 0 0 ˇ (t, T ) ↑ (r1) β ∈ X ↔ rˇ2 ∪ {(α(ˇr2), m(ˇr2), β)} ∈ σ.¯ 0 0 0 0 But since r1 ∪ r2 = r ∈ P2, we have that (t, T ) ↑ (r1) decides the value of 0 0 0 0 0 0 0 0 Γ(cf α(ˇr ))(m(ˇr )) to be γ := r (cf α(r ))(m(r )). So (t, T ) ↑ (r ) Xˇ ∈ Φˇ 0 0 . 2 2 1 2 2 1 cf α(ˇr2),γ Since X ∈ V and Φ 0 0 ∈ V we have that X ∈ Φ 0 0 . So take an cf α(ˇr2),γ cf α(ˇr2),γ 0 0 0 0 ∨ arbitrary β ∈ X. Then (t, T ) ↑ (r1) (r2 ∪ {(α(r2), m(r2), β)}) ∈ σ¯ which by the deﬁnition of R means that r0 ∪ {(α(r0), m(r0), β)} ∈ R and consequently 0 X ⊆ SucR(r ) ∈ Φα(r0),r0(cf0α(r0))(m(r0)).

0 0 0 0 Similarly for the other cases where α(r ) > κξ and cf α(r ) ≥ κξ, and α(r ) < κξ.

To conclude Subclaim 2 we want to show that the last condition for member- 0 ship in J is fulﬁlled, i.e., if r ∈ R, α ≥ κξ, and n < ω are such that (α, n) ∈ 2 0 2 0 0 dom (r ) \ dom (t) and cf α < α, then Φα,r0(cf0α)(n) is κξ-complete. Let q ⊆ r be such that for some q0 ∈ R, q = q0 ∪ {(α, n, β)}, α(q0) = α, and m(q0) = n. Note that r0(cf0α)(n) = q0(cf0α)(n). 16 ARTHUR W. APTER, IOANNA M. DIMITR´IOU, AND PETER KOEPKE

0 If cf α ≥ κξ then clearly Φα,q0(cf0α)(n) is κξ-complete.

0 0 If cf α < κξ then note that q (e ∩ κξ) ∈ T and (t, T ) ∈ E, i.e., for some ∗ 0 0 (s, S) ∈ J,(s, S) Ee∩κξ = (t, T ). So q (cf α)(n) must be high enough for the ultraﬁlter Φα,q0(cf0α)(n) to be κξ-complete. qed Subclaim 2

Lastly, we want to show that i((r, R)) ≤ ˙ ((t, T ), σ). Let i((r, R))) = ((r1,R1), ρ), E∗Q 0 and ρ = (ˇr2, ρ¯)ˇ. By the deﬁnition of R and of i we immediately get that (r1,R1) ≤E (t, T ). So it remains to show that (r1,R1) ρ¯ ⊆ σ¯, i.e., (r1,R1) ∀π(π ∈ ρ¯ → π ∈ σ¯), or equivalently that 0 0 0 0 0 0 ∀π ∈ V E∀(b, B) ≤ (r1,R1)∃(b ,B ) ≤ (b, B)((b ,B ) ¬π ∈ ρ¯ or (b ,B ) π ∈ σ¯). 0 0 So let π ∈ V E and (b, B) ≤ (r1,R1) be arbitrary. There is some (b ,B ) ≤ (b, B) 0 0 0 0 that decides “π ∈ ρ¯”. Assume that (b ,B ) 6 ¬π ∈ ρ¯. Then (b ,B ) π ∈ ρ¯. By the 0 0 0 0 ∨ deﬁnition ofρ ¯, there must be some r ∈ R such that (b ,B ) π = (r (e \ κξ)) 0 0 0 0 0 and (b ,B ) is compatible with (r1,R1) ↑ (r (e ∩ κξ)). Call r1 := r (e ∩ κξ), 0 0 00 00 0 0 0 r2 := r (e\κξ), and let (b ,B ) be stronger than both (b ,B ) and (r1,R1) ↑ (r1).

0 0 0 0 By Subclaim 1, (t, T ) ↑ (r1) rˇ2 ∈ σ¯. Then (r1,R1) ↑ (r1) rˇ2 ∈ σ¯ so 00 00 0 (b ,B ) rˇ2 ∈ σ¯. qed Claim 2 So we have shown that Q can indeed be seen as the top part of Ee, cut in κξ. qed ∗ Lemma A.3. (The Prikry lemma for Q) In V [G ], let τ1, . . . , τk 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(τ1, . . . , τn). 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

Φr := Φα(r). 0 If α(r) is of type 2 and cf α(r) ≥ κξ then call

Φr := Φα(r),r(cf0α(r))(m(r)). 0 0 S ∗ 0 If α(r) is of type 2 and cf α(r) < κξ then let γ ∈ cf α be such that G (cf α(r))(m(r)) = γ, and call Φr := Φα(r),γ .

For all r ∈ P2 ∩ (e \ κξ) × ω × κρ deﬁne

Φ¯ r := {X ⊆ α(r); ∃Y ∈ Φr(Y ⊆ X)}.

For each r ∈ S, the ultrafilter Φr is at least κξ complete. Since |E| < κξ, we can ∗ use arguments from the Lévy-Solovay theorem [LS67] to get that in V [G ], each Φ¯ r is at least κξ-complete as well. Also define the following sets.

S0 := {r ∈ S ; r ⊇ s and ∃R ⊆ S((r, R) ∈ Q and (r, R) Q φ(τ1, . . . , τk))}

S1 := {r ∈ S ; r ⊇ s and ∃R ⊆ S((r, R) ∈ Q and (r, R) Q ¬φ(τ1, . . . , τk))} S2 := {r ∈ S ; r ⊇ s and ∀R ⊆ S(if (r, R) ∈ Q then (r, R) does not decide

φ(τ1, . . . , τk))}. THE FIRST MEASURABLE CAN BE THE FIRST UNCOUNTABLE REGULAR 17

Clearly, S = S0 ∪ S1 ∪ S2. Let e \ κξ := {α0, . . . , αn−1}. We will now enumerate 2 the set ((e \ κξ) × ω) \ dom (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) = (αj, m) for some αj ∈ e \ κξ and m ∈ ω. If j < n − 1 then let

x(i + 1) := (αj+1, m), and if j = n − 1 then

x(i + 1) := (α0, m + 1). For every i ∈ ω deﬁne 2 2 F˜i := {r ∈ S ; dom (r) \ dom (s) = x“(i + 1)}.

Now for i ≤ j we will deﬁne recursively on j − i a set of functions Fi,j : F˜i → 3. For ` < 3 and r ∈ F˜i let Fi,i(r) = ` : ⇐⇒ r ∈ S`.

For i < j let Fi,j(r) = ` : ⇐⇒ the set

{β ; r ∪ {(α(r), m(r), β)} ∈ S and Fi+1,j(r ∪ {(α(r), m(r), β)}) = `} ˜0 ˜ is in the ultrafilter Φr. Define recursively on i < ω a subset Fi ⊂ Fi. Using the ¯ ˜0 ˜ definition and the ω-completeness of the ultrafilter Φs, we find (in V ), a set F0 ⊆ F0 which is homogeneous for all functions in the set {F0,j ; j < ω} and which is such that the set ˜0 {β ; s ∪ {(α(s), m(s), β)} ∈ F0} ˜0 is in Φs. By homogeneous here we mean that for all t1, t2 ∈ F0 and for all 0 ≤ j < ω, F0,j(t1) = F0,j(t2). For i > 0 we take ˜0 ˜ − ˜0 − Fi := {r ∈ Fi ; r ∈ Fi−1 and ∀i ≤ j(Fi,j(r) = Fi−1,j(r ))}, where r− is defined as in Definition 2.3(5). By the induction hypothesis, it follows ˜0 that Fi is homogeneous for all functions in the set {Fi,j ; i ≤ j < ω}. The definition ˜0 of the functions Fi,j implies that for every r ∈ Fi−1, ˜0 (2) {β ; r ∪ {(α(r), m(r), β)} ∈ Fi } ∈ Φr. Define the set ˜ [ ˜0 F := {s} ∪ {Fi ; i < ω}.

˜ Claim. If s1, s2 ∈ F ,(s1,A1), (s2,A2) ∈ Q, and (s1,A1), (s2,A2) ≤Q (s, S), then it is impossible to have that (s1,A1) Q φ(τ1, . . . , τk) and (s2,A2) Q ¬φ(τ1, . . . , τk).

Proof of claim. We have that for some i1, i2 < ω and for every j = 1, 2, 2 2 dom (sj) = dom (s) ∪ {x(`); ` < ij}.

Without loss of generality we may assume that i1 ≤ i2. If i1 < i2 then we can 2 increase the dom (s1), one step at a time until we get i1 = i2. We have that the set E := {β < α(s ); s ∪ {(α(s ), m(s ), β)} ∈ F˜0 } is in Φ 1 1 1 1 i1 s1 and since (s1,A1) is a condition in Q we have that also the set 0 E := {β < α(s1); s1 ∪ {(α(s1), m(s1), β)} ∈ A1} is in Φs1 . 18 ARTHUR W. APTER, IOANNA M. DIMITR´IOU, AND PETER KOEPKE

0 Let β ∈ E ∩ E , lets ¯1 := s1 ∪ {(α(s1), m(s1), β)}, and let A¯1 := {t ∈ A1 ; t ⊇ s¯ }. Then we have thats ¯ ∈ F˜0 ⊆ F˜ and that (¯s , A¯ ) ≤ (s ,A ). Therefore, 1 1 i1 1 1 Q 1 1 ¯ (¯s1, A1) Q φ(τ1, . . . , τn), and 2 2 dom (¯s1) = dom (s) ∪ {x(k); k < i1 + 1}.

This way we keep increasing i1 until we get i1 = i2. Denote i2 −i1 by i. If i = 0 then s1 = s2 = r, therefore (s1,A1) and (s2,A2) are compatible which is a contradiction. ˜0 If i > 0 then s1, s2 ∈ Fi−1. Because (s1,A1) ≤Q (s, S) and (s1,A1) Q φ(τ1, . . . , τn), we have that s1 ∈ S0, thus Fi−1,i−1(s1) = 0. Similarly we get that Fi−1,i−1(s2) = 1 ˜0 which contradicts the homogeneity of Fi−1 for Fi−1,i−1. qed Claim

So to finish the proof of this lemma, we first show that (s, F˜) is indeed a condition 0 ˜ 0 0 in Q. It suffices to show that for every s ∈ F , the set SucF˜ (s ) of successors of s in F˜ is in the ultrafilter Φs0 . We have that 0 0 0 0 0 ˜ SucF˜ (s ) ={β < α(s ); s ∪ {(α(s ), m(s ), β)} ∈ F } 0 0 0 0 ={β < α(s ); s ∪ {(α(s ), m(s ), β)} ∈ F˜i+1, } 0 ˜ 0 where i ∈ ω is such that s ∈ Fi. By (2) we get that SucF˜ (s ) ∈ Φs0 .

Finally, let (s1,A1) ∈ Q be any condition that decides φ(τ1, . . . , τn) and extends ˜ (s, F ). Without loss of generality assume that (s1,A1) Q φ(τ1, . . . , τn). By the ˜ definition of Q we can assume that dom(s1) = e \ κξ, and hence s1 ∈ F and A1 ⊆ F˜. Suppose that (s, F˜) does not decide φ(τ1, . . . , τn), then there must be ˜ some (s2,A2) ≤Q (s, F ) such that (s2,A2) Q ¬φ(τ1, . . . , τn). But this contradicts the Claim. Therefore (s, F˜) decides φ(τ1, . . . , τn). qed ∗ Lemma A.4. In V [G ], the partial order (Q, ≤Q) does not add bounded subsets to κξ. Proof. We work in V [G∗]. Consider the relation ≤∗ ⊆ ≤ , defined as (s, S)≤∗ (r, R) Q Q Q iff s = r and S ⊆ R.

Claim. ( , ≤∗ ) is κ -closed. Q Q ξ Proof of claim. Let γ < κ and let h(s ,S ); ζ < γi be a ≤∗ -descending sequence ξ ζ ζ Q of elements in Q. Since E is small forcing with respect to κξ, we will use the Lévy- Solovay theorem [LS67] to get that all ultrafilters involved in the definition of Q ∗ can be extended to κξ-complete ultrafilters in V [G ]. T Let s = sζ for all ζ < γ and define a set S ⊆ ζ<γ Sζ inductively (above s) as follows. If s0 ⊆ s then s0 ∈ S. Assume that s ⊆ s0 ∈ S. Since for every ζ < γ, 0 0 s ∈ Sζ , there is a set Φ, that is a κξ-complete ultrafilter over α(s ) in V , and such that for every ζ < γ, 0 0 0 0 SucSζ (s ) := {β ; s ∪ {(α(s ), m(s ), β)} ∈ Sζ } ∈ Φ. 0 In particular, we choose these Φ to fit with the definition of P3, i.e., if α(s ) is 0 0 0 of type 1 then Φ = Φα(s0), if α(s ) is of type 2 and cf α(s ) ≥ κξ then Φ = 0 0 0 Φα(s0),s0(cf0α(s0))(m(s0)), and if α(s ) is of type 2 and cf α(s ) < κξ then Φ = Φα(s0),S G∗(cf0α(s0))(m(s0)). By the Lévy-Solovay theorem [LS67], Φ∗ := {X ⊆ α(s0); ∃Y ⊆ X(Y ∈ Φ)} 0 T 0 ∗ is a κξ-complete ultrafilter over α(s ). So ζ<γ SucSζ (s ) ∈ Φ and consequently 0 0 0 there is some Ys ⊆ SucSζ (s ) such that Ys ∈ Φ. We define then 0 0 0 s ∪ {(α(s ), m(s ), β)} ∈ S : ⇐⇒ β ∈ Ys0 . THE FIRST MEASURABLE CAN BE THE FIRST UNCOUNTABLE REGULAR 19

Clearly (s, S) ∈ and for every ζ < γ,(s, S) ≤∗ (s ,S ). qed Claim Q Q ζ ζ

Now to show that (Q, ≤Q) does not add new bounded subsets to κξ, we proceed as usual with Prikry type forcings. Let (s, S) ∈ Q, let τ be a Q-name, γ < κξ, and assume that (s, S) Q τ ⊆ γˇ. Using the Prikry Lemma for Q (by Lemma A.3) and that ( , ≤∗ ) is κ -closed (by the claim), we get a ≤∗ -decreasing sequence Q Q ξ Q h(s, Sζ ); ζ < γi of conditions in Q such that for each ζ < γ,(s, Sζ ) decides the statement “ζˇ ∈ τ”. Again by using the claim we get a condition (s, S0) stronger than all (s, Sζ ), therefore one such that 0 0 ˇ (s, S ) τ =x, ˇ where x = {ζ ∈ γ ;(s, S ) ζ ∈ τ}. qed

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A. W. Apter, The Graduate Center of The City University of New York, Mathe- matics, 365 Fifth Avenue, New York, NY 10016, USA & Department of Mathematics, Baruch College, One Bernard Baruch Way, New York, NY 10010, USA E-mail address: [email protected] URL: http://faculty.baruch.cuny.edu/aapter/

I. M. Dimitr´ıou, Mathematisches Institut der Universitat¨ Bonn, Endenicher Allee 60, 53115 Bonn, Germany E-mail address: [email protected] URL: http://boolesrings.org/ioanna/

P. Koepke, Mathematisches Institut der Universitat¨ Bonn, Endenicher Allee 60, 53115 Bonn, Germany E-mail address: [email protected] URL: http://www.math.uni-bonn.de/people/koepke/index.shtml