TALL CARDINALS IN EXTENDER MODELS

GABRIEL FERNANDES* AND RALF SCHINDLER‡

Abstract. Assuming that there is no inner model with a Woodin cardinal, we obtain a characterization of 휆-tall cardinals in extender models that are iterable. In particular we prove that in such extender models, a cardinal 휅 is a tall cardinal if and only if it is either a strong cardinal or a measurable limit of strong cardinals.

1. Introduction Tall cardinals appeared in varying contexts as hypotheses in the work of Woodin and Gitik but they were only named as a distinct type of large cardinal by Hamkins in [Ham09]. Definition 1.1. Let 훼 be an ordinal and 휅 a cardinal. We say that 휅 is 훼-tall iff there is an elementary embedding 푗 : 푉 → 푀 such that the following holds: a) crit(푗) = 휅, b) 푗(휅) > 훼, c) 휅푀 ⊆ 푀. We say that 휅 is a tall cardinal iff 휅 is 훼-tall for every ordinal 훼. One can compare this notion with that of strong cardinals. Definition 1.2. Let 훼 be an ordinal and 휅 a cardinal. We say that 휅 is 훼-strong iff there is an elementary embedding 푗 : 푉 → 푀 such that the following holds: a) crit(푗) = 휅, b) 푗(휅) > 훼, c) 푉훼 ⊆ 푀. We say that 휅 is a strong cardinal iff 휅 is 훼-strong for every ordinal 훼. In this paper, working under the hypothesis that there is no inner model with a Woodin cardinal, we present a characterization of 휆-tall cardinals in ‘extender models’ (see Definition 2.5) that are ‘self-iterable’ (see Definition 3.10). Given a cardinal 휅, if 휅 is 훼-strong then 휅 is 훼-tall, and the existence of a strong cardinal is equiconsistent with the existence of a tall cardinal (see [Ham09]). We will prove that the following equivalence holds in extender models: Corollary A. Suppose that there is no inner model with a Woodin cardinal, 푉 is an extender model of the form 퐿[퐸] which is iterable. Then given a cardinal 휅 the following equivalence holds: 휅 is a tall cardinal iff 휅 is either a strong cardinal or a measurable limit of strong cardinals.

2010 Mathematics Subject Classification. Primary 03E55. Secondary 03E45. Key words and phrases. Tall cardinals, Strong cardinals, Extender models, Core model. *The author is funded by the European Research Council (grant agreement ERC-2018-StG 802756) as a postdoctoral fellow at Bar-Ilan University. ‡The author is funded by the Deutsche Forschungsgemeinschaft (DFG, German Research Foun- dation) under Germany’s Excellence Strategy EXC 2044 390685587, Mathematics M¨unster:Dy- namics - Geometry - Structure. 1 2 GABRIEL FERNANDES* AND RALF SCHINDLER‡

Remark 1.3. In contrast to Corollary A, Hamkins in [Ham09, Theorem 4.1] adapted Magidor’s results in [Mag76] to prove that it is consistent (assuming the consistency of 푍퐹 퐶 plus a strong cardinal) that there is a model of 푍퐹 퐶 where there exists a cardinal 휅 such that 휅 is a tall cardinal which is neither a strong cardinal nor a limit of strong cardinals. Hence the equivalence from Corollary A does not hold in such model. An extender1 is a means of encoding elementary embeddings of models of (frag- ments of) 푍퐹 퐶 in a set-size object. There are various ways to represent extenders. Extenders are a generalization of measures, in particular, notions such as a ‘critical point’ which are used in the context of measures can also be used when talking about extenders. Extender models are a generalization of G¨odel’sconstructible universe that can accommodate large cardinals. In general, given a predicate 퐸, which can be a set or a proper class, 퐿[퐸] is the smallest inner model2 closed under the operation 푥 ↦→ 퐸 ∩ 푥. Inner models of the form 퐿[퐸] can be stratified using the 퐽-hierarchy: 퐸 ∙ 퐽∅ = ∅, 퐸 퐸 퐸 ∙ 퐽훼+1 := rud퐸({퐽훼 } ∪ 퐽훼 ), 퐸 ⋃︀ 퐸 ∙ 퐽훾 := 휉<훾 퐽휉 for 훾 a limit ordinal, ⋃︀ 퐸 ∙ 퐿[퐸] = 휉∈푂푅 퐽휉 3 where rud퐸 is the closure under rudimentary functions and the function 푥 ↦→ 퐸 ∩ 푥. We are interested in the special case where 퐸 is such that 퐸 : 푂푅 → 푉 and for every ordinal 훼, either 퐸훼 = ∅ or 퐸훼 is a partial extender (see Definition 2.5). That is, 퐸 is a ‘sequence of extenders’. Convention. There are different ways of organizing sequences of extenders, we will use Jensen’s 휆-indexing (see Remark 2.3). Definition 1.4. Suppose 푉 is an extender model of the form 퐿[퐸]. Given a cardinal 4 휅 we define 표(휅) := otp({훽 | 푐푟푖푡(퐸훽) = 휅}) and 푂(휅) := 푠푢푝{훽 | 푐푟푖푡(퐸훽) = 휅}. Our main result, Theorem A, is a level-by-level version of Corollary A. The statement of Theorem A uses the notion of 휇-stable premouse which is introduced in Definition 3.11. Theorem A. Suppose that there is no inner model with a Woodin cardinal and that the universe 푉 is an iterable extender model 퐿[퐸]. Let 휅 < 휇 be regular cardinals. Suppose further that 퐿[퐸]|휇 is 휇-stable above 휅. Then 휅 is 휇-tall iff 표(휅) > 휇 or (︁ )︁ 표(휅) > 휅+ ∧ 푠푢푝{휈 < 휅 | 표(휈) > 휇} = 휅

We prove Theorem A in Section 4. The rest of this introduction gives a technical overview of our proof of Theorem A.

1For an introduction to the theory of extender we recommend [Kan09]. 2An inner model is a transitive proper class that models 푍퐹 . 3 See [SZ10] for the definition of 퐸rud . 4Note that our definitions of 푂(휅) and 표(휅) are not standard, because we do not only consider extenders which are total, but also consider partial extenders. For this reason, our definitions differ from those in other references such as[Zem02] and [GM96]. TALL CARDINALS IN EXTENDER MODELS 3

In order to prove Theorem A we will need some results from core model theory. Specifically, we shall need the core model 풦 below a Woodin cardinal. This model is an extender model5 that generalizes the covering, absoluteness, and definability properties of 퐿. The following result due to Jensen and Steel guarantees that such a model exists.

Theorem 1.5. ([JS13]) There are Σ2 formulae 휓풦(푣) and 휓Σ(푣) such that, if there is no inner model with a Woodin cardinal, then

(i) 풦 = {푣 | 휓풦(푣)} is an inner model satisfying 푍퐹 퐶, 푉 푉 [푔] 푉 푉 [푔] (ii) 휓풦 = 휓풦 , and 휓Σ = 휓Σ ∩ 푉 , whenever 푔 is 푉 -generic over a poset of set size, (iii) For every singular strong limit cardinal 휅, 휅+ = (휅+)풦, (iv) {푣 | 휓Σ(푣)} is an iteration strategy for 풦 for set-sized iteration trees, and moreover the unique such strategy,

(v) 풦|휔1 is Σ1 definable over 퐽휔1 (R). We now describe our strategy for proving Theorem A. One direction is due to Hamkins (see Theorem 4.26 and Theorem 4.27), so we start from the assumption that 휅 and 휇 are cardinals such that 휇 > 휅, 휅 is 휇-tall and 푗 witnesses that 휅 is 휇-tall. Note that this implies that 휅 is measurable and that 휇 < 푗(휅). We now consider two cases. Either 휅 is a limit of cardinals 훽 such that 표(훽) > 휇, in which we case we get the second alternative of the direction we are proving using Lemma 4.23. So, suppose that 휅 is not a limit of cardinals 훽 such that 표(훽) > 휇, and then we work towards proving that 표(휅) > 휇. As a first step, we combine Lemma 4.12 and Theorem 4.15 to obtain that 푗 is an 풯 iteration map coming from an iteration tree 풯 on 퐿[퐸] such that 푗 = 휋0,∞. This is Lemma 4.17. We will spend the main part of the proof of Theorem A analyzing the iteration tree 풯 . We shall prove that 표(휅) > 휇 by contradiction. That is, we shall start with the assumption that 표(휅) ≤ 휇. Then, we will find Θ and 훽* such that 훽* < 휅 ≤ Θ ≤ 휇 and 휏 ∈ (훽*, Θ] implies 표(휏) < Θ (see Lemma 3.24). Using the upper bounds that 풯 we obtain in Section3 we shall finally prove that 푗(휅) = 휋0,∞(휅) ≤ Θ ≤ 휇, which 풯 will contradict the fact that 휇 < 푗(휅) = 휋0,∞(휅). The following is essentially a reformulation of Theorem A which we also prove in Section 4. Theorem B. Suppose there is no inner model with a Woodin cardinal and 퐿[퐸] is an extender model that is self-iterable. Let 휅, 휇 be ordinals such that 휅 < 휇 and 휇 is a regular cardinal. If 퐿[퐸]|휇 is 휇-stable above 휅, then (휅 is 휇-tall)퐿[퐸] iff (표(휅)) > 휇)퐿[퐸] (1) or (표(휅) > 0 ∧ 휅 = sup{휈 < 휅 | 표(휈) > 휇})퐿[퐸] It follows that if 퐿[퐸] is weakly iterable and 퐿[퐸]|휇 is 휇-stable above 휅, then (휅 is 휇-tall)퐿[퐸] iff (1) holds.

2. Preliminaries In this section we summarize the notation that will be used in this paper. We follow closely the notation used in [Zem02].

5See [SW16] for an example where 풦 is not an extender model. 4 GABRIEL FERNANDES* AND RALF SCHINDLER‡

퐸 Definition 2.1. We say that ℳ := ⟨퐽훼 , ∈, 퐸  훼, 퐸훼⟩ is an acceptable 퐽-structure iff ℳ is a transitive amenable structure and for every 휉 < 훼 and 휏 < 훼휔 if (풫(휏) ∩ 퐸 퐸 퐸 퐸 퐽휉+1) ∖ 퐽휉 ̸= ∅, then there is 푓 : 휏 → 퐽휉 surjective such that 푓 ∈ 퐽휉+1. Acceptablity is a strong form of 퐺퐶퐻. We can define fine structure for accept- able 퐽-structures. Let ℳ be an acceptable 퐽-structure. We shall write (see [Zem02, Chapter 2] ): ∙ ht(ℳ) for the ordinal ℳ ∩ 푂푅, ∙ 휌푛(ℳ) for the 푛-th projectum of ℳ, ℳ ∙ 푃푛 for the set of good parameters (i.e., for the set of parameters witnessing 휌푛(ℳ) is the 푛-th projectum), ℳ ℳ ∙ 푝푛 for the 푛-th standard parameter of ℳ (i.e, the least element of 푃푛 where least refers to the canonical well-order of [푂푅]<휔), 푛,푝 푛,푝 ∙ ℎℳ for the canonical Σ1 Skolem function of ℳ , ˜푛 (푛−1) ∙ ℎℳ for the good uniformly Σ1 (ℳ) function with two parameters which is the result of iterated composition of the Skolem functions of the 푖-th reducts. 퐴 Definition 2.2. Let ℳ = ⟨퐽훼 , ∈, 퐹 ⟩ be an acceptable 퐽-structure. We say that ℳ is a coherent 퐽-structure iff there is an훼 ¯ < 훼 6 퐴 ∙ 퐹 is a whole extender in 퐽훼¯ where훼 ¯ < 훼, 퐴 ∙ 퐽훼¯ |= “ crit(퐹 ) is the largest cardinal” 퐴 퐴 ∙ 퐽훼 = 푈푙푡0(퐽훼¯ , 퐹 ). 퐴 퐴 Given 훽 < 훼 we define ℳ|훽 := ⟨퐽훽 , ∈, 퐸  휔훽⟩ and ℳ||훽 := ⟨퐽훽 , ∈, 퐸  7 휔훽, 퐸휔훽⟩ . 퐸 We say that ℳ := ⟨퐽훼 , ∈, 퐸  휔훼, 퐸휔훼⟩ is a premouse iff ∙ 퐸 is a set of triples ⟨휈, 푥, 푦⟩ for 휈 ≤ 훼 such that, setting

퐸휔휈 := {⟨푥, 푦⟩ | ⟨휈, 푥, 푦⟩ ∈ 퐸},

the structure ℳ||휈 is coherent whenever 퐸휔휈 ̸= ∅. ∙ For every 휈 ≤ 훼, if 퐸휔휈 ̸= ∅, then 퐸휔휈 is weakly amenable w.r.t. ℳ||휈. ∙ ℳ||휈 is sound for every 휈 < 훼. Remark 2.3 (Indexing). Notice that implicitly in our definition of a premouse we use 휆-indexing, also called Jensen indexing, which means that extenders are indexed at the successor of the image of their critical point under the ultrapower map, i.e., ℳ ℳ if ℳ is a premouse, 퐸 ̸= ∅, 풩 = 푈푙푡0(ℳ||훽, 퐸 ) and 휋 ℳ : ℳ||훽 → 풩 is the 훽 훽 퐸훽 ℳ +풩 ultrapower map, then 훽 = 휋 ℳ (crit(퐸 )) . 퐸훽 훽 Definition 2.4. Let 퐹 be an extender over a premouse ℳ. We denote by 휆(퐹 ) the image of the critical point of 퐹 under the ultrapower map, i.e. if 휋퐹 : ℳ → 푈푙푡0(ℳ, 퐹 ) is the ultrapower map, we let 휆(퐹 ) = 휋퐹 (crit(퐹 )). Definition 2.5. We say that 퐿[퐸] is an extender model iff 퐿[퐸] is a proper class premouse.

3. Upper bounds for the images of ordinals under iteration maps In this section we define iteration trees8 and prove general facts about upper bounds for the images of ordinals under iteration maps.

6See [Zem02, p.42] for the definition of extender and [Zem02, p. 53] for the definition of whole extender. 7Depending on the reference ℳ||훽 and ℳ|훽 may have their roles switched. We stick to the notation in [Zem02]. 8The reader interested in the intuition behind the definition of iteration trees is refered to [MS94]. TALL CARDINALS IN EXTENDER MODELS 5

Definition 3.1. A tree 푇 = ⟨휃, ≤푇 ⟩ on an ordinal 휃 is an iteration tree iff a) 0 is the root of 푇 and each successor ordinal 훼 < 휃 has an immediate 푇 -predecessor pred푇 (훼) < 훼; b) if 훼 < 휃 is a limit ordinal, then 훼 = 푠푢푝{휉 < 훼 | 휉 <푇 훼}

If 푇 ⊆ 휃 is an iteration tree, given 훼 < 훽 elements of 푇 we write (훼, 훽]푇 = {훾 | 훼 <푇 훾 ≤푇 훽} and similarly for (훼, 훽)푇 ,[훼, 훽]푇 . Definition 3.2. Let ℳ be a sound premouse and 휃 ∈ 푂푅. We say that 풯 is an iteration tree 풯 on ℳ with 푙ℎ(풯 ) = 휃 iff 풯 is a 6-tuple9:

풯 = ⟨⟨ℳ훼 | 훼 < 휃⟩, ⟨휈훽 | 훽 + 1 ∈ 휃⟩, ⟨휂훽 | 훽 + 1 < 휃⟩, ⟨휋훼,훽 | 훼 ≤푇 훽 < 휃⟩, 퐷, 푇 ⟩ satisfying: (a) 푇 is an iteration tree. (b) Each ℳ훼 is a premouse and ℳ0 = ℳ. (c) ⟨휋훼,훽 | 훼 ≤푇 훽⟩ is a commutative system of partial maps, where 휋훼,훽 : ℳ훼 → ℳ훽.

(d) Setting 휉훼 = pred푇 (훼 + 1), we have 휂훼 ≤ ℎ푡(ℳ휉훼 ) and for every 훽 < 휃 there

are only finitely many 훼 such that 휉훼 <푇 훽 and 휂훼 < ℎ푡(ℳ휉훼 ).

(e) 퐷 ⊆ 휃 and 훼 ∈ 퐷 iff 휂훼 < ht(ℳ휉훼 ). +ℳ ||휈 (f) If 훼 + 1 < 휃, setting 휅 := crit(퐸ℳ훼 ) and 휏 := 휅 훼 훼 , we have 휏 = 훼 휈훼 훼 훼 훼 +ℳ휉훼 ||휂훼 ℳ훼 ℳ 휅훼 , 퐸  휏훼 = 퐸 휉훼  휏훼, and 휋 : ℳ ||휂 −→* 푈푙푡*(ℳ ||휂 , 퐸ℳ훼 ). 휉훼,훼+1 휉훼 훼 퐸훼 휉훼 훼 휈훼

(g) If 훼 < 휃 is a limit ordinal, then ⟨푀훼, 휋훽,훼 | 훽 < 훼⟩ is the direct limit of the ¯ diagram ⟨ℳ훽, 휋훽,훽¯ | 훽 ≤푇 훽 <푇 훼⟩. Given an iteration 풯 , we denote the objects from the above definition related to 풯 풯 풯 풯 풯 풯 풯 by ℳ훼 , 휈훼 , 휂훼 ,퐷 ,푇 . We shall often write 푇 instead of 푇 . We also set: 풯 풯 ℳ훼 풯 ∙ 퐸훼 := 퐸휈훼 (the extender used to form ℳ훼+1), 풯 풯 ∙ 휅훼 := crit(퐸훼 ), 풯 풯 ∙ 휆훼 := 휆(퐸훼 ), 풯 풯 + ℳ풯 ||휈풯 ∙ 휏훼 := ((휅훼 ) ) 훼 훼 and ∙ 퐷풯 := {훼 + 1 | 휂풯 < ℎ푡(ℳ풯 )}. 훼 휉훼 If in addition 푙ℎ(풯 ) = 훾 + 1 for some 훾 ∈ 푂푅, we write 풯 풯 ∙ℳ ∞ for the last model ℳ훾 in the iteration tree 풯 , ∙ [0, 훾]푇 is called the main branch of 풯 , and 풯 풯 ∙ if 훽 ∈ 푙ℎ(풯 ) and 훽 <푇 훾, we let 휋훽,∞ := 휋훽,훾 . We say that 풯 is a normal iteration tree iff

∙ 휈훽 < 휈훼 whenever 훽, 훼 ∈ 푙ℎ(풯 ) and 훽 < 훼; ∙ 휉훼 = the least 휉 ∈ 퐵 such that 휅훼 < 휆휉; +ℳ풯 ||휂 휉훼 ∙ 휂훼= the maximal 휂 ≤ ℎ푡(푀휉훼 ) such that 휏훼 = 휅훼 . 풯 Remark 3.3. We stress that the maps 휋훽,훽¯ are partial functions. Fact 3.4. If 풯 is a normal iteration tree on a premouse ℳ, then the inductive application of the coherency condition yields: 풯 풯 ∙ℳ 훼 |휈훽 = ℳ훽 |휈훽 whenever 훽 ≤ 훼. 풯 풯 ∙ If 훽 < 훼, then 휈훽 is a successor cardinal in ℳ훼 , but not a cardinal in ℳ훽 풯 풯 when 휈훽 ∈ ℳ훽 .

9 When ℳ is a proper class premouse we allow 휂훼 = 푂푅, and formally we use 휂훼 = ∅. 6 GABRIEL FERNANDES* AND RALF SCHINDLER‡

Convention. In this paper all iteration trees that we will encounter are normal iteration trees. In what follows when we write iteration tree we mean normal iteration tree. Definition 3.5. Let ℳ be a premouse and 풯 an iteration tree on ℳ such that 푙ℎ(풯 ) is a limit ordinal. We say that 푏 is a cofinal wellfounded branch through 풯 iff ∙ 푏 is a branch through 푇 풯 cofinal in 푙ℎ(풯 ), ∙ 퐷풯 ∩ 푏 is finite, ∙ the direct limit along 푏 is well-founded. Remark 3.6. As we work under the hypothesis that there is no inner model with a Woodin cardinal, it follows that if 풯 is an iteration tree on a premouse ℳ and 푙ℎ(풯 ) is a limit ordinal then 풯 has at most one cofinal wellfounded branch through 풯 (see [Zem02][Corollary 9.4.7], [Ste10][Theorem 6.10]). In order to simplify notation we will avoid mentioning iteration strategies in our definition of iterability but we warn the reader that this is not how iterability is usually defined. In general, we would need a strategy to choose cofinal branches at limit stages, see [Zem02][p. 290]. For the reader familiar with iteration strategies, as we work under the hypothesis that there is no inner model with a Woodin cardinal, the strategy of any premouse we encounter will always choose the unique cofinal wellfounded branch. Definition 3.7 (Normal iterability). Let 훼 be a limit ordinal or the proper class 푂푅. We say that ℳ is normaly 훼-iterable iff for any normal iteration tree 풯 on ℳ of length < 훼 the following holds: (a) If 풯 is of limit length, then there is a cofinal wellfounded branch 푏 through 풯 . 풯 (b) If 풯 is of successor length 훾 and 훾+1 < 훼 and 휈 ≥ sup{휈훼 | 훼 < 훾} is such that 풯 ℳ훾−1 ′ 풯 ′ 퐸휈 ̸= ∅, then 풯 has a normal extension 풯 of length 훾 + 1 with 휈훾−1 := 휈. 풯 ′ 풯 * 풯 ′ 풯 ′ ℳ훾 In other words, setting 휈훾−1 = 휈, the ultrapower 푈푙푡 (ℳ 풯 ′ ||휂훾−1, 퐸 풯 ′ ) 휉훾−1 휈훾−1 풯 ′ 풯 ′ is well founded, where 휂훾−1 and 휉훾−1 are determined by the rules of normal iteration trees. When 훼 = 푂푅 and ℳ is normaly 푂푅-iterable we shall omit 푂푅 and write that ℳ is normaly iterable. We shall also need a stronger notion of iterability .

Definition 3.8. Let ℳ be a premouse and 푛 ∈ 휔. We say that 푇⃗ = ⟨풯푘 | 푘 ≤ 푛⟩ 0 is a stack of normal iteration trees on ℳ iff 풯0 is an iteration tree on ℳ0 = ℳ 푘 풯푘−1 푘 and for every 푘 < 푛, ℳ0 ▷ ℳ∞ and 풯푘 is a normal iteration tree on ℳ0 . Definition 3.9 (Iterability for stacks of normal trees). Let ℳ be a premouse. We say that ℳ is iterable iff

풯푛 (a) If 풯 = ⟨풯푘 | 푘 ≤ 푛⟩ is a stack of iteration trees on ℳ, then ℳ∞ is normaly iterable. (b) If 풯 = ⟨풯푛 | 푛 ∈ 휔⟩ is a stack of iteration trees on ℳ such that for each 푛 ∈ 휔, 풯푛 ℳ∞ is normaly iterable, then for all sufficiently large 푛 ∈ 휔 we have that 풯푛 풯푛 풯푛 푛 풯푛 퐷 ∩ 푏 = ∅, where 푏 is the main branch of 풯 so that 휏푛 : ℳ0 → ℳ∞ = 푛+1 ℳ0 is defined for all 푛 sufficiently large. Moreover, the direct limit of the 푛 ℳ0 ’s under the 휏푛’s is wellfounded. Definition 3.10 (Self-iterability). If 퐿[퐸] is an extender model, we say that 퐿[퐸] is self-iterable iff the following sentence in the language {∈, 퐸˙ } holds: 퐸 ˙ ˙ 퐿[퐸] (∀훼(훼 ∈ 푂푅 → ⟨퐽훼 , ∈, 퐸  휔훼, 퐸휔훼⟩ is iterable)) TALL CARDINALS IN EXTENDER MODELS 7

The following definition is a slight variation of the notion of stable premouse defined in [JS13]. Definition 3.11. Let ℳ be a premouse and 휇 a regular cardinal and 휅 ∈ 휇. If ℳ ∩ 푂푅 ≤ 휇 we say that ℳ is 휇-stable above 휅 iff one of the following holds: (1) ℳ ∩ 푂푅 < 휇, or (2) ℳ ∩ 푂푅 = 휇 and one of the following holds: (a) (There is no largest cardinal)ℳ, or (b) There is 훾 < 휇 such that (훾 is the largest cardinal ∧ cf(훾) < 휅)ℳ, or (c) There is 훾 < 휇 such that (훾 is the largest cardinal ∧cf(훾) ≥ 휅)ℳ and there ℳ ℳ is no 훽 such that 퐸훽 is a total measure on ℳ with critical point cf (훾). The next lemma shows that given a regular cardinal 휇, under very general con- ditions the ultrapower of a premouse ℳ of height ≤ 휇 by an extender 퐹 with 휆(퐹 ) < 휇, has height ≤ 휇. Lemma 3.12. Let 휇 be a regular cardinal and ℳ a sound10 premouse such that ℳ ∩ 푂푅 ≤ 휇. Let 휅 < 훼 ≤ 휇 and 퐹 be such that: ∙ 퐹 is an extender over ℳ||훼 and 훼 is the largest ordinal ≤ 휇 with this property. ∙ 푈푙푡푛(ℳ||훼, 퐹 ) is well founded, where 푛 is the largest 푘 ≤ 휔 such that crit(퐹 ) < 휌푘(ℳ). ∙ 휆(퐹 ) < 휇. ∙ (휅 ≤ crit(퐹 ) ∧ crit(퐹 )+ exists)ℳ||훼. Suppose further that if 훼 = 휇 and there is 훾 < 훼 such that (훾 is the largest cardinal)ℳ||훼 and (푐푓(훾) ≥ 휅)ℳ||훼, then (푐푟푖푡(퐹 ) ̸= 푐푓(훾))ℳ||훼. Then

푈푙푡푛(ℳ||훼, 퐹 ) ∩ 푂푅 ≤ 휇. Moreover if 훼 < 휇 the above inequality is strict and if 훼 = 휇 then equality holds. Proof. We split the analysis into two cases: 훼 < 휇 and 훼 = 휇 and the second case splits further into two subcases. ∙ Suppose 훼 < 휇. Notice that for 푛 > 0 the set (푛−1) {푓 : crit(퐹 ) → ℳ||훼 | 푓 ∈ Σ1 (ℳ||훼)} has cardinality ≤ |훼|. Therefore

|푈푙푡푛(ℳ||훼, 퐹 )| ≤ max{|훼|, |휆(퐹 )|} < 휇, which implies 푈푙푡푛(ℳ||훼, 퐹 ) ∩ 푂푅 < 휇. ∙ Suppose 훼 = 휇. Since ℳ is sound, 휌휔(ℳ) = 휇 and 푛 = 휔. Let 푖퐹 : ℳ → 푈푙푡0(ℳ, 퐹 ) be the ultrapower map derived from 퐹 . We split this case into two subcases: ℳ I Suppose there is 훾 such that (훾 is the largest cardinal) :

Claim 3.13. If 푖퐹 (훾) < 휇, then 푈푙푡0(ℳ, 퐹 ) ∩ 푂푅 ≤ 휇

Proof. For a contradiction suppose that 푖퐹 (훾) < 휇 and that there is 휉 ∈ 푈푙푡0(ℳ, 퐹 ) such that 휉 ≥ 휇. The ultrapower map 푖퐹 is cofinal in 푈푙푡0(ℳ, 퐹 ), therefore there is 훽 ∈ 휇 such that 푖퐹 (훽) ≥ 휉 ≥ 휇. Let 휙(훽, 훾, ℎ) := “(ℎ : 훾 → 훽) ∧ (ℎ is a surjection)”

10 푛 푛,ℳ Soundness implies that ℳ = ℎℳ(휌푛(ℳ) ∪ {푝 }) for all 푛 ∈ 휔. 8 GABRIEL FERNANDES* AND RALF SCHINDLER‡

The formula ∃ℎ휙(훽, 훾, ℎ) is Σ1 and therefore it is preserved by 푖퐹 and

(2) 푈푙푡0(ℳ, 퐹 ) |= ∃ℎ휙(푖퐹 (훽), 푖퐹 (훾), ℎ).

Fix ℎ ∈ 푈푙푡0(ℳ, 퐹 ) which witnesses (2). As 휙(푖퐹 (훽, 푖퐹 (훾)), ℎ) is Σ0 and 푈푙푡0(ℳ, 퐹 ) is transitive it follows that 휙(푖퐹 (훽, 푖퐹 (훾)), ℎ) holds in 푉 . Therefore ℎ is a surjection from 푖퐹 (훾) onto 푖퐹 (훽). As 푖퐹 (훾) < 휇 and 푖퐹 (훽) > 휇 this contradicts our hypothesis that 휇 is a cardinal. 

<휔 Next we verify 푖퐹 (훾) = 푠푢푝휉<훾 (휉). Let 휁 ∈ 푖퐹 (훾) and let 푓 ∈ ℳ, 푎 ∈ 휆(퐹 ) be such that 푓 : crit(퐹 ) → 훾 and [푎, 푓] represents 휁 in the ultrapower of ℳ by 퐹 . From the hypothesis in our lemma, if 푐푓 ℳ(훾) ≤ 휅 or 푐푓 ℳ(훾) > 휅 in both cases we have 푐푓 ℳ(훾) ̸= crit(퐹 ). Then (i) or (ii) below must hold: (i) 푐푓 ℳ(훾) > crit(퐹 ) implies that there is 휉 < 훾 such that 푠푢푝(푟푎푛(푓)) < 휉, (ii) 푐푓 ℳ(훾) < crit(퐹 ) implies that there is 휉 < 훾 such that {푢 ∈ crit(퐹 )|푎| | 푓(푢) ∈ 휉} ∈ 퐹푎.

Thus we can find 휉 < 훾 such that [푎, 푓] ∈ 푖퐹 (휉) ∈ 푖퐹 (훾). Hence 푖퐹 (훾) = 푠푢푝휉<훾 푖퐹 (휉). Claim 3.14. Given 휉 < 훾, it follows that

crit(퐹 ) ℳ +ℳ ℳ |푖퐹 (휉)| ≤ max{(|휉 |) , crit(퐹 ) , |휆(퐹 )| } ≤ 훾 < 휇. Proof. From our hypothesis that 휆(퐹 ) < 휇, since 훾 is the largest cardinal of ℳ, it follows that |휆(퐹 )|ℳ ≤ 훾. From our hypothesis that crit(퐹 )+ℳ exists in ℳ, it follows that crit(퐹 )+ ≤ 훾. crit(퐹 ) ℳ Notice that 휉 < 훾 and 훿 < 훾 imply |휉 | ≤ 훾.  Therefore from the above claim and the regularity of 휇 we have:

푖퐹 (훾) = 푠푢푝휉<훾 푖퐹 (휉) < 휇. ℳ I Suppose (there is no largest cardinal) . Since 푖퐹 is cofinal in 푈푙푡0(ℳ, 퐹 ) it will be enough to verify that 푖퐹 (휉) < 휇 for all 휉 < 휇. Given 휉 < 휇, similarly as in the crit(퐹 ) ℳ +ℳ proof of Claim 3.14 we get that |푖퐹 (휉)| ≤ max{(|휉 |) , crit(퐹 ) , |휆(퐹 )|} < 휇.  Using induction and Lemma 3.12 we can obtain the following: Lemma 3.15. ([JS13, Lemma 4.8]) Let 휇 be a regular cardinal in 푉 , 휅 an ordinal such that 휅 < 휇 and ℳ is a sound premouse that is 휇-stable above 휅. Let 풯 be an 풯 iteration tree on ℳ such that 푙ℎ(풯 ) < 휇 and, for all 훽 + 1 < 푙ℎ(풯 ), crit(퐸훽 ) ≥ 휅. 풯 Then 훽 ∈ 푙ℎ(풯 ) implies ℳ훽 ∩ 푂푅 ≤ 휇. Definition 3.16. Given a premouse ℳ, 휇 ∈ ℳ ∩ 푂푅 and a normal iteration tree

풯 풯 풯 풯 풯 풯 = ⟨⟨ℳ훼 | 훼 < 휃⟩, ⟨휈훽 | 훽 + 1 < 휃⟩, ⟨휂훽 | 훽 + 1 < 휃⟩, ⟨휋훼,훽 | 훼 ≤푇 훽 < 휃⟩, 푇 ⟩ on ℳ, we say that 풯 lives on ℳ|휇 iff

풰 풰 풰 풰 풰 풰 := ⟨⟨ℳ훼 | 훼 < 휃⟩, ⟨휈훽 | 훽 + 1 < 휃⟩, ⟨휂훽 | 훽 + 1 < 휃⟩, ⟨휋훼,훽 | 훼 ≤푇 풰 훽 < 휃⟩, 푇 ⟩ 풰 풯 풰 풰 풯 such that ℳ0 = ℳ|휇, 푇 = 푇 and ⟨휈훽 | 훽 < 휃⟩ = ⟨휈훽 | 훽 < 휃⟩ is a normal iteration tree on ℳ|휇. We will denote 풰 by 풯  (ℳ|휇) and call it the restriction of 풯 to ℳ|휇. 풰 풰 Remark 3.17. In Definition 3.16 the sequence ⟨휂훽 | 훽 ∈ 퐵 ⟩ is determined by the other parameters in 풰 and the requirement that 풰 is normal. TALL CARDINALS IN EXTENDER MODELS 9

Lemma 3.18. Let 휇 be a regular cardinal, 휅 an ordinal such that 휅 < 휇 and ℳ be a proper class premouse. Suppose that ℳ|휇 is 휇-stable above 휅. Let 풯 be an iteration 풯 tree on ℳ that lives on ℳ|휇 and for all 훽 + 1 ∈ 푙ℎ(풯 ) we have crit(퐸훽 ) ≥ 휅 and 푙ℎ(풯 ) < 휇. Let 풰 = 풯  ℳ|휇. Then for all 훽 ∈ 푙ℎ(풯 ), the following holds: 풯 (푎)훽 If 퐷 ∩ [0, 훽]푇 = ∅, then 풯 – 휋0,훽(휇) is defined, 풯 풯 – 휋0,훽(휇) = sup훾<휇(휋0,훽(훾)) = 휇 풯 풰 – ℳ훽 |휇 = ℳ훽 . 풯 풯 풰 (푏)훽 If 퐷 ∩ [0, 훽]푇 ̸= ∅, then ℳ훽 = ℳ훽 .

Proof. We proceed by induction. Suppose 훽 ∈ 푙ℎ(풯 ) and (푎)훾 and (푏)훾 holds for all 훾 < 훽. We split the analysis into two cases, 훽 is a successor ordinal or 훽 is a limit ordinal. ∙ Suppose 훽 = 훿 + 1 for some ordinal 훿. We again need to divide into two 풯 subcases based on whether 퐷 ∩ [0, 훽]풯 is empty or not. 풯 풯 I Suppose 퐷 ∩ [0, 훽]푇 = ∅. Then 휋0,훽(휇) is defined. Recalling our notation 풯 풯 풯 from Definition 3.2, 휉훿 = pred푇 (훽) and 휂훿 is the largest ordinal such that 퐸훿 is a 풯 풯 total extender over ℳ 풯 ||휂훿 . 휉훽 풯 풯 풯 As 퐷 ∩ [0, 훽]푇 = ∅, it follows that ℳ 풯 is a proper class and ℳ훽 is the 휉훿 풯 풯 Σ0-ultrapower of ℳ 풯 by 퐸훿 . 휉훿 풯 풯 <휔 풯 풯 |푎| Hence, given 휁 < 휋휉 ,훿(휇), there are 푎 ∈ 푙ℎ(퐸훿 ) and 푓 ∈ ℳ 풯 , 푓 : (crit(퐸훿 )) → 훿 휉훿 풯 풯 휇 such that [푎, 푓]퐸풯 represents 휁 in 푈푙푡0(ℳ 풯 , 퐸훿 ). 훿 휉훿 As 휇 is a regular cardinal there is Υ < 휇 such that sup(푟푎푛(푓)) < Υ < 휇 and therefore 휁 < 휋풯 (Υ) < 휋풯 (휇). 휉훿 ,훽 휉훿 ,훽 By our induction hypothesis we have 풯 풰 풯 풰 ℳ 풯 |휇 = ℳ휉 and 퐸훿 = 퐸훿 . 휉훿 훿 Therefore, ℳ풰 = 푈푙푡 (ℳ풰 , 퐸풯 ) = 푈푙푡*(ℳ풯 |휇, 퐸풯 ), 훽 0 휉훿 훿 휉훿 훿 and by Lemma 3.15 applied to 풰 훽 + 1 for all 훾 < 휇 we have 휋풰 (훾) < 휇. Thus  휉훿 ,훽 for 훾 = Υ we have 휁 < 휋풯 (Υ) < 휇. 휉훿 ,훽 Therefore, 휇 ≥ sup(휋풯 (훾)) ≥ 휋풯 (휇) ≥ 휇, 휉훿 ,훽 휉훿 ,훽 훾∈휇 and * 풯 풯 풯 푈푙푡 (ℳ 풯 |휇, 퐸훿 ) = ℳ훽 |휇. 휉훿 풯 I Suppose 퐷 ∩ [0, 훽]푇 ̸= ∅. We need to further subdivide into two subcases 풯 풯 depending on whether 퐷 ∩ [0, 휉훿 ] is empty or not. 풯 풯 풯 ⋆ If 퐷 ∩ [0, 휉훿 ]푇 = ∅, then 훽 ∈ 퐷 and by induction hypothesis we have 풯 풰 풯 풯 ℳ 풯 |휇 = ℳ 풯 and 퐸훿 is not a total extender on ℳ 풯 . 휉훿 휉훿 휉훿 풯 As ℳ 풯 is an acceptable 퐽-structure, it follows that 휉훿 풯 풯 풯 풯 ℳ 풯 ∩ 퐻휇 = ℳ 풯 |휇 ⊇ 풫(crit(퐸훿 )) ∩ ℳ 풯 휉훿 휉훿 휉훿 Hence, if 휂풯 ≥ 휇, it follows that 퐸풯 is a total extender on ℳ풯 and 훿 훿 휉훿 훽 ̸∈ 퐷풯 , which is a contradiction as we are assuming that 훽 ∈ 퐷풯 . 풯 풯 풰 Therefore 휂훿 < 휇, 휂훿 = 휂훿 and by our induction hypothesis 풯 풯 풰 풯 ℳ 풯 ||휂훿 = ℳ 풯 ||휂훿 . 휉훿 휉훿 10 GABRIEL FERNANDES* AND RALF SCHINDLER‡

Then 풯 * 풯 풯 풯 * 풰 풰 풰 풰 ℳ훽 = 푈푙푡 (ℳ 풯 ||휂훿 , 퐸훿 ) = 푈푙푡 (ℳ 풯 ||휂훿 , 퐸훿 ) = ℳ훽 . 휉훿 휉훿 풯 풯 풯 풰 ⋆ If 퐷 ∩ [0, 휉훿 ]푇 ̸= ∅, then by our induction hypothesis ℳ 풯 = ℳ 풯 , so 휉훿 휉훿 풯 * 풯 풯 * 풰 풰 풰 ℳ훽 = 푈푙푡 (ℳ 풯 , 퐸훿 ) = 푈푙푡 (ℳ 풯 , 퐸훿 ) = ℳ훽 . 휉훿 휉훿 ∙ Suppose 훽 is a limit ordinal. Again, we divide into two subcases based on 풯 whether 퐷 ∩ [0, 훽]풯 is empty or not. 풯 풯 I Suppose 퐷 ∩ [0, 훽]푇 = ∅. Given 훾 < 휋0,훽(휇), we have that 휁 < 훾 iff there are ¯ ¯ 풯 풯 ¯ 풯 훽 ∈ [0, 훽)푇 and 휁 < 훾¯ < 휇 = 휋0,훽¯(휇) such that 휋훽,훽¯ (휁) = 휁 and 휋훽,훽¯ (¯훾) = 훾. 풯 By a cardinality argument it follows that 훾 < 휇. Therefore 휋0,훽(휇) ≤ 휇 and 풯 풯 풯 풰 풰 풰 ℳ |휇 = 푑푖푟푙푖푚 ¯ (ℳ |휇, 휋 |휇) = 푑푖푟푙푖푚 ¯ (ℳ , 휋 ) = ℳ 훽 훽∈[0,훽]푇 훽¯ 훽,훽¯ 훽∈[0,훽]푇 훽¯ 훽,훽¯ 훽 풯 풯 I Suppose 퐷 ∩ [0, 훽]푇 ̸= ∅, let 휁 be the largest element in 퐷 ∩ [0, 훽]풯 . Then by induction hypothesis 풯 풯 풯 풰 풰 풰 ℳ = 푑푖푟푙푖푚 ¯ (ℳ , 휋 ) = 푑푖푟푙푖푚 ¯ (ℳ , 휋 ) = ℳ 훽 훽∈[휁,훽]푇 훽¯ 훽,훽¯ 훽∈[휁,훽]풰 훽¯ 훽,훽¯ 훽  Remark 3.19. Given a proper class premouse 퐿[퐸] and a regular cardinal 휇 such that 퐿[퐸]|휇 is 휇-stable above 휅, our next result, Lemma 3.20, gives a sufficient condition on iteration trees 풯 on 퐿[퐸] for 풯 to live on 퐿[퐸]|휇. Lemma 3.20. Let 휇 be a regular cardinal. Suppose ℳ is a proper class premouse and 풯 is a finite 11 iteration tree on 퐿[퐸] such that for all 훽 + 1 ∈ 푙ℎ(풯 ) we have 풯 crit(퐸훽 ) > 휅 and 푙ℎ(풯 ) < 휇. Suppose further that 퐿[퐸]|휇 is 휇-stable above 휅. If 풯 for all 훽 ∈ 푙ℎ(풯 ) we have 휈훽 < 휇, then 풯 lives on ℳ|휇. Proof. We define recursively an iteration tree 풰 on ℳ|휇 such that 푙ℎ(풰) ≤ 푙ℎ(풯 ) 풰 풯 풰 풯 풯 풯 as follows: If 풰|훽 is defined, 훽 ∈ 풯 , ℳ훽 ∩ 푂푅 ≥ 휈훽 and ℳ훽 ||휈훽 = ℳ훽 ||휈훽 , then 풰 풯 풰 we let 휈훽 = 휈훽 , otherwise we let 휈훽 be undefined and 푙ℎ(풰) = 훽. Suppose 훽 ∈ 푙ℎ(풯 ), 훽 + 1 < 푙ℎ(풯 ) and 훽 < 푙ℎ(풰), we will verify that 훽 + 1 < 푙ℎ(풰). 풯 풯 풰 I If 퐷 ∩[0, 훽]푇 = ∅, by Lemma 3.18 applied to 풯  훽+1 we have ℳ훽 |휇 = ℳ훽 . 풯 풰 풰 풯 풯 풯 Hence 휈훽 < 휇 = ℳ훽 ∩ 푂푅 and ℳ훽 ||휈훽 = ℳ훽 ||휈훽 . 풯 I If 퐷 ∩ [0, 훽]푇 ̸= ∅, then by Lemma 3.18 applied to 풯  훽 + 1 we have 풯 풰 ℳ훽 = ℳ훽 . 풯 풰 풰 풯 풰 풯 Therefore 휈훽 ∈ ℳ훽 ∩ 푂푅 and ℳ훽 ||휈훽 = ℳ훽 ||휈훽 .  Remark 3.21. (a) Suppose that 푉 is an extender model 퐿[퐸], 휆 is a cardinal and 풯 + 풯 is an iteration tree on 퐿[퐸] such that 푠푢푝훼∈푙ℎ(풯 )휈훼 < 휆 . Then by Lemma 3.20 for 휇 := 휆++퐿[퐸] we have that 풯 lives on 퐿[퐸]|휇. Notice that 퐿[퐸]|휇 and 풯 are in the hypothesis of Lemma 3.15. So in particular if 풰 = 풯  (퐿[퐸]|휇) is the 풰 restriction of 풯 to 퐿[퐸]|휇, then for all 훽 ∈ 푙ℎ(풰) it follows that ℳ훽 ∩ 푂푅 ≤ 휇. (b) Our hypotheses on Lemma 3.18 are optimal in the following sense: suppose that 휇 is a regular cardinal and there is 훾 < 휇 such that (훾 is the largest cardinal ∧ cf(훾) > 휅)퐿[퐸]|휇. 퐿[퐸] If there is 훽 ∈ 휇 such that (푐푟푖푡(퐸훽) = 푐푓(훾)) and 퐸훽 is a total measure in 퐿[퐸] cf (훾), then 푈푙푡0(퐿[퐸]|휇, 퐸훽) ∩ 푂푅 > 휇.

11This lemma remains true if we drop the hypothesis that 풯 is finite. TALL CARDINALS IN EXTENDER MODELS 11

(c) In Lemma 4.36 we show that we can not drop the hypothesis that 퐿[퐸]|휇 is 휇-stable above 휅 in the statement of Theorem A. In the proof of Lemma 4.36 we use Remark 3.21 (b). 퐸 Lemma 3.22. [Sch02, Lemma 1.1] Let ℳ = ⟨퐽훼 , ∈, 퐸, 퐹 ⟩ be an iterable premouse, where 퐹 ̸= ∅ . Suppose that for no 휇 ≤ ℳ ∩ 푂푅 do we have ℳ 풥휇 |= 푍퐹 퐶 + there is a Woodin cardinal. +ℳ Set 휅 = 푐푟푖푡(퐹 ) and let 휉 ∈ (휅, 휌1(ℳ)). Then there is some 휈˜ ∈ (휉, 휉 ) with 푐푟푖푡(퐸휈˜) = 푐푟푖푡(퐹 ). Lemma 3.23. Suppose that 푉 is a proper class premouse 퐿[퐸] that is iterable. Let 휅 < 휇 be cardinals. Then 푂(휅) > 휇 ←→ 표(휅) > 휇. Moreover, 푂(휅) = 휇 ←→ 표(휅) = 휇. Proof. Notice that if 푋 ⊆ 푂푅, then 표푡푝(푋) ≤ sup(푋). From this general fact it follows that 표(휅) > 휇 implies 푂(휅) > 휇. For the other direction we split the analysis into two cases. I Suppose first that 휇 is a limit cardinal and 푂(휅) > 휇. We will verify that 표(휅) > 휇. For that we show that given a regular cardinal 휒 < 휇 such that 휅 < 휒 the following holds:

(3) 푠푢푝({훽 < 휒 | 푐푟푖푡(퐸훽) = 휅}) = 휒. 퐿[퐸] Let 훼 > 휒 be such that 푐푟푖푡(퐸훼) = 휅 and let ℳ = 풥훼 . Then 휌1(ℳ) ≥ 휒. Given 휉 < 휒, by Lemma 3.22 there is 휉˜ ∈ (휉, 휉+ℳ) such that 푐푟푖푡(퐸ℳ) = 휅. As 휉˜ 휉 was arbitrary the equality in (3) follows. Thus (3) holds for any regular cardinal 휒 such that 휅 < 휒 < 휇. Therefore

|{훽 < 휇 | 푐푟푖푡(퐸훽) = 휅}| = 휇.

Then otp({훽 < 휇 | 푐푟푖푡(퐸훽) = 휅}) ≥ 휇. Notice that

표(휅) = otp({훽 < 휇 | crit(퐸훽) = 휅} ∪ {훽 ≥ 휇 | crit(퐸훽) = 휅}) =

otp({훽 < 휇 | crit(퐸훽) = 휅}) ⊕ otp({훽 ≥ 휇 | crit(퐸훽) = 휅}) where ⊕ represents the ordinal sum. As 푂(휅) > 휇 we have

otp({훽 ≥ 휇 | 푐푟푖푡(퐸훽) = 휅}) > 0 Therefore 표(휅) ≥ 휇 + 1 > 휇. + I Suppose 휇 = 휃 and suppose 푂(휅) > 휇. Fix 훼 such that 훼 > 휇 with ℳ 퐿[퐸] crit(퐸훼 ) = 휅 and consider ℳ = 풥훼 . We have 휌1(ℳ) ≥ 휇 and given 휉 ∈ (휅, 휇), by Lemma 3.22, there is 휉˜ ∈ (휉, 휉+ℳ) such that 푐푟푖푡(퐸ℳ)) = 휅. Hence 휉˜

(4) 푠푢푝({훽 < 휇 | 푐푟푖푡(퐸훽) = 휅}) = 휇. Therefore (4) with 푂(휅) > 휇 implies that 표(휅) > 휇. For the second part, suppose 표(휅) = 휇, then 푂(휅) ≤ 휇, otherwise by the first part we would have 표(휅) > 휇. Hence 푂(휅) = 휇. For the other direction again we split the analysis into two cases: I Suppose 푂(휅) = 휇 and 휇 is a limit cardinal. Given 휒 a regular cardinal, such that 휅 < 휒 < 휇 we have by the first part of the lemma that 표(휅) > 휒. Hence 표(휅) ≥ 휇 and thus 표(휅) = 휇. + I Suppose 푂(휅) = 휇 and 휇 = 휃 for some cardinal 휃. Since 휇 is a regular cardinal it follows that 표(휅) = 휇.  12 GABRIEL FERNANDES* AND RALF SCHINDLER‡

Lemma 3.24. Suppose 푉 is an extender model 퐿[퐸] which is iterable. Let 휅 and 휇 be cardinals, such that 휅 < 휇. Suppose that 푂(휅) ≤ 휇, {훼 < 휅 | 푂(훼) > 휇} is bounded in 휅 and 휅 < 푂(휅) 12 . Let (5) 훽* = 푠푢푝{ 훼 < 휅 | 푂(훼) > 휇 }. and (6) Θ = 푠푢푝{ 푂(훼) | 훽* < 훼 ≤ 휅 } ≤ 휇 Then there is no 휂 ∈ (훽*, Θ] such that 푂(휂) > Θ. * Proof. Suppose otherwise. Let 퐸훼 be such that 훼 > Θ and 휂 := 푐푟푖푡(퐸훼) ∈ (훽 , Θ] * and let ℳ be the largest initial segment of 퐿[퐸] where we can apply 퐸훼. Let * * 풩 := 푈푙푡0(ℳ , 퐸훼) and 휋 : ℳ → 풩 the ultrapower map. We split the analysis into two cases and the second case will split into two subcases. ∙ Suppose 휂 = Θ. In this case13 퐿[퐸]|훼 |= “Θ is a cardinal” and hence 퐸Θ = ∅, as extenders are not indexed at cardinals. Therefore Θ = 푠푢푝({푂(훼) | 훽* < 훼 ≤ 휅} ∩ Θ) * and Θ is a limit of ordinals 훾 such that crit(퐸훾 ) ∈ (훽 , 휅]. Let 휙(Θ, 훽*, 휅) be the following formula: ′ ′ * ∀훾 < Θ ∃훾 < Θ( 훾 < 훾 ∧ 퐸˙ 훾′ ̸= ∅ ∧ 푐푟푖푡(퐸˙ 훾 ) ∈ (훽 , 휅]). * Note that 휙(Θ, 훽 , 휅) is a Σ0-formula in the language {∈, 퐸˙ }. We have 퐿[퐸] |= 휙(Θ, 훽*, 휅), and therefore ℳ* |= 휙(Θ, 훽*, 휅).

Notice that 푂(휅) > 휅 implies Θ ≥ 푂(휅) > 휅. Hence by Σ1-elementarity of 휋퐸훼 we have: * 풩 |= 휙(휋퐸훼 (Θ), 훽 , 휅 ). ⏟ ⏞ ⏟ ⏞ * =휋퐸 (휅) =휋퐸훼 (훽 ) 훼

Thus, in 풩 , 휋퐸훼 (Θ) is a limit of indexes of extenders with critical points in the * interval (훽 , 휅]. Notice that as crit(퐸훼) = Θ then Θ < 휋퐸훼 (Θ). Therefore there is 풩 * 훾 such that Θ < 훾 < 휋퐸훼 (Θ) and crit(퐸훾 ) ∈ (훽 , 휅]. * +풩 풩 As 풩 |훼 = ℳ |훼 and 훼 = 휋퐸훼 (Θ) , it follows that 퐸훾 = 퐸훾 . But Θ < 훾 and * crit(퐸훾 ) ∈ (훽 , 휅] contradict the definition of Θ. ∙ Suppose 휂 < Θ. We split this case into two subcases (see Figures2 and3): I Suppose 휂 < Θ and 휆(퐸훼) ≤ Θ. Then

휋퐸훼 (Θ) ≥ 휋퐸훼 (휆(퐸훼)) > 훼, 풩 and by Σ0-elementarity there is a 훾 ∈ (훼, 휋퐸훼 (Θ)) such that 퐸훾 ̸= ∅ and 풩 * 푐푟푖푡(퐸훾 ) ∈ (훽 , 휅]. 풩 풩 From (훼 is a cardinal) , it follows that 휌1(풥훾 ) ≥ 훼 for any 훾 ∈ (훼, 휋퐸훼 (Θ)). +풩 Since 훼 = 휆(퐸훼) and 훼 > Θ, we have that +풩 휌1(풩 ||훾) ≥ 훼 = Θ > Θ,

12For example, when there exists a total measure indexed on 퐸 with critical point 휅 we have 휅+ < 푂(휅). 13See Figure1. TALL CARDINALS IN EXTENDER MODELS 13

훼

휋퐸훼 (휂) = 휆(퐸훼)

휂 = Θ 휂 = Θ

휅 휅

훽* 훽*

Figure 1. Case 1, Lemma 3.24

휋퐸훼 (Θ) 훾 + 훼 Θ = 훼 ≤ 휌1(풩 ||훾)

Θ 휆(퐸훼) = 휋퐸훼 (휂) ≤ Θ

휂 휅 휅

훽* 훽*

Figure 2. Case 2 (i), Lemma 3.24

휋퐸훼 (Θ) 훼 훾

휆(퐸훼) = 휋퐸훼 (휂) > Θ

Θ Θ 휂 휅 휅 훽* 훽*

Figure 3. Case 2 (ii), Lemma 3.24

′ +풩 풩 it follows by Lemma 3.22 that there is 훾 ∈ (Θ, Θ ) such that 퐸훾′ ̸= ∅ and 풩 풩 +풩 * 풩 푐푟푖푡(퐸훾′ ) = 푐푟푖푡(퐸훾 ). As Θ ≤ 훼 and ℳ |훼 = 풩 |훼, we have 퐸훾′ = 퐸훾′ , which contradicts the definition of Θ.

I Suppose 휂 < Θ and 휆(퐸훼) > Θ. In this case 휋퐸훼 (Θ) > 휆(퐸훼) > Θ. Then for all 훾 such that 훾 ∈ (휆(퐸훼), 휋퐸훼 (Θ)) we have 풩 (7) 휌1(풥훾 ) ≥ 휆(퐸훼) > Θ. Then like in case 2 (i) we can find an extender in the sequence of 퐿[퐸] that is * indexed in the interval (Θ, 휆(퐸훼)) with critical point in the interval (훽 , 휅], contra- dicting the definition of Θ.

 14 GABRIEL FERNANDES* AND RALF SCHINDLER‡

4. Equivalence In this section we prove Theorem A and Theorem B. We will need some results from core model theory before we start with the proofs of Theorems A and B. Remark 4.1. Suppose there is no inner model with a Woodin cardinal and let 풦 be the core model. If 휅 is an ordinal and Ξ is a singular strong limit cardinal above 휅, then by (iii) of Theorem 1.5 풦|Ξ+ |= “Ξ is the largest cardinal”. By Definition 3.12 we have: (a) 풦|Ξ+ is Ξ+-stable above 휅, or + 풦 (b) there is 훽 ∈ Ξ such that 훽 is the least ordinal such that 퐸훽 is a total measure + 풦 풦 풦 over 풦|Ξ with crit(퐸훽 ) = cf (Ξ) and 푐푓 (Ξ) ≥ 휅. Definition 4.2. Suppose there is no inner model with a Woodin cardinal and let 풦 be the core model. If 휅 is an ordinal and Ξ is a singular strong limit cardinal above 휅 we say that 풲 is the (Ξ+, 휅)-stabilization of 풦 iff (a) 풲 = 풦|Ξ+ and 풦|Ξ+ is Ξ+-stable above 휅, or 풦 + (b) 풲 = 푈푙푡0(풦, 퐸훽 )|Ξ where 훽 is the least measure indexed in the sequence of + 풦 풦 풦 풦 that is total in 풦|Ξ with crit(퐸훽 ) = cf (Ξ) and cf (Ξ) ≥ 휅. Definition 4.3. Let ℳ be a premouse and 휅 an ordinal such that 휅 ∈ ℳ. We say that 휅 is a strong cutpoint of ℳ iff for all 훼 ≤ ℳ ∩ 푂푅 such that 훼 > 휅 we have ℳ ℳ that either 퐸훼 = ∅ or crit(퐸훼 ) > 휅. The following is a slight variation of [JS13, Proposition 4.4]. Lemma 4.4. [JS13, Proposition 4.4] Let Ω be a regular cardinal and 휅 ∈ Ω. Sup- pose that 풲 is Ω-stable above 휅, (Ω + 1)-iterable, 휅 is a strong cutpoint of 풲 and 풲 has a largest cardinal. Then for every sound premouse ℳ such that: ∙ℳ is (Ω + 1)-iterable, ∙ℳ∩ 푂푅 < Ω, ∙ ℳ||휅 = 풲||휅, ∙ 휅 is a strong cutpoint of ℳ, we have that there are14 iteration trees 풯 and 풰 on 풲 and ℳ respectively such that (a) for 푏풰 the main branch of 풰 we have 퐷풰 ∩ 푏풰 = ∅, 풰 (b) ℳ∞ is sound, 풰 풯 (c) ℳ∞ ▷ ℳ∞, 풯 풰 (d) for every 훼 ∈ 풯 and for every 훽 ∈ 풰 we have 휈훼 , 휈훽 > 휅. The following lemma is standard but we include it for the reader’s convenience. Lemma 4.5. Suppose there is no inner model with a Woodin cardinal and let 풦 be the core model. Let 휅 be an ordinal and ℳ a sound iterable premouse. Suppose ∙ 휅 is a strong cutpoint of 풦 and ℳ, ∙ ℳ||휅 = 풦||휅, and ∙ 휌휔(ℳ) ≤ 휅. Then ℳ ▷ 풦.

14These iteration trees are obtained by the so called Comparison Lemma, see [Ste10, Section 3.2] or [Zem02, Lemma 9.1.8]. TALL CARDINALS IN EXTENDER MODELS 15

Proof. Let 휅 be as in the hypotheses of the lemma and let Ξ be a singular strong limit cardinal such that cf(Ξ) > 휅. Let 풲 be the (Ξ+, 휅)-stabilization of 풦|Ξ+. We can apply Lemma 4.4 to ℳ, 풲 and 휅. Let 풯 and 풰 be iteration trees given by Lemma 4.4 where 풯 is on 풲 and 풰 is on ℳ. 풰 Claim 4.6. For all 훼 ∈ 푙ℎ(풰) we have crit(퐸훼 ) ≥ 휅 and 휅 is a strong cutpoint of 풰 ℳ훼 . Proof. We prove the claim by induction on 훼 ∈ 푙ℎ(풰). Suppose that 훼 ∈ 푙ℎ(풰) and 풰 풰 that for all 훽 < 훼 we have that 휅 is a strong cutpoint of ℳ훽 and crit(퐸훽 ) > 휅. I Suppose 훼 is a limit ordinal. Let 훾 ∈ [0, 훼]푇 풰 be large enough such that 풰 풰 풰 풰 퐷 ∩ (훾, 훼]푇 풰 = ∅ and hence dom(휋훾,훼) = ℳ훾 , so 휋훾,훼 is not a partial map but has full domain. 풰 풰 As 휅 < crit(퐸훽 ) for all 훽 < 훼, it follows that 휋훼,훽(휅) = 휅. As 휅 is a strong 풯 풰 cutpoint of ℳ훾 it follows, by the Σ1-elementarity of 휋훾,훼, that 휅 is a strong cutpoint 풰 풰 풰 of ℳ훼 . Since 휈훼 > 휅, it then follows that crit(퐸훼 ) > 휅. I Suppose 훼 = 훾 + 1 for some 훾 ∈ 푂푅. From our induction hypothesis we have 풰 풰 풰 that 휅 is a strong cutpoint of ℳ 풰 , therefore 휅 is a strong cutpoint of ℳ ||휂훼 . 휉훼 휉훼 By the Σ -elementarity of 휋풰 (ℳ풰 ||휂풰 ) it follows that 휅 is a strong cutpoint 1 휉훼,훼  휉훼 훼 풰 풰 풰 of ℳ훼 . As 휈훼 > 휅, it follows that crit(퐸훼 ) > 휅.  풰 Claim 4.7. ℳ = ℳ∞. 풰 풰 Proof. Suppose not, and let 퐸훼 be the first extender applied to ℳ0 such that 훼 + 1 ∈ 푏풰 , where 푏풰 is the main branch of the iteration tree 풰. By Claim 4.6 풰 it follows that crit(퐸훼 ) > 휅. Let 푛 ∈ 휔 be the least such that 휌푛(ℳ) ≤ 휅. By 풰 standard arguments 휌푛(ℳ훼+1) = 휅, but 풰 풰 풯 ˜푛 풰 ℳ훼 ˜푛 풰 ℳ훼 crit(퐸 ) ̸∈ ℎ 풰 (crit(퐸훼) ∪ {휋 (푝푛 )}) ⊇ ℎ 풰 (휅 ∪ {휋 (푝푛 )}). 훼 ℳ훼 0,훼+1 ℳ훼 0,훼+1 풰 Therefore ℳ훼+1 is not 푛-sound, which contradicts the fact that every element in 풰 풰 푏 is sound. Hence 풰 is trivial and ℳ = ℳ∞.  Notice that if 퐷풯 ∩푏풯 = ∅ then 풲 ∩푂푅 ≥ Ξ+ and if 퐷풯 ∩푏풯 ̸= ∅ by arguments 풯 similar to the proof of Claim 4.7 it follows that ℳ∞ is not sound. In both cases 풯 we cannot have ℳ = ℳ∞ as ℳ ∩ 푂푅 < Ξ and ℳ is sound. 풯 Therefore ℳ is a proper initial segment of ℳ∞. 풯 Claim 4.8. 풲 = ℳ∞. 풰 풯 풯 Proof. Suppose 퐸0 ̸= ∅. Then 휈0 ≤ ℳ ∩ 푂푅 and 휈0 is a successor cardinal in 풯 ˜푛 ℳ 풯 ℳ∞. On the other hand ℳ = ℎℳ(휅 ∪ {푝푛 }) ▷ ℳ∞ and as ℳ is a proper initial 풯 segment of ℳ∞ it follows that 풯 ℳ∞ |= |ℳ ∩ 푂푅| ≤ 휅, contradicting that 풯 풯 풯 ℳ∞ |= “휈0 > 휅 and 휈0 is a cardinal”. 풯 Thus, 풯 must be trivial and ℳ∞ = 풲.  + 풦 + 풦 If 풲 = 풦|Ξ we are done, so suppose 풲 = 푈푙푡0(풦, 퐸훽 )|Ξ where 퐸훽 is the least total measure indexed on the sequence 퐸풦 with critical point 푐푓 풲 (Ξ) ≥ 휅. As ℳ is a proper initial segment of 풲 and 풲 |= “|ℳ ∩ 푂푅| ≤ 휅 and 훽 is a cardinal”, it follows that ℳ ∩ 푂푅 < 훽. As 풲|훽 = 풦|훽, it follows that ℳ ▷ 풦.  16 GABRIEL FERNANDES* AND RALF SCHINDLER‡

푉 Remark 4.9. Theorem 4.10 implies Lemma 4.4 for 휅 > ℵ2 . We will use Lemma 4.4 and Theorem 4.10 to prove Lemma 4.12.

Theorem 4.10. [GSS02, Lemma 3.5] If there is no inner model with a Woodin 푉 cardinal, 풦 is the core model and 휅 ≥ ℵ2 is a 풦-cardinal, then for every sound iterable mouse ℳ such that ℳ||휅 = 풦||휅 and 휌휔(ℳ) ≤ 휅 it holds that ℳ ▷ 풦.

Definition 4.11. We define the following hypothesis: (︁ (∆) ←→ (︀there is no inner model with a Woodin cardinal)︀ ∧ )︁ (︀푉 = 퐿[퐸])︀ ∧ (︀퐿[퐸] is iterable)︀

Lemma 4.12 (Steel). Assume (∆). Then 푉 = 풦.

Proof. By Theorem 1.5 we can build 풦, the core model. We prove by induction on the cardinals 휅 of 푉 that 풦||휅 = 퐿[퐸]||휅.

Claim 4.13. 풦||ℵ2 = 퐿[퐸]||ℵ2

Proof. Because of acceptability and soundness there are cofinally many 훼 < 휔1 such that 휌휔(퐿[퐸]||훼) = 휔. Fix such an 훼 < 휔1, and let ℳ = 퐿[퐸]||훼. We have that ℳ is a sound iterable premouse such that ℳ||휔 = 풦||휔. Hence, by Lemma 4.5 it follows that ℳ ▷ 풦. Thus

풦|ℵ1 = 퐿[퐸]|ℵ1.

Again by acceptability and soundness there are unboundedly many 훽 < ℵ2 such that 휌휔(퐿[퐸]||훽) = 휔1. We fix such a 훽 and consider 풩 = 퐿[퐸]||훽. As 풦|휔1 = 퐿[퐸]|휔1, it follows that 풩 |휔1 = 풦|휔1 and as 휔1 is a cardinal it follows that 풩 ||휔1 = 풦||휔1.

Subclaim 4.14. 휔1 is a strong cutpoint of 퐿[퐸] and 풦.

Proof. We start by verifying this for 퐿[퐸]. Suppose 훾 > 휔1 and 퐸훾 ̸= ∅. Then 퐿[퐸]||훾 퐿[퐸]||훾 휔1 = 휔1. From (crit(퐸훾 ) is a limit cardinal) it follows that crit(퐸훾 ) > 휔1. Next we verify it for 풦. From Claim 4.13 we have 풦||휔1 = 퐿[퐸]||휔1. There- 풦 풦 풦||훾 fore 휔1 = 휔1. Suppose 훾 > 휔1 and crit(퐸훾 ) ̸= ∅. Then 휔1 = 휔1. From 풦 풦||훾 풦 (crit(퐸훾 ) is a limit cardinal) it follows that crit(퐸훾 ) > 휔1. 

Thus, by Lemma 4.5 it follows that 풩 ▷ 풦. Therefore we have 풦|ℵ2 = ℳ|ℵ2. Since extenders are not indexed at cardinals we have 퐸풦 = ∅ = 퐸 . Then ℵ2 ℵ2 풦||ℵ2 = 퐿[퐸]||ℵ2. 

+ Now suppose 휅 > ℵ2 is a successor cardinal in 푉 , say 휅 = 휇 and 풦||휇 = 퐿[퐸]||휇. Then by Theorem 4.10 for every 휉 ∈ (휇, 휅) such that 휌휔(퐿[퐸]||휉) ≤ 휇 we have 퐿[퐸]||휉 ▷ 풦. Thus as there are unboundedly many such 휉 below 휅 it follows that 풦 풦|휅 = 퐿[퐸]|휅. As 퐸휅 = ∅ = 퐸휅 it follows that 풦||휅 = 퐿[퐸]||휅. Lastly, suppose 휅 is a limit cardinal in 푉 and for every cardinal 휇 < 휅 we have 풦 퐿[퐸]|휇 = 풦|휇, then 퐿[퐸]|휅 = 풦|휅. As 휅 is a cardinal, it follows that 퐸휅 = ∅ = 퐸휅 . Therefore 풦||휅 = 퐿[퐸]||휅. This concludes the induction and verifies the lemma.  TALL CARDINALS IN EXTENDER MODELS 17

Theorem 4.15. [Sch06, Theorem 2.1] If there is no inner model15 with a Woodin cardinal and 푗 : 푉 → 푀 is an elementary embedding and 푀 휔 ⊆ 푀, then there is an iteration tree 풯 on 풦푉 which does not drop along the main branch such that 풯 푀 풯 ℳ∞ = 풦 and 푗|풦 = 휋0,∞. Definition 4.16. Suppose there is no inner model with a Woodin cardinal. Given 푗 : 푉 −→ 푀 an elementary embedding, let 풯 and 풰 be the iteration trees obtained by comparing16 풦푉 and 풦푀 respectively. Then we say that 풯 is the iteration tree induced by 푗. The next lemma makes it clear how we would like to combine Theorem 4.15 and Lemma 4.12. Lemma 4.17. Assume (∆). Suppose that 푗 : 푉 → 푀 and 푀 휔 ⊆ 푀. Then there is 풯 an iteration tree on 퐿[퐸] induced by 푗 such that: (a) There is no drop along the main branch 푏풯 of 풯 . 풯 풯 풯 풯 풯 (b) 휋0,∞ : ℳ0 −→ ℳ∞, i.e., 푑표푚(휋0,∞) = ℳ0 . 풯 (c) 휋0,∞ = 푗. 풯 (d) ℳ∞ = 푀. Proof. Apply Theorem 4.15 to 푗 and 푀 and let 풯 be the iteration tree on 풦 induced 풯 풯 by 푗. Apply lemma 4.12 to obtain 풦 = 푉 and hence 휋0,∞ = 푗. Thus 풯 and 휋0,∞ are as sought. Let us verify (d). Let 휓풦(푥) be as in Theorem 1.5 such that 푥 ∈ 풦 if and only if 퐿[퐸] 휓풦(푥). By Lemma 4.12 we have (∀푥 휓풦(푥)) , then by the elementarity of 푗 we 푀 푀 풯 have (∀푥 휓풦(푥)) . Therefore 푀 = 풦 = ℳ∞, where we get the last equality by Theorem 4.15.  Lemma 4.18. Let ℳ be a premouse and 풯 be an iteration tree on ℳ. If 푙ℎ(풯 ) ≥ 풯 휔 풯 휔 + 1, then (ℳ∞) ̸⊆ ℳ∞.

Proof. Suppose 푙ℎ(풯 ) ≥ 휔 + 1 and 푏 = [0, 휔]푇 is the cofinal branch on 휔. Let ⟨휅푛 | 푛 ∈ 휔 ∩ 푏⟩ be such that

(︁ 풯 )︁ ∀푛 ∈ (푏 ∖ 푛0) 휅푛 = 푐푟푖푡(휋푛,휔) ,

풯 where 푛0 is large enough such that 퐷 ∩ (푛0, 휔)푇 = ∅, and for 푛 ∈ 푏 ∩ 푛0 we set 휅푛 = ∅. Denote ⃗휅 := ⟨휅푛 | 푛 ∈ 휔 ∩ 푏⟩. 풯 Claim 4.19. ⟨휅푛 | 푛 ∈ 휔 ∩ 푏⟩ ̸∈ ℳ휔 . 풯 풯 Proof. For a contradiction suppose ⃗휅 ∈ ℳ휔 , and let 푚 ∈ 휔 ∩ 푏 and 푥 ∈ ℳ푚 such that 풯 휋푚,휔(푥) = ⃗휅. Then 풯 풯 풯 푐푟푖푡(휋푚,휔) = 휋푚,휔(푥)(푚) ∈ 푟푎푛(휋푚,휔). 풯 풯 풯 This is a contradiction since 푐푟푖푡(휋푚,휔) ̸∈ 푟푎푛(휋푚,휔). Therefore ⃗휅 ̸∈ ℳ휔 . 

15 We have omitted the hypothesis that 풫(R) ⊆ 푀 since we start from the hypothesis that there is no inner model with a Woodin cardinal which is stronger than the hypothesis in [Sch06, Theorem 2.1]. 16Given two iterable premice ℳ and 풩 the comparison between (or coiteration of) ℳ and 풩 풯 is the process of iterating ℳ and 풩 so that at each sucessor step 훼 + 1 the extenders 퐸훼+1 and 풰 풯 풰 퐸훼+1 are the least extender in the sequences of ℳ훼 and ℳ훼 where there is a disagreement. The 풯 풰 last models of these iterations should line up, i.e., ℳ∞ ◁ ℳ∞ or vice versa. See [Ste10, Section 3.2] or [Zem02, Lemma 9.1.8] . 18 GABRIEL FERNANDES* AND RALF SCHINDLER‡

The lemma will follow from our next claim:

풯 풯 Claim 4.20. If ⃗휅 ∈ ℳ∞, then ⃗휅 ∈ ℳ휔 . 풯 풯 Proof. If 푙ℎ(풯 ) = 휔 + 1, then 푀∞ = ℳ휔 and there is nothing to do in this case. Let us assume 푙ℎ(풯 ) > 휔 + 1. The normality of 풯 implies the following: 풯 풯 (8) 푠푢푝푛∈휔휅푛 ≤ sup{휈푚 | 푚 + 1 ∈ (휔 ∩ 푏)} ≤ 휈휔 . 풯 풯 풯 By Fact 3.4, it follows that 휈휔 is a successor cardinal in ℳ휔+1. Since sup{휈푚 | 풯 푚 + 1 ∈ (휔 ∩ 푏)} is clearly a limit cardinal in ℳ휔+1, we have that the second inequality in (8) is in fact a strict inequality. * 풯 Write 휈 := sup{휈푚 | 푚 + 1 ∈ (휔 ∩ 푏)}, then 풯 * +ℳ휔+1 풯 {휅푛 | 푛 ∈ 휔} ⊆ 푠푢푝푛∈휔휅푛 < (휈 ) ≤ 휈휔 . In particular, 풯 풯 * + 풯 ℳ휔+1|휈휔+1 |= “푠푢푝푛∈휔휅푛 < (휈 ) ≤ 휈휔 ”. Since 풯 풯 풯 풯 ℳ휔+1|휈휔+1 = ℳ∞|휈휔+1, we have 풯 풯 * + 풯 (9) ℳ∞|휈휔+1 |= “푠푢푝푛∈휔휅푛 < (휈 ) ≤ 휈휔 ”. 풯 풯 풯 By (9) and ⃗휅 ∈ ℳ∞, we have ⃗휅 ∈ ℳ∞|휈휔 . We also have 풯 풯 풯 풯 풯 풯 ℳ휔+1|휈휔 = ℳ∞|휈휔 = ℳ휔 |휈휔 , and hence 풯 ⃗휅 ∈ ℳ휔 .  풯 휔 풯 풯 Hence if (ℳ∞) ⊆ 푀∞ and 푙ℎ(풯 ) ≥ 휔+1, by Claim 4.20 it follows that ⃗휅 ∈ ℳ휔 which by Claim 4.19 is a contradiction.  Lemma 4.21. Assume (∆). If 휅 < 훼 are ordinals, 푗 : 퐿[퐸] −→ 푀 witnesses that 휅 is 훼-tall, and 풯 is the iteration tree induced by 푗, then 풯 is finite.

풯 Proof. Let 풯 be the iteration tree induced by 푗. Lemma 4.17 implies that 휋0,∞ = 푗 풯 휔 and ℳ∞ = 푀. We have that 푀 ⊆ 푀 as 푗 : 푉 → 푀 witnesses that 휅 is 훼-tall. Hence, by Lemma 4.18, 풯 is finite.  Lemma 4.22. Assume (∆). Let 휅 be a cardinal. Suppose 푗 : 푉 → 푀 is an elementary embedding such that crit(푗) = 휅 and 푀 휅 ⊆ 푀. If 풯 is the iteration 풯 tree induced by 푗, then for 훼 ∈ 푙ℎ(풯 ) we have that crit(퐸훼 ) ≥ 휅. 풯 Proof. Suppose for a contradiction that there is an 훼 ∈ 푙ℎ(풯 ) such that crit(퐸훼 ) < 풯 풯 풯 풯 풯 풯 휅. Let 휋 풯 : ℳ ||휈 → 푈푙푡 (ℳ ||휈 , 퐸 ) be the ultrapower map and 휏 = 퐸훼 훼 훼 0 훼 훼 훼 훼 풯 +ℳ풯 ||휈풯 풯 crit(퐸 ) 훼 훼 . Since ℳ is a premouse, it follows that 푟푎푛(휋 풯 휏 ) is cofinal 훼 훼 퐸훼  훼 풯 in 휈훼 . 풯 휅 Notice that 휏훼 ≤ 휅. Hence by 푀 ⊆ 푀 it follows that 풯 풯 풯 +ℳ훼 ||휈훼 풯 푀 |= “ cf(휈훼 ) ≤ crit(퐸훼) < 휈훼 ”. 풯 On the other hand by Lemma 4.17 we have ℳ∞ = 푀 and by Fact 3.4, 풯 풯 ℳ∞ |= “휈훼 is a successor cardinal”. 풯 This is a contradiction. Therefore for all 훼 ∈ 푙ℎ(풯 ) we have crit(퐸훼 ) ≥ 휅.  TALL CARDINALS IN EXTENDER MODELS 19

We remind the reader that our definitions of 표(휅) and 푂(휅) are different from the usual ones. For the usual definitions the following lemma would be false. Lemma 4.23. Assume (∆). If 휅 is a measurable cardinal17 , then 표(휅) > 휅+. If 휇 is a cardinal, 푐푓(휇) > 휅 and 휅 is 휇-strong, then 표(휅) > 휇. Proof. Let 푗 : 퐿[퐸] → 푀 witness either that 휅 is measurable or that 휅 is 휇-strong. Because 푐푓(휇) > 휅 in both cases we have that 푀 휅 ⊆ 푀. 풯 Let 풯 be the iteration tree induced by 푗 and let 퐸훼 be the first extender applied 풯 풯 풯 to ℳ0 such that 훼 + 1 ∈ 푏 , the main branch of 풯 . Since crit(푗) = crit(휋0,∞) = 휅 풯 and 풯 is a normal iteration tree, it follows that crit(퐸훼 ) = 휅. 풯 Claim 4.24. If 푗 witnesses that 휅 is 휇-strong, then 휈0 > 휇. 풯 Proof. Suppose not. Then 휈0 ≤ 휇, and since we do not index extenders on cardinals it follows that 휈0 < 휇. Hence,

풯 +퐿[퐸]|휈0 (퐿[퐸]|휇) |= “푐푓(휈0) ≤ (푐푟푖푡(퐸0 ) ) < 휈0 < 휇”. Since

풯 퐿[퐸] 푀|휇 = ℳ∞|휇 ⊇ 푉휇 ⊇ 퐿[퐸]|휇, it follows that

풯 +퐿[퐸]|휈0 푀 |= “푐푓(휈0) ≤ (푐푟푖푡(퐸0 ) ) < 휈0 < 휇”.

풯 풯 풯 But ℳ∞ = 푀 and ℳ∞ |= “휈0 is regular cardinal”, which is a contradiction. 

I Suppose 훼 = 0. We split our analysis into two cases depending on whether 푗 witnesses that 휅 is a measurable cardinal or 푗 witnesses that 휅 is 휇-strong. 풯 Suppose 푗 witnesses that 휅 is a measurable cardinal. As 훼 = 0, it follows that 퐸0 풯 + 풯 is a total measure over 퐿[퐸] which implies that 휈0 > 휅 . Therefore 퐸0 witnesses that 푂(휅) > 휅+. By Lemma 3.23 we have 표(휅) > 휅+. Suppose 푗 witnesses that 휅 is 휇-strong. In this case, by Claim 4.24 it follows 풯 that 퐸0 witnesses 푂(휅) > 휇. Therefore by Lemma 3.23 it follows that 표(휅) > 휇. I Suppose we are in the case where 훼 > 0. 풯 Claim 4.25. If 푗 witnesses that 휅 is measurable, then for every 훾 ≥ 휈훼 we have 풯 + 풯 휌1(ℳ훼 ||훾) > 휅 . If 푗 witnesses that 휅 is 휇-strong, then for every 훾 ≥ 휈훼 we have 풯 휌1(ℳ훼 ||훾) > 휇. Proof. Suppose 푗 witnesses that 휅 is a measurable cardinal. By Lemma 4.22 it 풯 풯 풯 풯 풯 풯 +ℳ풯 |휈풯 follows that 휅 ≤ crit(퐸0 ) < 휈0 . As ℳ0 |휈0 = ℳ훼 |휈0 it follows that 휅 0 0 = +ℳ풯 풯 풯 풯 풯 휅 훼 < 휈0 . Since crit(퐸훼 ) = 휅, 훼 + 1 ̸∈ 퐷 and 퐸훼 is the first extender used + +ℳ훼 + 풯 along the main branch of 풯 , altogether we have that 휅 = 휅 . Hence 휅 < 휈0 . 풯 풯 + As every initial segment of ℳ0 is sound, it follows that 휌1(ℳ0 |훾) ≥ 휅 for 풯 any 훾 ≥ 휈0 . Hence one can verify by induction along the branch (0, 훼]푇 that 풯 + 풯 휌1(ℳ훽 ||훾) > 휅 for any 훽 ∈ (0, 훼]푇 and any 훾 ≥ 휈0 . 풯 풯 Next suppose 푗 witnesses that 휅 is a 휇-strong cardinal. Then 휈훼 > 휈0 > 휇, 풯 where the second inequality is Claim 4.24 and we have 휌1(ℳ0 ||훾) ≥ 휇 for any 풯 풯 훾 ≥ 휈0 . Again, by induction along (0, 훼]푇 one can verify that 휌1(ℳ훽 ||훾) > 휇 for 풯 any 훽 ∈ (0, 훼]푇 and any 훾 ≥ 휈0 . 

17The fact that the first part of this lemma holds under much weaker hypothesis isdue Schlutzenberg, see [Sch13]. Here we are working with the hypothesis that (Δ) holds which makes it easy to verify the lemma. 20 GABRIEL FERNANDES* AND RALF SCHINDLER‡

We again split our analysis depending on whether 푗 witnesses that 휅 is a mea- surable cardinal or 푗 witnesses that 휅 is 휇-strong. Suppose 푗 witnesses that 휅 is a measurable cardinal. By Claim 4.25 we have 풯 풯 풯 ++ℳ훼||휈 풯 풯 풯 휌1(ℳ훼 ||휈훼 ) ≥ 휅 훼 , hence we can apply Lemma 3.22 to ℳ훼 ||휈훼 and 퐸훼 to 풯 풯 풯 풯 풯 풯 풯 ℳ훼 ℳ훼 ||휈훼 + ++ℳ ||휈 풯 ++ℳ ||휈 find 퐸훾 with 푐푟푖푡(퐸훾 ) = 휅 and 훾 ∈ (휅 , 휅 훼 훼 ). As 휈0 > 휅 훼 훼 풯 ℳ훼 + it follows that 퐸훾 = 퐸훾 . Therefore 푂(휅) > 휅 and by Lemma 3.23 we have 표(휅) > 휅+. Suppose 푗 witnesses that 휅 is a 휇-strong cardinal. By Claim 4.25 we have 풯 풯 풯 +ℳ훼||휈 풯 풯 풯 휌1(ℳ훼 ||휈훼 ) ≥ 휇 훼 , hence we can apply Lemma 3.22 to ℳ훼 ||휈훼 and 퐸훼 풯 풯 풯 풯 풯 ℳ훼 ℳ훼 ||휈훼 +퐿[퐸] +ℳ ||휈 풯 to find 퐸훾 with 푐푟푖푡(퐸훾 ) = 휅 and 훾 ∈ (휅 , 휇 훼 훼 ). As 휈0 ≥ 풯 풯 풯 +ℳ ||휈 ℳ훼 휇 훼 훼 it follows that 퐸훾 = 퐸훾 . Thus 푂(휅) > 휇 which by Lemma 3.23 implies 표(휅) > 휇.  We now have all the technical tools we need to prove Theorem A. We shall use the following two results due to Hamkins which establish one direction of Theorem A. Theorem 4.26 (Theorem 2.10 in [Ham09]). Suppose 푉 is an extender model 퐿[퐸] that is normaly iterable. If 휇 is a cardinal, 푐푓(휇) > 휅 and 표(휅) > 휇 then 휅 is 휇-tall. Theorem 4.27 (Corollary 2.7 in [Ham09]). Suppose 푉 is an extender model 퐿[퐸] that is normaly iterable. If 표(휅) > 휅+ and sup{훼 < 휅 | 표(훼) > 휇} = 휅 then 휅 is 휇-tall. Moreover, if 표(휅) > 휅+ and sup{훼 < 휅 | 훼 is a strong cardinal} = 휅, then 휅 is a tall cardinal. We now prove the main theorem. Theorem A. Assume (∆). Let 휅 < 휇 be regular cardinals. Suppose further that 퐿[퐸]|휇 is 휇-stable above 휅. Then 휅 is 휇-tall iff 표(휅) > 휇 or (︁ )︁ 표(휅) > 휅+ ∧ 푠푢푝{휈 < 휅 | 표(휈) > 휇} = 휅

Proof. (⇐) It follows from 4.26 and 4.27. (⇒) Since 휅 is 휇-tall, we have that 휅 is measurable and hence by Lemma 4.23 we also have that 표(휅) > 휅+. Suppose that 휅 > 푠푢푝({훼 < 휅 | 표(훼) > 휇}). We will verify that 표(휅) > 휇. Towards a contradiction, suppose that 표(휅) ≤ 휇. By 3.23, 푠푢푝({훽 < 휅 | 표(훽) > 휇}) < 휅 implies that 푠푢푝({훽 < 휅 | 푂(훽) > 휇}) < 휅. Let 훽* := 푠푢푝({훽 < 휅 | 푂(훽) > 휇}), and set Θ := 푠푢푝({푂(훽) | 훽* < 훽 ≤ 휅}). Notice that Θ ≤ 휇: by Lemma 3.23, 표(휅) ≤ 휇 implies 푂(휅) ≤ 휇 and the definition of 훽* implies that 푠푢푝({푂(훽) | 훽* < 훽 < 휅}) ≤ 휇. Let 푗 : 푉 → 푀 witnesses the 휇-tallness of 휅 and let 풯 be the iteration tree 풯 풯 푀 induced by 푗. Lemma 4.17 gives that 푗 = 휋0,∞ and ℳ∞ = 풦 = 푀 and Lemma 4.21 implies that 풯 is finite. TALL CARDINALS IN EXTENDER MODELS 21

Let 푏 be the main branch of 풯 . We know by Theorem 4.15 (or Lemma 4.17) that there is no drop along 푏 so we can define

(︁ 풯 풯 )︁ 푡0 = 푚푖푛 {푚 ∈ 푏 | 휋0,푚(휅) = 휋0,∞(휅)} . For Υ ∈ 푂푅 and 푘 ∈ 푙ℎ(풯 ) let 휓(푘, Υ) denote the following statement: 풯 ℳ푘 |= “∀휁 > Υ(푐푟푖푡(퐸휁 ) ̸∈ (휇, Υ))”. 풯 풯 Claim 4.28. Let 푛 ≤ 푡0. Then 휈푛 < 휇 and whenever 휋0,푛(Θ) is defined we have 풯 풯 휈푛 ≤ 휋0,푛(Θ) ≤ 휇 . Before we start with the proof of Claim 4.28 we observe that Lemma 3.20 and Claim 4.28 imply that 풯 lives on 퐿[퐸]|휇. Our hypothesis that 퐿[퐸]|휇 is 휇-stable 풯 above 휅 together with Lemma 3.15 imply that 휋0,∞(휅) ≤ 휇. This give us a contra- 풯 diction since 휇 < 푗(휅) = 휋0,∞(휅) and therefore we have 표(휅) > 휇. Hence Theorem A will follow once we prove Claim 4.28.

Proof of Claim 4.28. We prove this by induction on 푛 ≤ 푡0. We start with the base case.

풯 풯 Subclaim 4.29. For 푛 = 0, we have that 휈0 ≤ Θ and 휈0 < 휇. 풯 Proof. For a contradiction suppose that 휈0 > Θ. We will prove that for all 푘 + 1 < 푙ℎ(풯 ) we have crit(퐸풯 ) > Θ. This will imply that if 퐸풯 is the first extender 푘 푘0 풯 applied to ℳ0 with 푘0 + 1 ∈ [0, 푡0]푇 , then crit(휋풯 ) = crit(퐸풯 ) > Θ > 휅 = crit(휋풯 ), 0,∞ 푘0 0,∞ which will give us a contradiction. Suppose 휓(푘, Θ) holds. That is, 풯 ℳ푘 |= “∀휁 > Θ(푐푟푖푡(퐸휁 ) ̸∈ (휇, Θ))”. 풯 풯 풯 * 풯 As 풯 is normal we have 휈푘 ≥ 휈0 > Θ, and therefore, crit(퐸푘 ) ≤ 훽 or crit(퐸푘 ) > 풯 Θ. By Lemma 4.22, it follows that crit(퐸푘 ) > Θ. Thus given 푘 ∈ 푙ℎ(풯 ), 휓(푘, Θ) 풯 implies crit(퐸푘 ) > Θ. We will verify by induction that 휓(푘, Θ) holds for all 푘 ∈ 푙ℎ(풯 ). For 푘 = 0, Lemma 3.24 implies 휓(0, Θ). Suppose now that 푘 ∈ 푙ℎ(풯 ) and 휓(푙, Θ) holds for all 푙 ≤ 푘. We will prove that 휓(푘 + 1, Θ) holds. 풯 As observed above, 휓(푘, Θ) implies crit(퐸푘 ) > Θ. By the induction hypothesis, 휓(푘, Θ) holds, and hence we have18 풯 풯 휂푘 > 푐푟푖푡(퐸푘 ) > Θ. 풯 By induction hypothesis we also have 휓(휉푘 , Θ), therefore the following holds: (ℳ풯 )||휂풯 |= “∀훾 > Θ(푐푟푖푡(퐸 ) ̸∈ (휇, Θ))”. 휉푘 푘 훾 Thus by the Σ -elementarity of 휋풯 it follows that 휓(푘 + 1, Θ) holds. This 1 휉푘+1,푘 concludes the proof that 휓(푘, Θ) holds for all 푘 ∈ 푙ℎ(풯 ). 풯 As observed above, this implies that crit(휋0,∞) > Θ, which is a contradiction. Hence 풯 휈0 ≤ Θ ≤ 휇, 풯 and as 휇 is a cardinal we also have 휈0 < 휇. This concludes the case 푛 = 0 of Claim 4.28 and the proof of subclaim 4.29. 

18 풯 풯 풯 Recall 휂푘+1 is the height of the model we apply 퐸푘 to form ℳ푘+1. 22 GABRIEL FERNANDES* AND RALF SCHINDLER‡

We now perform the inductive step of the proof of Claim 4.28. We shall need to split this into two cases. 풯 I Suppose 푛 = 푘 + 1 and 휋0,푘+1(Θ) is not defined. By our induction hypothesis 풯 휈푙 < 휇 for all 푙 < 푘 + 1, hence by Lemma 3.20, the iteration tree 풯 |(푘 + 2) lives on 퐿[퐸]|휇. Therefore by Lemma 3.18, 풯 풰 (10) ℳ푘+1 = ℳ푘+1 풰 where 풰 = (풯  푘 + 2)  (퐿[퐸]|휇). By Lemma 3.15 we have ℳ푘+1 ∩ 푂푅 ≤ 휇. 풯 Therefore 휈푘+1 < 휇. 풯 I Suppose 푛 = 푘 + 1 and 휋0,푘+1(Θ) is defined. By our induction hypothesis 풯 휈푙 < 휇 for all 푙 < 푘 + 1, hence by Lemma 3.20 the iteration tree 풯  (푘 + 2) lives on 퐿[퐸]|휇. Therefore by Lemma 3.18 풯 (11) 휋0,푘+1(Θ) ≤ 휇. 풯 풯 Thus we only have to verify that 휈푘+1 ≤ 휋0,푘+1(Θ). From our induction hypothesis 풯 풯 풯 풯 풯 풯 +ℳ푘+1||휈푘 we have 휈 풯 ≤ 휋 풯 (Θ). Let 휏푘 = 푐푟푖푡(퐸푘 ) , then 휉푘 0,휉푘 풯 풯 풯 휏푘 < 휈 풯 ≤ 휋 풯 (Θ), 휉푘 0,휉푘 which implies the following: 풯 풯 풯 풯 풯 풯 휋0,푘+1(Θ) ≥ 휋 풯 (휈 풯 ) > 휋 풯 (휏푘 ) = 휈푘 . 휉푘 ,푘+1 휉푘 휉푘 ,푘+1 Therefore, 풯 풯 (12) 휈푘 < 휋0,푘+1(Θ). 풯 풯 Suppose for a contradiction that 휈푘+1 > 휋0,푘+1(Θ). Using Lemma 3.24, one can 19 풯 verify by induction along [0, 푘 + 1]풯 that 휓(푘 + 1, 휋0,푘+1(Θ)) holds. Recall that 풯 휓(푘 + 1, 휋0,푘+1(Θ)) denotes 풯 풯 * 풯 ℳ푘+1 |= “∀휈 > 휋0,푘+1(Θ) ( 푐푟푖푡(퐸휈 ) ̸∈ (훽 , 휋0,푘+1(Θ)))”. Subclaim 4.30. For 푙 ∈ 푙ℎ(풯 ), if 푘 + 1 < 푙 then 풯 (13) (푘 + 1 <푇 푙) ∧ 휓(푙, 휋0,푘+1(Θ)). Proof. We will verify this by induction, similar to what we did for the case 푛 = 0. 풯 풯 We shall start by verifying that 푘 + 1 <푇 푘 + 2. As 휈푘+1 > 휋0,푘+1(Θ) and 풯 풯 풯 휓(푘 + 1, 휋0,푘+1(Θ)) holds, we must have, by (12), that crit(퐸푘+1) > Θ > 휈푘 . Thus 풯 휉푘+1 = 푘 + 1 and 푘 + 1 <푇 푘 + 2. 풯 풯 Next we verify that 휓(푘 + 2, 휋0,푘+1(Θ)) holds. As 휓(푘 + 1, 휋0,푘+1(Θ)) holds, it follows that 풯 풯 풯 * 풯 (ℳ 풯 ) ||휂 |= “∀휁 > 휋 (Θ) (푐푟푖푡(퐸휁 ) ̸∈ (훽 , 휋 (Θ)))”. 휉푘+1 푘+1 0,푘+1 0,푘+1 풯 풯 Then by the Σ1-elementarity of 휋푘+1,푘+2 it follows that 휓(푘 + 2, 휋0,푘+1(Θ)) holds. This concludes the base step of this induction. We now prove the inductive step. Suppose (13) holds for all 푙 such that 푘 + 1 < 풯 풯 푙 ≤ 푚, let us verify that (13) holds for 푚+1. As 풯 is normal, we have 휈푚 > 휈푘 . By 풯 our induction hypothesis 휓(푚, 휋0,푘+1(Θ)) holds and we have only two possibilities: 풯 * 풯 풯 푐푟푖푡(퐸푚) ≤ 훽 or 푐푟푖푡(퐸푚) > 휋0,푘+1(Θ). The first possibility is excluded by Lemma 4.22, and so it must be true that 풯 풯 풯 풯 풯 풯 푐푟푖푡(퐸푚) > 휋0,푘+1(Θ). By (12), 푐푟푖푡(퐸푚) > 휋0,푘+1(Θ) implies 푐푟푖푡(퐸푚) > 휈푘 .

19 We use induction like in case 푛 = 0, there may be drops in model along [0, 푘 + 1]풯 but by 풯 hypothesis 휋0,푘+1(Θ) is defined for all 푚 ∈ [0, 푘 + 1]풯 . TALL CARDINALS IN EXTENDER MODELS 23

풯 풯 Therefore 푘 + 1 ≤ 휉푚 = pred푇 ((푚 + 1)) and thus we have either 푘 + 1 = 휉푚 or 풯 푘 + 1 < 휉푚. 풯 If the former is true, then we can appeal to 휓(푘 + 1, 휋0,푘+1(Θ)), and if the latter is true, then we can appeal to the induction hypothesis, and in either case, we can 풯 conclude that 휓(휉푚, 휋0,푘+1(Θ)) holds and 풯 풯 풯 * 풯 (ℳ 풯 ) ||휂 |= “∀휁 > 휋 (Θ) (푐푟푖푡(퐸) ) ̸∈ (훽 , 휋 (Θ)))”. 휉푚 푚 0,푘+1 휁 0,푘+1 풯 풯 풯 By Σ1-elementarity of 휋 풯 , we have 휓(푚+1, 휋 (Θ)). Now if 푘 +1 = 휉푚, 휉푚,푚+1 0,푘+1 풯 then by the definition of 휉푚, we have 푘 + 1 <푇 푚 + 1. On the other hand, if 풯 풯 푘 + 1 < 휉푚, then by the induction hypothesis, 푘 + 1 <푇 휉푚, and so again we have that 푘 + 1 <푇 푚 + 1. This concludes the inductive step and the induction and verifies Subclaim 4.30.  풯 Given 푙 ∈ 푙ℎ(풯 ) such that 푙 ≥ 푘 + 1, we have that 휓(푙, 휋0,푘+1(Θ)) implies 풯 풯 crit(퐸푙 ) > 휋0,푘+1(Θ). Therefore Subclaim (4.30) gives that 풯 풯 풯 휋푘+1,∞  (휋0,푘+1(Θ) + 1) = 푖푑  (휋0,푘+1(Θ) + 1). Hence 풯 풯 풯 풯 풯 휋0,∞(휅) = 휋푘+1,∞ ∘ 휋0,푘+1(휅) = 푖푑 ∘ 휋0,푘+1(휅) ≤ 휋0,푘+1(Θ) ≤ 휇, where the last inequality is given by (11). This contradicts the fact that 풯 휋0,∞(휅) = 푗(휅) > 휇. 풯 풯 Thus 휈푘+1 ≤ 휋0,푘+1(Θ) ≤ 휇. This verifies the case 푛 = 푘 + 1.  As observed before the proof of Claim 4.28, we have by Lemma 3.20 and Claim 풯 4.28 that 휇 < 푗(휅) = 휋0,∞(휅) ≤ 휇 which is a contradiction. Thus 표(휅) > 휇. This concludes the proof of Theorem A.  Definition 4.31. (Hamkins) A cardinal 휅 is < 훼-tall if and only if for all 훽 < 훼 휅 is 훽-tall. Corollary 4.32. Assume (∆). Suppose that 훼 is a limit cardinal and 푐푓(훼) > 휅. Then 휅 is < 훼-tall iff “표(휅) ≥ 훼 or 표(휅) > 휅+ ∧ 휅 = 푠푢푝{훽 < 휅 | 표(훽) ≥ 훼}” Proof. (⇐) It follows from 4.26 and 4.27. ++ (⇒) Let ⟨휇휉 | 휉 < 푐푓(훼)⟩ be a cofinal sequence in 훼. Note that for 휇 := 휇휉 ++ ++ we are in hypothesis of Theorem A. If for each 휇휉 > 휅 we have 표(휅) > 휇휉 then 표(휅) ≥ 훼 and we are done. Suppose 표(휅) < 훼. By Lemma 4.23, we have 표(휅) > 휅+, therefore we will be done if we find 퐵 ⊆ 휅 such that 퐵 is cofinal in 휅 and for all 훽 ∈ 퐵 we have 표(훽) ≥ 훼. ++ ++ Fix 휉 < 푐푓(훼) such that 휇휉 > 표(휅). As 휅 is 훼-tall, it follows that 휅 is 휇휉 -tall. ++ ++ Applying Theorem A to 휅 and 휇휉 , as 표(휅) < 휇휉 , this gives us a set 푌 ⊆ 휅 ++ cofinal in 휅 such that for all 훽 ∈ 푌 we have 표(훽) > 휇휉 . For each 휉 < 푐푓(훼), let ++ 퐵휉 = {훽 < 휅 | 표(훽) > 휇휉 }.

Then ⟨퐵휉 | 휉 < 푐푓(훼)⟩ is a sequence of cofinal subsets of 휅 such that (︁ )︁ ∀휉 휉 < 휁 < 푐푓(훼) −→ 퐵휁 ⊆ 퐵휉 . 24 GABRIEL FERNANDES* AND RALF SCHINDLER‡

From the fact that 푐푓(훼) > 휅, it follows that the sequence ⟨퐵휉 | 휉 < 푐푓(훼)⟩ is eventually constant, i.e., there is some 퐵 ⊆ 휅 such that for all sufficiently large 휉 < 푐푓(훼) we have 퐵 = 퐵휉. We have that 퐵 is cofinal in 휅 and for all 훽 ∈ 퐵 we have 표(훽) ≥ 훼, which verifies the corollary.  Corollary A. Assume (∆). 휅 is tall if and only if 휅 is a strong cardinal or a measurable limit of strong cardinals. We will need one further notion of iterability for Theorem B. Definition 4.33. We say that a premouse ℳ is weakly iterable if every countable 20 premouse ℳ¯ that elementarily embeds into ℳ is (휔1 + 1, 휔1)-iterable . The following lemma together with Theorem A imply Theorem B. Lemma 4.34. Suppose there is no inner model with a Woodin cardinal. Let 퐿[퐸] be a proper class premouse that is weakly iterable. Then 퐿[퐸] is self-iterable and 퐿[퐸] |= (∆). Proof Sketch. Let 풯 ∈ 퐿[퐸] be an iteration on 퐿[퐸] of limit lenght. By our hy- pothesis that there is no inner model with a Woodin cardinal it follows that21 퐿[ℳ(풯 )] |= “훿(풯 ) is not a Woodin cardinal”. Let 휂 be the least ordinal such that (퐿[ℳ(풯 )]|휂 + 휔) |= “훿 is not a Woodin cardinal”, ℳ and set 풬(풯 ) := 퐿[ℳ(풯 )]||휂. Let 푋 be a countable set such that 푋 ≺Σ휔 퐻Θ , let 푋¯ be the Mostowski collapse of 푋 and 휋 : 푋¯ → 푋 be the inverse of the Mostowski collapse. For each 푤 ∈ 푋 we will denote by푤 ¯ the pre-image of 푤 under 휋. Since 퐿[퐸] is countably iterable, it follows that there exists 푏 a cofinal wellfounded ¯ 풯¯ ¯ 22 branch of 풯 such that ℳ∞ ◁ 풬(푏, 풯 ) = 풬(풯 ) and such 푏 is unique . 푋¯ Let 퐺 be a 퐶표푙(휔, 휈)-generic over 푋¯, where (휈 = |풯¯ |) . As 푋¯[퐺] ≺ 1 푉 , it Σ1 follows that 푏 ∈ 푋¯[퐺] and by homogeneity of 퐶표푙(휔, 휈) it follows that 푏 ∈ 푋. Therefore ℳ 퐻Ω |= ∃푐 (푐 is a cofinal well founded branch of 풯 )  Theorem B. Suppose there is no inner model with a Woodin cardinal and 퐿[퐸] is an extender model that is self-iterable. Let 휅, 휇 be ordinals such that 휅 < 휇 and 휇 is a regular cardinal. If 퐿[퐸]|휇 is 휇-stable above 휅, then (휅 is 휇-tall)퐿[퐸] iff (표(휅)) > 휇)퐿[퐸] (14) or 표(휅) > 휅+ ∧ 휅 = sup{휈 < 휅 | 표(휈) > 휇})퐿[퐸]

In particular, if 퐿[퐸] is weakly iterable, then (휅 is 휇-tall)퐿[퐸] iff (14) holds. Proof. The first part follows from the fact that 퐿[퐸] |= (∆) and Theorem A. The second part follows from Lemma 4.34 and Theorem A. 

20 See [Zem02, p.309] for the definition of (휔1 + 1, 휔1)-iterable. 21 ⋃︀ 풯 ℳ(풯 ) denotes the common part model, see [Ste10, Definition 6.9]. For 훿(풯 ) = 훼∈푙ℎ(풯 ) 휈훼 풯 풯 ⋃︀ ℳ훼 휈훼 E and E := 훼∈푙ℎ(풯 ) 퐸 , ℳ(풯 ) := 퐽훿 . 22See [Ste10, Definition 6.11] for the definition of 풬(푏, 풯¯). TALL CARDINALS IN EXTENDER MODELS 25

Corollary B. Suppose there is no inner model with a Woodin cardinal and 퐿[퐸] is an extender model that is self-iterable. Let 휅 be an ordinal. Then (휅 is tall)퐿[퐸] iff (휅 is a strong cardidnal (15) or 휅 is a measurable limit of strong cardinals)퐿[퐸]. In particular, if 퐿[퐸] is weakly iterable, then (휅 is tall)퐿[퐸] iff (15) holds. Next we prove that we can not remove the hypothesis that 퐿[퐸]|휇 is 휇-stable above 휅 in Theorem A. For that we will use the following lemma. Lemma 4.35 (Lemma 2.3 in [Ham09]). If 휅 < 휃 are ordinals and 푗 : 푉 −→ 푀 is an elementary embedding such that 휅푀 ⊆ 푀 and 푗(휅) ≥ 휃, then 휅 is 휃-tall. Proof sketch. By elementarity of 푗 it follows that 푗(휅) is measurable in 푀, let 푈 ∈ 푀 be a total measure on 푀 with 푐푟푖푡(푈) = 푗(휅) and 푖 the ultrapower embedding from 푈, then 푖 ∘ 푗 witness that 휅 is 휃-tall.  Lemma 4.36. Suppose (∆) . Let 휅 and 휆 be cardinals. If (I) 1. 휅 < 휆, 2. 표(휅) ∈ [휆, 휆+), 3. for some 휇, 휅 < 푐푓(휇) and 4. cf(휇) is a measurable cardinal. or (II) 1. 휅 is a measurable cardinal, 2. 휆 > 휅 and cf(휆) = 휅, 3. 휅 = sup({훽 < 휅 | 표(훽) > 휆}), Then 휅 is 휆+-tall.

Proof. I Suppose (I) holds. By Lemma 4.23 there is 퐸훽 a total measure with crit(퐸훽) = cf(휆). Consider 풩 = 푈푙푡(푉, 퐸훽) and let 휋 be the ultrapower map. We have 휋(휆) ≥ 휆+, which implies 풩 |= 표(휅) ∈ [휋(휆), 휋(휆+푉 )] ⊆ [휆+푉 , 휋(휆)+풩 ) 풩 +푉 풩 풩 Let 퐸훼 be such that 훼 > 휆 and crit(퐸훼 ) = 휅. Let 풲 := 푈푙푡(풩 , 퐸훼 ) and let 푖 : 풩 → 풲 be the ultrapower map. It follows that 푖 ∘ 휋(휅) ≥ 휆+푉 and 풲휅 ⊆ 풲. Hence by Lemma 4.35 휅 is 휆+-tall. I Suppose (II) holds. By Lemma 4.23 there is 퐸훽 a total measure with crit(퐸훽) = 휅. Consider 풩 = 푈푙푡(푉, 퐸훽) and let 휋 : 푉 → 풩 be the ultrapower map. We have 풩 |= sup({훽 < 휋(휅) | 표(훽) ≥ 휋(휆)}) = 휋(휅)(16) Notice that 휋(휆) > 휆+푉 . By (16) and Lemma 3.22 there is 훾 ∈ (휆+푉 , 휋(휆)) such 풩 that crit(퐸훾 ) ∈ (휅, 휋(휅)). + +푉 +푉 휅 If we consider 퐹 := 퐸훾  휆 , as cf(휆 ) = 휆 > 휅 and 풩 ⊆ 풩 , it follows that 풲 := 푈푙푡(풩 , 퐹 ) is 휅-closed. We also have for 푖 : 풩 → 풲, the ultrapower map, that 푖 ∘ 휋(휅) ≥ 휆+푉 . Hence by Lemma 4.35 휅 is 휆+-tall.  Remark 4.37. Notice that in Lemma 4.36 setting 휇 = 휆+ it follows that 퐿[퐸]|휇 is not 휇-stable above 휅, as 휆 is the largest cardinal of 퐿[퐸]|휇 and 푐푓(휆) is a measurable > 휅. 26 GABRIEL FERNANDES* AND RALF SCHINDLER‡

Acknowledgements The authors express their gratitude to Tanmay Inamdar for reading earlier ver- sions of this paper and providing many helpful comments and suggestions. The authors express their gratitude to the referee for a thorough reading and valuable report.

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Department of Mathematics, Bar-Ilan University, Ramat-Gan 5290002, Israel. URL: http://u.math.biu.ac.il/~zanettg Email address: [email protected]

Institut fur¨ mathematische Logik und Grundlagenforschung, Universitat¨ Munster,¨ Einsteinstr. 62, 48149, Munster,¨ Germany Email address: [email protected] URL: https://ivv5hpp.uni-muenster.de/u/rds/