Ordinal Definable Subsets of Singular Cardinals 11

ORDINAL DEFINABLE SUBSETS OF SINGULAR CARDINALS

JAMES CUMMINGS, SY-DAVID FRIEDMAN, MENACHEM MAGIDOR, ASSAF RINOT, AND DIMA SINAPOVA

Abstract. A remarkable result by Shelah states that if 휅 is a singular strong limit cardinal of uncountable cofinality then there is a subset 푥 of 휅 such that HOD푥 contains the power set of 휅. We develop a version of diagonal extender- based supercompact Prikry forcing, and use it to show that singular cardinals of countable cofinality do not in general have this property, and in factitis consistent that for some singular strong limit cardinal 휅 of countable cofinality + 휅 is supercompact in HOD푥 for all 푥 ⊆ 휅.

1. Introduction It is a familiar phenomenon in the study of singular cardinal combinatorics that singular cardinals of countable cofinality can behave very differently from those of uncountable cofinality. For example in the area of cardinal arithmetic we may contrast Silver’s theorem [11] with the many consistency results producing models where GCH first fails at a singular cardinal of countable cofinality [2]. The results in this paper show that there is a similar sharp dichotomy involving questions about the definability of subsets of a cardinal rather than the size of its powerset. Shelah [12] proved that if 휅 is a singular strong limit cardinal of uncountable cofinality then there is a subset 푥 ⊆ 휅 such that 푃 (휅) ⊆ HOD푥. In this paper we show that the hypothesis is essential, by proving: Theorem. Suppose that 휅 < 휆 where cf(휅) = 휔, 휆 is inaccessible and 휅 is a limit of 휆-supercompact cardinals. There is a forcing poset Q such that if 퐺 is Q-generic then: ∙ The models 푉 and 푉 [퐺] have the same bounded subsets of 휅. ∙ Every infinite cardinal 휇 with 휇 ≤ 휅 or 휇 ≥ 휆 is preserved in 푉 [퐺]. ∙ 휆 = (휅+)푉 [퐺]. ∙ For every 푥 ⊆ 휅 with 푥 ∈ 푉 [퐺], (휅+)HOD푥 < 휆.

From stronger assumptions we can use Q to obtain a model in which 휅 is a + singular strong limit cardinal of coﬁnality 휔, and 휅 is supercompact in HOD푥 for all 푥 ⊆ 휅.

Cummings was partially supported by the National Science Foundation, grants DMS-1101156 and DMS-1500790. Friedman would like to thank the Austrian Science Fund for its generous support through the research project P25748. Rinot was partially supported by the Israel Science Foundation, grant 1630/14. Sinapova was partially supported by the National Science Foundation, grants DMS-1362485 and Career-1454945. 1 2 J. CUMMINGS, S.-D. FRIEDMAN, M. MAGIDOR, A. RINOT, AND D. SINAPOVA

Acknowledgements. The results in this paper were conceived and proved during three visits to the American Institute of Mathematics, as part of the Institute’s SQuaRE collaborative research program. The authors are deeply grateful to Brian Conrey, Estelle Basor and all the staﬀ at AIM for providing an ideal working envi- ronment. Notation. Our notation is mostly standard. We write ℓ(푠) for the length of a sequence 푠. When Q is a forcing poset and 푞 ∈ Q we write Q ↓ 푞 for {푟 ∈ Q : 푟 ≤ 푞}, with the partial ordering inherited from Q.

2. A proof of Shelah’s theorem For the reader’s benefit we give a short and self-contained proof of the result by Shelah quoted in the Introduction. Let 휅 be a strong limit singular cardinal of uncountable cofinality 휇. Fix a set 푥 ⊆ 휅 such that 퐿[푥] contains an enumeration (푡휂)휂<휅 of 퐻휅, together with a sequence of ordinals (훼푖)푖<휇 which is increasing and cofinal in 휅. To each set 푋 ⊆ 휅 we associate a function 푓푋 : 휇 → 휅, where 푓푋 (푖) = 휂 for the unique 휂 such that 푋 ∩ 훼푖 = 푡휂. It is easy to see that if 푋 ̸= 푌 then 푓푋 (푖) ̸= 푓푌 (푖) for all large 푖. We now impose a strict partial ordering on 푃 (휅) by defining 푋 C 푌 if and only if 푓푋 (푖) < 푓푌 (푖) for all large 푖; since 휇 = cf(휇) > 휔, the ordering C is well-founded. Let 푅훼 be the set of elements of 푃 (휅) with C-rank 훼, and note that if 푋 and 푌 are distinct elements of 푅훼 then there are unboundedly many 푖 < 휇 with 푓푋 (푖) > 푓푌 (푖), since otherwise 푓푋 (푖) < 푓푌 (푖) for all large 푖 and it would follow that 푋 C 푌 . We 휇 claim that |푅훼| ≤ 2 , so that in particular |푅훼| < 휅. Supposing for contradiction 휇 that |푅훼| > 2 we fix a sequence (푋푗)푗<(2휇)+ of distinct elements of 푅훼, and then define a colouring 푐 of pairs from (2휇)+ by setting 푐(푗, 푗′) = 푖 for the least 푖 < 휇 with 푓푋푗 (푖) > 푓푋푗′ (푖). By the Erd˝os-Radotheorem 푐 has a homogeneous set of order type 휇+, which is impossible because it would yield an infinite decreasing sequence of ordinals. ⋃︀ We now define a well-ordering ≺훼 of 푅훼. To do this we let 퐴훼 = {range(푓푋 ): 푋 ∈ 푅훼}, note that |퐴훼| < 휅, and define 푔훼 to be the order-isomorphism between 퐴훼 and ot(퐴훼). If 푋 ∈ 푅훼 then 푔훼 ∘ 푓푋 ∈ 퐻휅, and we define 푋 ≺훼 푌 if and only if the index of 푔훼 ∘ 푓푋 is less than that of 푔훼 ∘ 푓푌 in the enumeration (푡휂)휂<휅 of 퐻휅. Forming the sum of the orderings ≺훼 in the natural way, we obtain a well- ordering ≺ of 푃 (휅) which is clearly definable from 푥. Now every element of 푃 (휅) is ordinal definable from 푥 as “the 휂th element of 푃 (휅) in the ordering ≺”, so that 푃 (휅) ⊆ HOD푥.

3. The main theorem

3.1. Setup. Let 휅 < 휆 with 휆 strongly inaccessible and 휅 = sup푛 휅푛, where (휅푛)푛<휔 is a strictly increasing sequence of 휆-supercompact cardinals. Fix 푈푛 a

휅푛-complete ﬁne normal ultraﬁlter on 푃휅푛 휆, and for 휅 ≤ 훼 < 휆 let 푈푛,훼 be the projection of 푈푛 to 푃휅푛 훼 via the map 푥 ↦→ 푥 ∩ 훼. We need a form of diagonal intersection lemma.

Lemma 1. Let 푘 < 푙 < 휔, let (훼푖)푘≤푖<푙 be a ≤-increasing sequence of elements of ∏︀ [휅, 휆), and let 푆 ⊆ 푘≤푖<푙 푃휅푖 훼푖 be a set of ⊆-increasing sequences. ORDINAL DEFINABLE SUBSETS OF SINGULAR CARDINALS 3

Let 푚 ∈ [푙, 휔), let 훼 ∈ [훼푙−1, 휆), and let 퐴⃗ = (퐴푠)푠∈푆 be an 푆-indexed family with 퐴푠 ∈ 푈푚,훼 for each 푠 ∈ 푆. Then the following set is in 푈푚,훼:

퐷 = {푥 ∈ 푃휅푚 훼 : ∀(푥푘, . . . , 푥푙−1) ∈ 푆 [푥푙−1 ⊆ 푥 =⇒ 푥 ∈ 퐴(푥푘,...푥푙−1)]}.

Proof. Let 푗 : 푉 → 푀 be the ultrapower map computed from 푈푚. To show that 퐷 ∈ 푈푚,훼 it suﬃces to show that 푗“훼 ∈ 푗(퐷). Let 푡 = (푦푘, . . . 푦푙−1) ∈ 푗(푆) with 푦푙−1 ⊆ 푗“훼. Write 퐵⃗ = 푗(퐴⃗). We need to show that 푗“훼 ∈ 퐵푡. For each 푖 with 푘 ≤ 푖 < 푙, 푦푖 ⊆ 푗“훼푖 and |푦푖| < 휅푖 < 휅푚 = crit(푗), so that

푦푖 = 푗“푥푖 = 푗(푥푖) for some 푥푖 ∈ 푃휅푖 훼푖. Write 푠 = (푥푘, . . . 푥푙−1). Then 푡 = 푗(푠) and hence 푠 ∈ 푆. In particular 퐴푠 ∈ 푈푚,훼, and so 푗“훼 ∈ 푗(퐴푠) = 퐵푡. In the setting of Lemma1 we will refer to 퐷 as the diagonal intersection of 퐴⃗ and we will write “푠 b 푥” as shorthand for “푠 = (푥푘, . . . 푥푙−1) with 푥푙−1 ⊆ 푥”. With a view to the proof of Lemma 12 below we also need a technical lemma about measure one sets:

Lemma 2. Let 푖 < 휔 and let 훼푖 ≤ 훽푖 ≤ 훽푖+1 be ordinals Let 퐵 ∈ 푈푖,훽푖 and let ′ 퐶 ∈ 푈푖+1,훽푖+1 . Let 퐶 be the set of 푦 ∈ 퐶 such that for every 푥 ∈ 퐵 with 푥 ∩ 훼푖 ⊆ 푦 ′ ′ ′ there is 푥 ∈ 퐵 ∩ 푃 (푦) with 푥 ∩ 훼푖 = 푥 ∩ 훼푖. Then 퐶 ∈ 푈푖+1,훽푖+1 .

Proof. Let 푗 : 푉 → 푀 be the ultrapower map computed from 푈푖+1. It suﬃces to ′ prove that 푗“훽푖+1 ∈ 푗(퐶 ).

Since 퐶 ∈ 푈푖+1,훽푖+1 , we have 푗“훽푖+1 ∈ 푗(퐶). Suppose that 푋 ∈ 푗(퐵) satisﬁes ′ ′ 푋∩푗(훼푖) ⊆ 푗“훽푖+1; we shall ﬁnd 푋 ∈ 푗(퐵∩푗“훽푖+1) such that 푋 ∩푗(훼푖) = 푋∩푗(훼푖).

Since 퐵 ∈ 푃휅푖 훽푖, 휅푖 < 휅푖+1 = crit(푗) and 푋 ∈ 푗(퐵), it follows that 푋 ∩ 푗(훼푖) = ′ ′ 푗(푡) = 푗“푡 for some 푡 ∈ 푃휅푖 훼푖. By elementarity there is 푥 ∈ 퐵 such that 푥 ∩훼푖 = 푡. ′ ′ ′ ′ Now 푗(푥 ) ∈ 푗(퐵), 푗(푥 ) = 푗“푥 ⊆ 푗“훽푖+1, and 푗(푥 ) ∩ 푗(훼푖) = 푗(푡) = 푋 ∩ 푗(훼푖), so ′ ′ 푋 = 푗(푥 ) is as required. 3.2. The poset Q. The poset Q is a diagonal extender-based forcing of the general type introduced by Gitik and Magidor [4]. For the experts, we note the main innovations:

∙ The extender used here on level 푛 is formed from the approximations 푈푛,훼 to the supercompactness measure 푈푛, rather than from measures approxi- mating some short extender. ∙ The supports of the components of Q are quite large. This is necessary because the proofs below require long diagonalisations, which are only possible because the coordinates of the extender on level 푛 are linearly ordered and all the measures 푈푛,훼 are normal. ∙ Although the supports of the components are large (potentially much bigger than the completeness of the relevant measures) we can avoid some of the complexities of the original extender based forcing [5]. In that context the projections between measures do not commute everywhere, and this complicates matters in several respects: in our context when 훼 < 훼′ the natural projection from 푈푛,훼′ to 푈푛,훼 is just restriction 푥 ↦→ 푥 ∩ 훼, so that the measures 푈푛,훼 and the natural projections form a linear Rudin-Keisler directed system with everywhere commutative projection maps. The poset Q is designed so that (Lemma 13) every subset 푥 of 휅 in the generic extension 푉 [퐺] lies in some intermediate extension by an explicitly described forcing 4 J. CUMMINGS, S.-D. FRIEDMAN, M. MAGIDOR, A. RINOT, AND D. SINAPOVA poset of size less than 휆. This guarantees that 휆 is preserved by Q (Lemma 14), 푉 [퐺] and makes possible the analysis of HOD푥 (Lemma 15). A note on notation: The definitions of the poset Q and its ordering ≤ con- tain quite a number of clauses. We will write for example “b푞” as shorthand for “Clauseb in the definition of 푞 ∈ Q” and “6푞,푟” as shorthand for “Clause6 in the definition of 푞 ≤ 푟”. 푞 Conditions in Q are sequences (푞푘)푘<휔 such that for some 푚 < 휔 we have 푞푘 = 푓푘 푞 푞 푞 for 푘 < 푚 and 푞푘 = (푎푘, 퐴푘, 푓푘 ) for 푘 ≥ 푚, where: 푞 푞 푞 푞 a) 푓푘 is a function with dom(푓푘 ) ⊆ [휅, 휆), |dom(푓푘 )| < 휆, and 푓푘 (휂) ∈ 푃휅푘 휂 푞 for all 푘 < 휔 and all 휂 ∈ dom(푓푘 ). 푞 푞 푞 푞 b) For 푘 ≥ 푚 we have 푎푘 ⊆ [휅, 휆), |푎푘| < 휆, 푎푘 and dom(푓푘 ) are disjoint sets 푞 푞 and 푎푘 has a maximum element 훼푘. 푞 푞 c) For 푘 ≥ 푚 we have 퐴푘 ∈ 푈푘,훼 . 푞 푘 d)( 푎푘)푚≤푘<휔 is ⊆-increasing with 푘. We call the integer 푚 the length of 푞 and write 푚 = ℓ(푞). Remark 1. We note that there is very little interaction among the “components” 푞푘 of the condition 푞: it is onlyd 푞 that connects entries in 푞푘 on different levels 푘. 푞 푞 푞 푞 Remark 2. Note that byb 푞 andd 푞, if ℓ(푞) ≤ 푘 < 푙 then 훼푘 ∈ 푎푙 and 훼푘 ≤ 훼푙 . Given 푟 and 푞 in Q, 푟 ≤ 푞 if and only if: (1) ℓ(푟) ≥ ℓ(푞). 푟 푞 푟 푟 푟 푞 푞 (2) For all 푘, 푓푘 ⊇ 푓푘 (that is, dom(푓푘 ) ⊇ dom(푓푘 ) and 푓푘 dom(푓푘 ) = 푓푘 ). 푞 푟 푟 푞 푞 푟 (3) For 푘 with ℓ(푞) ≤ 푘 < ℓ(푟), 푎푘 ⊆ dom(푓푘 ), 푓푘 (훼푘) ∈ 퐴푘, and 푓푘 (휂) = 푟 푞 푞 푓푘 (훼푘) ∩ 휂 for all 휂 ∈ 푎푘. 푟 푞 (4)( 푓푘 (훼푘))ℓ(푞)≤푘<ℓ(푟) is ⊆-increasing. 푞 푟 푞 푞 푟 (5) For 푘 ≥ ℓ(푟), we have 푎푘 ⊆ 푎푘, and 푥 ∩ 훼푘 ∈ 퐴푘 for all 푥 ∈ 퐴푘. 푟 푞 푟 (6) For 푘 ≥ ℓ(푟), if ℓ(푞) < ℓ(푟), then 푓ℓ(푟)−1(훼ℓ(푟)−1) ⊆ 푥 for all 푥 ∈ 퐴푘. 푟 푞 Remark 3. Note that if 푟 ≤ 푞 then 훼푘 ≥ 훼푘 for all 푘 ≥ ℓ(푟). Lemma 3. The ordering ≤ on Q is transitive.

Proof. Let 푟 ≤ 푞 ≤ 푝 be conditions in Q. Note that1 푟,푝,2 푟,푝 and5 푟,푝 are immediate. To verify3 푟,푝, let ℓ(푝) ≤ 푘 < ℓ(푟) and distinguish two cases: 푟 푝 푞 푝 푞 푝 푝 I ℓ(푝) ≤ 푘 < ℓ(푞): In this case 푓푘 (훼푘) = 푓푘 (훼푘) by2 푟,푞 and 푓푘 (훼푘) ∈ 퐴푘 푟 푝 푝 푝 푞 푞 푝 by3 푞,푝, so 푓푘 (훼푘) ∈ 퐴푘. Moreover if 휂 ∈ 푎푘 then 푓푘 (휂) = 푓푘 (훼푘) ∩ 휂 푟 푞 푟 푝 푞 푝 by3 푞,푝 for 푞 ≤ 푝. Also 푓푘 (휂) = 푓푘 (휂) and 푓푘 (훼푘) = 푓푘 (훼푘) by2 푟,푞, so 푟 푟 푝 푓푘 (휂) = 푓푘 (훼푘) ∩ 휂. 푝 푞 푟 푝 푟 푞 푝 I ℓ(푞) ≤ 푘 < ℓ(푟): We have 훼푘 ∈ 푎푘 by5 푞,푝, so 푓푘 (훼푘) = 푓푘 (훼푘) ∩ 훼푘 by 푟 푞 푞 푟 푞 푝 푝 3푟,푞. Now 푓푘 (훼푘) ∈ 퐴푘 by3 푟,푞, and so 푓푘 (훼푘) ∩ 훼푘 ∈ 퐴푘 by5 푞,푝, hence 푟 푝 푝 푓푘 (훼푘) ∈ 퐴푘. Since we will use this information again, we summarise it: 푟 푝 푟 푞 푝 (*) 푓푘 (훼푘) = 푓푘 (훼푘) ∩ 훼푘 whenever ℓ(푞) ≤ 푘 < ℓ(푟). 푝 푝 푞 푟 푟 푞 푟 푝 Moreover if 휂 ∈ 푎푘 then 휂, 훼푘 ∈ 푎푘 by5 푞,푝, 푓푘 (휂) = 푓푘 (훼푘)∩휂 and 푓푘 (훼푘) = 푟 푞 푝 푟 푟 푝 푓푘 (훼푘) ∩ 훼푘 by3 푟,푞, so 푓푘 (휂) = 푓푘 (훼푘) ∩ 휂. 푞 푝 To verify4 푟,푝, ﬁrst note that (푓푘 (훼푘))ℓ(푝)≤푘<ℓ(푞) is ⊆-increasing with 푘 by4 푞,푝, 푟 푞 and also that (푓푘 (훼푘))ℓ(푞)≤푘<ℓ(푟) is ⊆-increasing by4 푟,푞. We now distinguish three cases: ORDINAL DEFINABLE SUBSETS OF SINGULAR CARDINALS 5

푞 푝 푟 푝 I For ℓ(푝) ≤ 푘 < ℓ(푞), by2 푟,푞, we have 푓푘 (훼푘) = 푓푘 (훼푘) so that the sequence 푟 푝 (푓푘 (훼푘))ℓ(푝)≤푘<ℓ(푞) is ⊆-increasing. For ℓ(푞) ≤ 푘 < 푘 < ℓ(푟), we have that 푓 푟 (훼푞 ) ⊆ 푓 푟 (훼푞 ) and also (by I 0 1 푘0 푘0 푘1 푘1 Remark2) that 훼푝 ≤ 훼푝 , so using (*) we get 푘0 푘1 푓 푞 (훼푝 ) = 푓 푟 (훼푞 ) ∩ 훼푝 ⊆ 푓 푟 (훼푞 ) ∩ 훼푝 = 푓 푞 (훼푝 ). 푘0 푘0 푘0 푘0 푘0 푘1 푘1 푘1 푘1 푘1 푞 푝 It follows that (푓푘 (훼푘))ℓ(푞)≤푘<ℓ(푟) is ⊆-increasing. 푟 푞 푞 I For ℓ(푝) < ℓ(푞) < ℓ(푟), by3 푟,푞 we have 푓ℓ(푞)(훼ℓ(푞)) ∈ 퐴ℓ(푞), so by6 푞,푝 푞 푝 푟 푞 푝 we have 푓ℓ(푞)−1(훼ℓ(푞)−1) ⊆ 푓ℓ(푞)(훼ℓ(푞)) ∩ 훼ℓ(푞)−1. By (*) with 푘 = ℓ(푞) we 푟 푝 푟 푞 푝 푞 푝 have 푓ℓ(푞)(훼ℓ(푞)) = 푓ℓ(푞)(훼ℓ(푞)) ∩ 훼ℓ(푞), and we also have 푓ℓ(푞)−1(훼ℓ(푞)−1) = 푟 푝 푝 푝 푓ℓ(푞)−1(훼ℓ(푞)−1) by2 푟,푞. Since 훼ℓ(푞)−1 ≤ 훼ℓ(푞) by Remark2, we see that

푟 푝 푞 푝 푓ℓ(푞)−1(훼ℓ(푞)−1) = 푓ℓ(푞)−1(훼ℓ(푞)−1) ⊆ 푟 푞 푝 푟 푞 푝 푟 푝 ⊆ 푓ℓ(푞)(훼ℓ(푞)) ∩ 훼ℓ(푞)−1 ⊆ 푓ℓ(푞)(훼ℓ(푞)) ∩ 훼ℓ(푞) = 푓ℓ(푞)(훼ℓ(푞)). 푟 푝 We thus conclude that (푓푘 (훼푘))ℓ(푝)≤푘<ℓ(푟) is ⊆-increasing. 푟 Finally, to verify6 푟,푝, suppose that ℓ(푝) < ℓ(푟) ≤ 푘 and let 푥 ∈ 퐴푘. We distinguish two cases: 푞 푞 I ℓ(푝) < ℓ(푞) = ℓ(푟). In this case, 푥 ∩ 훼푘 ∈ 퐴푘 by5 푟,푞. By6 푞,푝 and the ob- 푝 푝 푞 푞 푝 푞 푝 servation that 훼ℓ(푟)−1 ≤ 훼ℓ(푟) ≤ 훼ℓ(푟) ≤ 훼푘, 푥 ∩ 훼ℓ(푟)−1 ⊇ 푓ℓ(푟)−1(훼ℓ(푟)−1). 푞 푝 푟 푝 푟 푝 Finally 푓ℓ(푟)−1(훼ℓ(푟)−1) = 푓ℓ(푟)−1(훼ℓ(푟)−1) by2 푟,푞, and so 푓ℓ(푟)−1(훼ℓ(푟)−1) ⊆ 푝 푥 ∩ 훼ℓ(푟)−1. 푟 푞 푞 I ℓ(푝) ≤ ℓ(푞) < ℓ(푟). In this case, by6 푟,푞, 푓ℓ(푟)−1(훼ℓ(푟)−1) ⊆ 푥 ∩ 훼ℓ(푟)−1. The 푝 푞 푞 푟 푝 ordinals 훼ℓ(푟)−1 and 훼ℓ(푟)−1 lie in 푎ℓ(푟)−1 by5 푞,푝. Now 푓ℓ(푟)−1(훼ℓ(푟)−1) = 푟 푞 푝 푟 푝 푞 푓ℓ(푟)−1(훼푙ℎ(푟)−1) ∩ 훼ℓ(푟)−1 by3 푟,푞, and so 푓ℓ(푟)−1(훼ℓ(푟)−1) ⊆ 푥 ∩ 훼ℓ(푟)−1 ∩ 푝 푝 푞 푝 훼ℓ(푟)−1 = 푥 ∩ 훼푟−1, using the fact that 훼푟−1 ≥ 훼푟−1 by Remark3. This completes the proof. It is easy to see that for each 푚 < 휔 and each 훼 with 휅 ≤ 훼 < 휆, the set 푞 of conditions 푞 with ℓ(푞) > 푚 is dense, as is the set of 푞 with 훼 ∈ dom(푓푚). It follows that we may view Q as adding a matrix of sets (푧푚,훼)푚<휔,휅≤훼<휆 such that

푧푚,훼 ∈ 푃휅푚 훼 for all 푚 and 훼. Lemma 4. The forcing poset Q collapses all cardinals 휇 with 휅 < 휇 < 휆 and preserves all cardinals 휇 with 휇 > 휆.

Proof. It is easy to see that |Q| = 휆, so that Q trivially has 휆+-cc and preserves cardinals greater than 휆. Given 휇 with 휅 < 휇 < 휆, we will do a density argument to show that 휇 is collapsed. Let 푝 ∈ Q arbitrary, and extend 푝 to obtain 푞 so that there is 훼 ≥ 휇 ⋃︀ 푞 ⋃︀ with 훼 ∈ ℓ(푞)≤푘<휔 푎푘. We claim that 푞 휇 ⊆ 푚 푧˙푚,훼; to see this let 훽 < 휇 be 푞 arbitrary, choose 푦 ∈ 퐴 푞 with 훽 ∈ 푦, and extend 푞 to a condition 푟 such that 훼ℓ(푞) 푟 푞 푟 ℓ(푟) = ℓ(푞) + 1 and 푓ℓ(푞)(훼ℓ(푞)) = 푦, so that 훽 ∈ 푓ℓ(푞)(훼) = 푦 ∩ 훼. The argument that the cardinal 휆 and cardinals 휇 with 휇 ≤ 휅 are preserved will require a deeper analysis of the forcing poset Q, which will be given in subsection 3.4. The intuition is that the 휔-sequences (푧푚,훼)푚<휔 for diﬀering values of 훼 are related, so that no unwanted information can be computed by comparing them. 6 J. CUMMINGS, S.-D. FRIEDMAN, M. MAGIDOR, A. RINOT, AND D. SINAPOVA

3.3. Basic properties of Q. Let us consider a secondary ordering on Q. Let 푟 ≤* 푞 (푟 is a direct extension of 푞) if and only if 푟 ≤ 푞 and ℓ(푟) = ℓ(푞).

* Lemma 5. Let ⃗푞 = (푞푖)푖<휇 be a ≤ -decreasing sequence of conditions where 휇 < * 휅ℓ(푞0). Then the sequence ⃗푞 has a ≤ -lower bound.

Proof. We define a lower bound 푟 with ℓ(푟) = ℓ(푞0) as follows: 푟 ⋃︀ 푞푖 I For all 푘 < 휔, let 푓푘 = 푖<휇 푓푘 . ⋃︀ 푞푖 푞푖 I Choose some ordinal 훽 < 휆 such that 푖<휇(푎푘 ∪ dom(푓푘 )) ⊆ 훽 for all 푟 ⋃︀ 푞푖 푟 푘 ≥ ℓ(푟), and define 푎푘 = 푖<휇 푎푘 ∪ {훽} for all such 푘, so that 훼푘 = 훽. 푟 푞푖 푞푖 I For 푘 ≥ ℓ(푟), set 퐴푘 = {푥 ∈ 푃휅푘 훽 : ∀푖 < 휇 푥 ∩ 훼푘 ∈ 퐴푘 }. * It is routine to check that 푟 is a condition with 푟 ≤ 푞푖 for all 푖. Note that Clauses3,4 and6 from the definition of the ordering ≤ are irrelevant here. * * We will also need a stronger version of ≤ . For 푖 < 휔, write 푟 ≤푖 푞 if and only * 푟 푟 푞 푞 if 푟 ≤ 푞 and in addition (푎푘, 퐴푘) = (푎푘, 퐴푘) whenever ℓ(푞) ≤ 푘 < ℓ(푞) + 푖. Note * * * that ≤ =≤0 and ≤푖 is transitive for all 푖. * Lemma 6 (Fusion). Let (푞푖)푖<휔 be a sequence of conditions such that 푞푖+1 ≤푖 푞푖 * for all 푖 < 휔. Then there exists a condition 푞∞ such that 푞∞ ≤푖 푞푖 for all 푖.

Proof. Define 푞∞ with ℓ(푞∞) = ℓ(푞0) as follows: 푞∞ ⋃︀ 푞푖 I 푓푘 = 푖<휔 푓푘 for all 푘 < 휔; 푞∞ 푞∞ 푞푘+1 푞푘+1 I (푎푘 , 퐴푘 ) = (푎푘 , 퐴푘 ) for 푘 ≥ ℓ(푞0). * It is routine to verify that 푞∞ is a condition and 푞∞ ≤푖 푞푖 for all 푖. Remark 4. Of course Lemma5 already implies that the decreasing sequence from * * Lemma6 has a ≤ -lower bound, the point here is that 푞∞ ≤푖 푞푖 rather than just * 푞∞ ≤ 푞푖. Definition 1. For conditions 푟 ≤ 푞, we let stem(푟, 푞) denote the finite sequence 푟 푞 (푓푖 (훼푖 ))ℓ(푞)≤푖<ℓ(푟). ∏︀ 푞 Note that by3 푟,푞 and4 푟,푞, stem(푟, 푞) ∈ ℓ(푞)≤푖<ℓ(푟) 퐴푖 , and is ⊆-increasing. ∏︀ 푞 Definition 2. Let 푞 be a condition, and let 푠 ∈ ℓ(푞)≤푖<푙 퐴푖 be a ⊆-increasing sequence for some 푙 ∈ (ℓ(푞), 휔). We define 푞 + 푠 as the 휔-sequence (푟푘)푘<휔 such that: 푞 ∙ For 푘 < ℓ(푞), 푟푘 = 푓푘 . 푞 푞 ∙ For ℓ(푞) ≤ 푘 < 푙, 푟푘 is the function with domain dom(푓푘 ) ∪ 푎푘 such that 푞 푞 푞 푟푘(휂) = 푓푘 (휂) for 휂 ∈ dom(푓푘 ) and 푟푘(휂) = 푠푘 ∩ 휂 for 휂 ∈ 푎푘. 푞 푞 푞 ∙ For 푘 ≥ 푙, 푟푘 = (푓푘 , 푎푘, 퐵푘) where 퐵푘 = {푥 ∈ 퐴푘 : 푠푙−1 ⊆ 푥}. By convention we also define 푞 + ⟨⟩ = 푞. We shall soon establish that 푞 + 푠 is a condition, but let us first point out that 푞 + 푠 may be obtained by adding in one entry of 푠 at a time.

Lemma 7. Let 푞 and 푠 = (푠푖)ℓ(푞)≤푖<푙 be as in Definition2. Let 푛 = 푙 − ℓ(푞). Define a decreasing sequence of conditions (푟푖)푖≤푛 as follows: 푟0 = 푞, and then 푟푖+1 = 푟푖 + ⟨푠푖⟩ for 푖 < 푛. Then 푟푛 = 푞 + 푠.

Proof. The proof is straightforward by induction on the length of 푠. ORDINAL DEFINABLE SUBSETS OF SINGULAR CARDINALS 7

Lemma 8. Let 푞 and 푠 be as in Definition2. Then 푞 + 푠 is a condition extending 푞. Moreover 푟 ≤* 푞 + stem(푟, 푞) for all 푟 ≤ 푞.

푞+푠 푞 Proof. By definition ℓ(푞 + 푠) = ℓ(푞) + ℓ(푠), so1 푞+푠,푞 holds. Clearly 푓푘 ⊇ 푓푘 for all 푘, so2 푞+푠,푞 holds. 푞+푠 푞 푞+푠 푞 푞 By construction 푓푘 (훼푘) = 푠푘 for ℓ(푞) ≤ 푘 < ℓ(푞) + ℓ(푠), so 푓 (훼푘) ∈ 퐴푘 by 푞+푠 푞+푠 푞 our assumptions on 푠. Also by definition 푓푘 (휂) = 푠푘 ∩ 휂 = 푓푘 (훼푘) ∩ 휂 for all 푞+푠 푞 such 푘, so that3 푞+푠,푞 holds. The sequence (푓푘 (훼푘))ℓ(푞)≤푘<ℓ(푞+푠) is equal to 푠, which is ⊆-increasing by our assumptions on 푠, so4 푞+푠,푞 holds. 푞+푠 푞 푞+푠 푞 We have 푎푘 = 푎푘 and 퐴푘 ⊆ 퐴푘 for all 푘 ≥ ℓ(푞 + 푠), so5 푞+푠,푞 holds. Finally 푞+푠 푞 we have by definition that 퐴푘 = {푥 ∈ 퐴푘 : 푠ℓ(푞+푠)−1 ⊆ 푥} for 푘 ≥ ℓ(푞 + 푠), and 푞+푠 푞 푓ℓ(푞+푠)−1(훼ℓ(푞+푠)−1) = 푠ℓ(푞+푠)−1, so6 푞+푠,푞 holds. Now suppose that 푟 ≤ 푞 and 푠 = stem(푟, 푞). If ℓ(푟) = ℓ(푞) then 푠 = ⟨⟩ and 푟 ≤* 푞 = 푞+푠, so we may as well assume that ℓ(푟) > ℓ(푞). We have ℓ(푟) = ℓ(푞+푠), so 푟 푞+푠 1푟,푞+푠 holds. By the definition of 푞+푠 and3 푟,푞, 푓푘 ⊇ 푓푘 whenever ℓ(푞) ≤ 푘 < ℓ(푟). 푞+푠 푞 푟 푞 For 푘 ≥ ℓ(푟), 푓푘 = 푓푘 and 푓푘 ⊇ 푓푘 by2 푟,푞, so2 푟,푞+푠 holds. Since ℓ(푟) = ℓ(푞 + 푠),3 푟,푞+푠,4 푟,푞+푠 and6 푟,푞+푠 hold vacuously. For 푘 ≤ ℓ(푞 + 푠) 푞+푠 푞 푞 푟 we have 푎푘 = 푎푘 by definition, and also 푎푘 ⊆ 푎푘 by5 푟,푞. Finally for all 푘 ≥ ℓ(푟) = 푟 푞+푠 푞 푞 ℓ(푞 + 푠) and all 푥 ∈ 퐴푘, 푥 ∩ 훼푘 = 푥 ∩ 훼푘 ∈ 퐴푘, and additionally 푠ℓ(푞+푠)−1 ⊆ 푞 푞 푞+푠 푥 ∩ 훼ℓ(푞+푠)−1; it follows from the definition that 푥 ∩ 훼ℓ(푞+푠) ∈ 퐴ℓ(푞+푠).

* Lemma 9. Let 푞 ≤푛 푝 and let 푟 ≤ 푞 with ℓ(푟) = ℓ(푞) + 푛. Then stem(푟, 푞) = stem(푟, 푝), and 푞 + stem(푟, 푞) ≤* 푝 + stem(푟, 푝).

푟 푞 * Proof. Let 푠 = stem(푟, 푞), that is 푠푘 = 푓푘 (훼푘) for ℓ(푞) ≤ 푘 < ℓ(푟). Since 푞 ≤푛 푝 푞 푝 푟 푝 we have that 훼푘 = 훼푘 for such 푘, and so 푠푘 = 푓푘 (훼푘) and hence stem(푟, 푝) = 푠 = stem(푟, 푞). If 푛 = 0 then 푞 = 푞 + 푠 ≤* 푝 = 푝 + 푠, so we may assume that 푛 > 0. Now ℓ(푞 + 푠) = ℓ(푟) = ℓ(푝 + 푠), so Clauses1,3,4 and6 are easily seen to hold. 푞+푠 푞 푝 푝+푠 For 푘 < ℓ(푝) or 푘 ≥ ℓ(푟), we have 푓푘 = 푓푘 ≤ 푓푘 = 푓푘 by the deﬁnitions and 푞+푠 푝+푠 푞 푝 2푞,푝. For ℓ(푝) ≤ 푘 < ℓ(푟), we have that 푓푘 ≤ 푓푘 because 푓푘 ≤ 푓푘 by2 푞,푝, and 푞+푠 푝+푠 푞 푝 푓푘 (휂) = 푠푘 ∩ 휂 = 푓푘 (휂) for every 휂 ∈ 푎푘 = 푎푘. 푝+푠 푝 푞 푞+푠 푞 푝 푝 For 푘 ≥ ℓ(푟) we have 푎푘 = 푎푘 ⊆ 푎푘 = 푎푘 and 푥 ∈ 퐴푘 =⇒ 푥 ∩ 훼푘 ∈ 퐴푘 푞+푠 푞 푝 푝 by5 푞,푝. If 푥 ∈ 퐴푘 then 푥 ∈ 퐴푘 and 푥 ⊇ 푠ℓ(푟)−1, so 푥 ∩ 훼푘 ∈ 퐴푘 and also 푝 푝 푝 푥 ∩ 훼푘 ⊇ 푠ℓ(푟)−1 using the fact that 훼푘 ≥ 훼ℓ(푟)−1. 3.4. Main technical lemma. If휏 ˙ is a name for an object in the ground model and 푟 is a condition, then we say that 푟 decides 휏˙ if and only if there is 푥 ∈ 푉 such that 푟 휏˙ =푥 ˇ. We write 푟 ‖ 휏˙ for “푟 decides휏 ˙”. We now state and prove the key technical lemma about Q. Lemma 10. Let 푝 ∈ Q and let 휏˙ be a name for an element of 푉 . Then there is a direct extension 푞 ≤* 푝 such that for every 푟 ≤ 푞 with 푟 ‖ 휏˙, we have 푞 +stem(푟, 푞) ‖ 휏˙.

Proof. We will build an 휔-sequence of conditions (푝푛)푛<휔 such that:

∙ 푝0 = 푝. * ∙ 푝푛+1 ≤푛 푝푛 for all 푛. ∙ For every 푟 ≤ 푝푛+1 with ℓ(푟) = ℓ(푝)+푛, if 푟 ‖ 휏˙, then 푝푛+1 +stem(푟, 푝푛+1) ‖ 휏˙. 8 J. CUMMINGS, S.-D. FRIEDMAN, M. MAGIDOR, A. RINOT, AND D. SINAPOVA

Before giving the construction we verify that building (푝푛)푛<휔 is suﬃcient. Ap- * pealing to Lemma6, we ﬁnd 푝∞ such that 푝∞ ≤푛 푝푛 for all 푛. Let 푟 ≤ 푝∞ * * with 푟 ‖ 휏˙ and ℓ(푟) = ℓ(푝) + 푛. Then 푝∞ ≤푛 푝푛+1, and so 푝∞ + stem(푟, 푝∞) ≤ 푝푛+1 +stem(푟, 푝푛+1) by Lemma9. By construction 푝푛+1 +stem(푟, 푝푛+1) ‖ 휏˙, and so 푝∞ + stem(푟, 푝∞) ‖ 휏˙. It follows that 푝∞ will serve as a witness to the conclusion.

To begin the construction we set 푝0 = 푝, and then ask whether there is a direct extension of 푝 which decides휏 ˙, and set 푝1 equal to such an extension if one exists, * * otherwise 푝1 = 푝. If there is 푟 ≤ 푝1 with 푟 ‖ 휏˙ then 푟 ≤ 푝, and 푝1 = 푝1 + stem(푟, 푝1) ‖ 휏˙. 푛 Suppose now that we have constructed 푝푛 for some 푛 > 0. Let (푠푗 )푗<휇 be an enumeration of all sequences 푠 such that ℓ(푠) = 푛 and 푝푛 + 푠 is well defined, that ∏︀ 푝푛 is to say all ⊆-increasing sequences in ℓ(푝)≤푘<ℓ(푝)+푛 퐴푘 . Since 휆 is inaccessible, 휇 < 휆. 푗 푗 푗 푗 Note to the reader. The objects 푓푘 , 푎푘, 훼푘, 퐴푘 built during the construction also depend on 푛, but we have suppressed the index 푛 to lighten the notation. We will construct by recursion for each 푗 < 휇: 푗 ∙ Functions 푓푘 for 푘 < 휔. 푗 푗 ∙ Sets 푎푘 and 퐴푘 for ℓ(푝) + 푛 ≤ 푘 < 휔. We will maintain the following hypotheses: 푝푛 0 ∙ 푎ℓ(푝)+푛−1 ⊆ 푎ℓ(푝)+푛. ′ ′ ′ 푗 푗 푗 푗 ∙ For all 푘 and all 푗 < 푗 , 푓푘 ⊇ 푓푘 and 푎푘 ⊆ 푎푘 . 푗 푗 ∙ For all 푘 and 푗, 푎푘 ⊆ [휅, 휆) with |푎푘| < 휆. 푗 ∙ For all 푗, the sequence (푎푘)ℓ(푝)+푛≤푘 is ⊆-increasing with 푘. 푗 푗 푗 ∙ 푎 has a maximum element 훼 and 퐴 ∈ 푈 푗 . 푘 푘 푘 푘,훼푘 푗 푝푛 ∙ For all 푘 with ℓ(푝) ≤ 푘 < ℓ(푝) + 푛 and all 푗, dom(푓푘 ) is disjoint from 푎푘 . 푗 푗 ∙ For all 푘 with ℓ(푝) + 푛 ≤ 푘 and all 푗, dom(푓푘 ) is disjoint from 푎푘. 0 푝푛 0 0 푝푛 푝푛 To begin the construction we set 푓푘 = 푓푘 for all 푘, and (푎푘, 퐴푘) = (푎푘 , 퐴푘 ) for 푘 ≥ ℓ(푝) + 푛. 푗 푗 푗 Suppose that we have constructed 푓푘 , 푎푘 and 퐴푘. We define various auxiliary conditions in Q: 푗 푗 푗 푗 푝푛 푝푛 푗 ∙ 푝 is the condition such that 푝푘 = 푓푘 for 푘 < ℓ(푝), 푝푘 = (푎푘 , 퐴푘 , 푓푘 ) for 푗 푗 푗 푗 ℓ(푝) ≤ 푘 < ℓ(푝) + 푛, and 푝푘 = (푎푘, 푃휅푘 훼푘, 푓푘 ) for ℓ(푝) + 푛 ≤ 푘. 푗 푗 푛 ∙ 푞 = 푝 + 푠푗 . * 푗 ∙ If there is 푟 ≤ 푞푗 such that 푟 ‖ 휏˙, then 푟 is some such condition 푟, otherwise 푟푗 = 푞푗. We now define: 푗+1 푟푗 I 푓푘 = 푓푘 for 푘 < ℓ(푝) or 푘 ≥ ℓ(푝) + 푛. 푗 푗 푗+1 푟 푟 푝푛 I 푓푘 = 푓푘 dom(푓푘 ) ∖ 푎푘 for ℓ(푝) ≤ 푘 < ℓ(푝) + 푛. 푗+1 푗+1 푟푗 푟푗 I (푎푘 , 퐴푘 ) = (푎푘 , 퐴푘 ) for 푘 ≥ ℓ(푝) + 푛. For 푗 limit we proceed roughly as in the proof of Lemma5, with the important 푗 caveat that we do not aim to make the sets 퐴푘 decrease with 푗: 푗 ⋃︀ 푖 I 푓푘 = 푖<푗 푓푘 for all 푘. I For 푘 ≥ ℓ(푝) + 푛: ORDINAL DEFINABLE SUBSETS OF SINGULAR CARDINALS 9

푗 ∙ We choose 푎푘 as in the final stage of the proof of Lemma5, that is to 푗 푗 ⋃︀ 푖 푗 say we choose some large enough 훽 and set 푎푘 = 푖<푗 푎푘 ∪ {훽 } for 푗 푗 all 푘, so that 훼푘 = 훽 . 푗 푗 ∙ We set 퐴푘 = 푃휅푘 훽 . 푗 푗 푗 After defining 푓푘 , 푎푘 and 퐴푘 for 푗 < 휇, we are ready to define 푝푛+1. 푝푛+1 ⋃︀ 푖 I 푓푘 = 푖<휇 푓푘 for all 푘. 푝푛+1 푝푛+1 푝푛 푝푛 I (푎푘 , 퐴푘 ) = (푎푘 , 퐴푘 ) for ℓ(푝) ≤ 푘 < ℓ(푝) + 푛. I For 푘 ≥ ℓ(푝) + 푛: ∙ As at the limit stages below 휇, we choose some large enough 훽* and 푝푛+1 ⋃︀ 푖 * 푝푛+1 * set 푎푘 = 푖<휇 푎푘 ∪ {훽 }, so that 훼푘 = 훽 for all 푘. 푝푛+1 * ∙ We define 퐴푘 by diagonal intersection, as the set of those 푥 ∈ 푃휅푘 훽 푝푛 푝푛 푛 푖+1 such that 푥 ∩ 훼푘 ∈ 퐴푘 and, for all 푖 < 휇 if 푠푖 b 푥 then 푥 ∩ 훼푘 ∈ 푖+1 퐴푘 . * 푖+1 푖+1 푖+1 * 푖+1 Note that {푥 ∈ 푃휅푘 훽 : 푥 ∩ 훼 ∈ 퐴 } ∈ 푈푘,훽 , because 퐴 ∈ 푈 and 푘 푘 푘 푘,훼푘 푖+1 푝푛+1 this is the projection of 푈푘,훽* under the map 푥 ↦→ 푥 ∩ 훼푘 . So 퐴푘 ∈ 푈푘,훽* by Lemma1. * It is routine to verify that 푝푛+1 is a condition and that 푝푛+1 ≤푛 푝푛. To finish the proof we must verify that for every 푟 ≤ 푝푛+1 such that ℓ(푟) = ℓ(푝) + 푛 and 푟 ‖ 휏˙, 푝푛+1 + stem(푟, 푝푛+1) ‖ 휏˙. Let 푠 = stem(푟, 푝푛+1), and pick some 푗 < 휇 such 푛 that 푠 = 푠푗 . * 푗 푝푗 푗 We claim that 푝푛+1 ≤푛 푝 . This is a routine verification: note that 퐴푘 = 푃휅푘 훼푘 푗 for ℓ(푝) + 푛 ≤ 푘, so that there is no problem with Clause5 for 푝푛+1 ≤ 푝 . * 푗 By Lemma8 we have 푟 ≤ 푝푛+1+푠, and by Lemma9 we have that stem( 푟, 푝 ) = 푠 * 푗 푗 * 푗 푗 * 푗 and 푝푛+1 + 푠 ≤ 푝 + 푠 = 푞 . So 푟 ≤ 푞 and 푟 ‖ 휏˙, hence we chose 푟 ≤ 푞 such 푗 푗 푗+1 푗+1 푗+1 that 푟 ‖ 휏˙ and used 푟 in the definitions of 푓푘 , 푎푘 , and 퐴푘 . 푗 We claim that 푝푛+1 + 푠 ≤ 푟 . The verification is fairly straightforward, the main point is to check the second part of Clause5. So let 푘 ≥ ℓ(푝) + 푛, and recall that 푝푛+1+푠 푝푛+1 by definition 퐴푘 is the set of 푥 ∈ 퐴푘 such that 푠 ⊆ 푥. By the construction 푝푛+1 푝푛+1 of 퐴푘 as a diagonal intersection, for every 푥 ∈ 퐴푘 such that 푠 b 푥 we have 푗+1 푗+1 푗+1 푟푗 that 푥 ∩ 훼푘 ∈ 퐴푘 . Since 퐴푘 = 퐴푘 , this is exactly what is needed for Clause 5.

3.5. Prikry lemma.

Lemma 11 (Prikry lemma). For every condition 푝 and every sentence 휑 in the forcing language there is 푞 ≤* 푝 such that 푞 ‖ 휑.

Proof. Let휏 ˙ name an ordinal which is 0 if 휑 is true and 1 if 휑 is false. Appealing * to Lemma 10 we find 푞0 ≤ 푝 which is such that for all 푟 ≤ 푞0, if 푟 decides 휑 then 푞0 + stem(푟, 푞) decides 휑. Let 푆 be the set of sequences 푠 such that 푞0 + 푠 is well-defined, and for each 푠 ∈ 푆 define 퐹 (푠) as follows: ⎧ 0, if 푞 + 푠 휑 ⎨⎪ 0 퐹 (푠) = 1, if 푞0 + 푠 ¬휑 ⎩⎪2, otherwise. 10 J. CUMMINGS, S.-D. FRIEDMAN, M. MAGIDOR, A. RINOT, AND D. SINAPOVA

푞0 For each 푡 ∈ 푆 we may find a measure one set 퐵푡 ⊆ 퐴 and 퐺(푡) such that ℓ(푞0)+ℓ(푡) 퐹 (푡a⟨푥⟩) = 퐺(푡) for all 푥 ∈ 퐵푡. Now for each 푘 ≥ ℓ(푞0), let 퐶푘 be the diagonal intersection of 퐵푡 for all sequences 푞0 푡 of length 푘 − ℓ(푞0), that is the set of 푥 ∈ 퐴푘 such that 푥 ∈ 퐵푡 for all 푡 b 푥. Let 푞0 푞1 be obtained from 푞0 by replacing 퐴푘 by 퐶푘 for all 푘 ≥ ℓ(푞0), and note that 푞1 푞0 훼푘 = 훼푘 for all 푘 ≥ ℓ(푞0) = ℓ(푞1). Let 푞2 ≤ 푞1 be a condition deciding 휑 with ℓ(푞2) chosen minimal. * We claim that ℓ(푞2) = ℓ(푞1), so that 푞2 ≤ 푝 and we are done. Suppose for a * contradiction that ℓ(푞2) > ℓ(푞1), and let 푠 = stem(푞2, 푞1). By construction 푞1 +푠 ≤ 푞0 + 푠 and 푞0 + 푠 decides 휑, so 푞1 + 푠 decides 휑. We will assume that it forces 휑; the argument for the case when 푞1 + 푠 forces ¬휑 is identical. Let 푥 = 푠(ℓ(푠) − 1) and let 푡 = 푠 ℓ(푠) − 1, so that 푠 = 푡a⟨푥⟩. By construction 푥 ∈ 퐵푡, and so by definition 퐺(푡) = 퐹 (푠) = 1. By Lemma8 it is easy to see that every extension of 푞1 + 푡 is compatible with 푞1 + 푡a⟨푦⟩ for some 푦 ∈ 퐵푡, and since 퐺(푡) = 1 we have that 푞1 + 푡a⟨푦⟩ 휑 for all 푦 ∈ 퐵푡, so that 푞1 + 푡 forces 휑. This contradicts the minimal choice of ℓ(푞2). Corollary 1. The forcing poset Q adds no bounded subsets of 휅. Proof. Let 푝 푥˙ ⊆ 휇 with 휇 < 휅, and find 푞 ≤ 푝 such that ℓ(푞) = 푘 for some 푘 large * enough that 휅푘 > 휇. Appealing to Lemmas 11 and5 we may build a ≤ -decreasing sequence (푞푖)푖<휇 such that 푞0 = 푞 and 푞푖+1 decides whether 푖 ∈ 푥˙ for all 푖. By Lemma5 again there is a condition 푟 such that 푟 ≤ 푞푖 for all 푖, and 푟 푥˙ =푦 ˇ for some 푦 ∈ 푉 .

3.6. Projected forcing. Let 푘 < 휔 and let ⃗훼 = (훼푖)푘≤푖<휔 be a ≤-increasing sequence from [휅, 휆). We deﬁne a forcing poset Q⃗훼. Conditions in Q⃗훼 are sequences

(푞푖)푘≤푖<휔 such that some 푙 ≥ 푘: 푞푖 ∈ 푃휅푖 훼푖 for 푖 < 푙 and 푞푖 ∈ 푈푖,훼푖 for 푖 ≥ 푙, and (푞푖)푘≤푖<푙 is ⊆-increasing. We say that 푙 is the length of 푞 and write 푙 = ℓ(푞). If 푟, 푞 ∈ Q⃗훼 then 푟 ≤ 푞 if and only if: ∙ ℓ(푟) ≥ ℓ(푞). ∙ 푟푖 = 푞푖 for 푘 ≤ 푖 < ℓ(푞). 푞 ∙ 푟푖 ∈ 퐴푖 for ℓ(푞) ≤ 푖 < ℓ(푟). 푟 푞 ∙ 퐴푖 ⊆ 퐴푖 for ℓ(푟) ≤ 푖 < 휔. We note that Q⃗훼 has the 휆-cc, because there are fewer than 휆 possible “stems” (푞푖)푘≤푖<ℓ(푞) and any two conditions with the same stem are compatible. It is also not hard to prove that Q⃗훼 obeys a version of the Prikry Lemma, but we will not need this (it actually follows from the next lemma). It is useful to note that the generic object added by Q⃗훼 is a ⊆-increasing sequence (푧푖)푘≤푖<휔 such that 푧푖 ∈ 푃휅푖 훼푖.

Lemma 12. Let 푞 ∈ Q, and let ⃗훼 = (훼푖)ℓ(푞)≤푖<휔 be a ≤-increasing sequence such 푞 that 훼푖 ∈ 푎푖 for all 푖. Let 휋 : Q ↓ 푞 → Q⃗훼 be the map defined by: 푟 ∙ 휋(푟)푖 = 푓푖 (훼푖) for ℓ(푞) ≤ 푖 < ℓ(푟). 푟 ∙ 휋(푟)푖 = {푥 ∩ 훼푖 : 푥 ∈ 퐴푖 } for 푖 ≥ ℓ(푟). Then 휋 is order-preserving and has the following property: for every 푟 ≤ 푞 there is * * ′ ′ 푟 ≤ 푟 such that for all 푞0 ≤ 휋(푟 ) there is 푟 ≤ 푟 with 휋(푟 ) ≤ 푞0. Proof. It is routine to check that 휋 is order-preserving. To verify the rest of the conclusion let 푟 ≤ 푞. We obtain a condition 푟* ≤* 푟 by shrinking measure one sets in 푟 as follows: ORDINAL DEFINABLE SUBSETS OF SINGULAR CARDINALS 11

∙ ℓ(푟*) = ℓ(푟). 푟* 푟 ∙ 푓푖 = 푓푖 for all 푖. 푟* 푟 ∙ 푎푖 = 푎푖 for all 푖 ≥ ℓ(푟). 푟* 푟* 푟 ∙ We deﬁne 퐴푘 for 푘 ≥ ℓ(푟) by recursion on 푘, arranging that 퐴푘 ⊆ 퐴푘 and 푟* 퐴 ∈ 푈 푟 . 푘 푘,훼푘 푟* 푟 푟* We start by setting 퐴ℓ(푟) = 퐴ℓ(푟). For 푖 > 0 we set 퐴ℓ(푟)+푖 to be the set of 푦 ∈ 푟 푟* ′ 푟* 퐴ℓ(푟)+푖 such that for all 푥 ∈ 퐴ℓ(푟)+푖−1 with 푥∩훼푖 ⊆ 푦 ∩훼푖+1 there is 푥 ∈ 퐴ℓ(푟)+푖−1 ′ ′ 푟* with 푥 ∩ 훼푖 = 푥 ∩ 훼푖 and 푥 ⊆ 푦. By Lemma2 we have 퐴ℓ(푟)+푖 ∈ 푈ℓ(푟)+푖,훽푖+1 . * 푟* Let 푞0 ≤ 휋(푟 ), and choose 푦푘 ∈ 퐵 푘 such that (푞0)푘 = 푦푘 ∩ 훼푘 for ℓ(푟) ≤ * 푘 < ℓ(푞 ). Using the deﬁnition of 퐵푟 we will choose 푦′ for 0 < 푖 ≤ ℓ(푟) by 0 푘 ℓ(푞0)−푖 * induction on 푖 as follows: 푦′ = 푦 , and 푦′ ∈ 퐵푟 with ℓ(푞0)−1 ℓ(푞0)−1 ℓ(푞0)−(푖+1) ℓ(푞0)−(푖+1) 푦′ ⊆∈ 푦 and 푦′ ∩ 훼 = 푦 ∩ 훼 = ℓ(푞0)−(푖+1) ℓ(푞0)−푖 ℓ(푞0)−(푖+1) ℓ(푞0)−(푖+1) ℓ(푞0)−(푖+1) ℓ(푞0)−(푖+1)

(푞0)ℓ(푞0)−(푖+1). ′ Let 푡 = (푦푘)ℓ(푟)≤푘<ℓ(푞0), and form 푟 + 푡. By shrinking measure one sets appro- ′ ′ * priately we obtain a condition 푟 ≤ 푟 + 푡 ≤ 푟 such that 휋(푟 ) ≤ 푞0.

The conclusion of Lemma 12 is a weakening of the standard property of being a projection between forcing posets, and was ﬁrst isolated by Foreman and Woodin [1]. The following corollary of Lemma 12 is routine:

Corollary 2. With the same hypotheses as in Lemma 12, if 퐺 is Q-generic with 푞 ∈ 퐺 then 휋[퐺] generates a Q⃗훼-generic filter. Remark 5. It is possible to give an alternative argument for Corollary2 by ﬁrst proving a characterisation of Q⃗훼-generic sequences in the style of Mathias’ theorem on Prikry forcing [8]. This characterisation, which we will not prove, states that an increasing sequence (푧푖)푘≤푖<휔 is generic if and only if for every sequence (퐴푖)푘≤푖<휔 in 푉 with 퐴푖 ∈ 푈푖,훼푖 for all 푖, we have 푧푖 ∈ 퐴푖 for all large 푖. It is straightforward to verify that the induced ﬁlter 퐺⃗훼 corresponds to a sequence (푧푖)푘≤푖<휔 with this property.

Lemma 13. Let 퐺 be Q-generic and let 푥 ⊆ 휅 with 푥 ∈ 푉 [퐺]. Then there exists ⃗훼 such that 푥 ∈ 푉 [퐺⃗훼].

Proof. Note that by Corollary1 we have 푥 ∩ 휅푛 ∈ 푉 for all 푛 < 휔. Let 푝 ∈ Q and * let휏 ˙푛 be a name for 푥 ∩ 휅푛. Appealing to Lemmas 10 and5 we may ﬁnd 푞 ≤ 푝 such that for all 푟 ≤ 푞, if 푟 decides휏 ˙푛 then then 푞 + stem(푟, 푞) decides 휏푛. Now let 푞 ⃗훼 = (훼푘)ℓ(푞)≤푘<휔. It is routine to verify that 푞 forces that 푥 can be computed from 퐺⃗훼. Explicitly, if 푞 ∈ 퐺 and (푧푖)푘≤푖<휔 is the generic sequence added by 퐺⃗훼 then ¯ 훽 ∈ 푥 ⇐⇒ ∃푘 ≥ 푘 푞 + (푧푖)푘≤푖<푘¯ 훽 ∈ 푥˙.

Lemma 14. Let 퐺 be Q-generic, then 휆 = (휅+)푉 [퐺]. Proof. Suppose for a contradiction that 휆 is not a cardinal in 푉 [퐺]. By Lemma4, cardinals 휇 with 휅 < 휇 < 휆 are all collapsed in 푉 [퐺], so that if 휆 is also collapsed then there is a set 푥 ⊆ 휅 in 푉 [퐺] coding a well-ordering of 휅 with order type 휆. By Lemma 13 푥 ∈ 푉 [퐺⃗훼] for some ⃗훼, which gives an immediate contradiction since Q⃗훼 is 휆-cc. 12 J. CUMMINGS, S.-D. FRIEDMAN, M. MAGIDOR, A. RINOT, AND D. SINAPOVA

3.7. Definability. Let 퐺 be Q-generic. The last step in the proof of our main result is to analyse the class of sets HOD푥 where 푥 ⊆ 휅 with 푥 ∈ 푉 [퐺]. This will require a rather technical result which appears below as Lemma 16. In order to motivate the statement of Lemma 16, we will state a lemma which contains the necessary information about HOD푥 and prove it modulo one missing fact, which will then be provided by Lemma 16.

푉 [퐺] Lemma 15. Let 푥 ⊆ 휅 with 푥 ∈ 푉 [퐺], then there exists ⃗훼 such that (HOD푥) ⊆ 푉 [퐺] + (HOD푥) + 푉 [퐺⃗훼]. In particular (휅 ) < 휅 . 푞 Proof of Lemma 15, Part One. By Lemma 13 find 푞 ∈ 퐺, ⃗훼 with 훼푘 ∈ 푎푘 for all 푉 [퐺] 푘 and푥 ˙ a Q⃗훼-name such that 푖퐺⃗훼 (푥 ˙) = 푥. Since HOD푥 is a model of ZFC, to 푉 [퐺] show that HOD푥 ⊆ 푉 [퐺⃗훼] it will be sufficient to show that every set of ordinals 푉 [퐺] 푦 ∈ HOD푥 lies in 푉 [퐺⃗훼]. Let 푦 be such a set and fix a definition of 푦 from 푥 and some ordinal parameter 훿, say 푌 = {훾 : 푉 [퐺] |= 휓(훾, 훿, 푥)}.

In order to define 푦 in 푉 [퐺⃗훼], we define an auxiliary set 퐻 ⊆ Q. Let 푘 = ℓ(푞) and let (푧푖)푘≤푖<휔 be the generic sequence corresponding to the filter 퐺⃗훼. We define 퐻 to be the set of conditions 푟 ∈ Q such that 푟 ≤ 푞, and for every 푘¯ ≥ 푘 there exists 푟1 ¯ 푟1 ≤ 푟 such that 푓푖 (훼푖) = 푧푖 for 푘 ≤ 푖 < 푘. Clearly 퐺 ⊆ 퐻 and 퐻 ∈ 푉 [퐺⃗훼]. ¯ We claim that 푌 = {훾 : ∃푟 ∈ 퐻 푟 휓(훾, 훿, 푥˙)}. If 푌 is the set defined on the right hand side of this equation then clearly 푌 ⊆ 푌¯ and 푌¯ ∈ 푉 [퐺훼], so we need only show that 푌¯ ⊆ 푌 . Suppose for a contradiction that this is not the case, fix ¯ 훾 ∈ 푌 ∖ 푌 , and then choose conditions 푟 ∈ 퐻 and 푠 ∈ 퐺 such that 푟 휓(훾, 훿, 푥˙) and 푠 ¬휓(훾, 훿, 푥˙). We may extend 푠 to find 푠* ≤ 푠, 푞 such that 푠* ∈ 퐺 and ℓ(푠*) ≥ ℓ(푟), and then use the fact that 푟 ∈ 퐻 to find 푟* ≤ 푟 (not necessarily lying in 퐻) such that ℓ(푟*) = ℓ(푠*) 푟* 푠* * * and 푓푖 (훼푖) = 푧푖 = 푓푖 (훼푖) for all 푖 with ℓ(푞) = 푘 ≤ 푖 < ℓ(푟 ) = ℓ(푠 ). At this point we will be done if we can show that 푟* and 푠* can not force contradictory information about 휓(훾, 훿, 푥˙), and this is exactly the point of Lemma 16. Accordingly we interrupt the proof of Lemma 15 to state and prove Lemma 16.

Lemma 16. Let 푞 ∈ Q, and let ⃗훼 = (훼푖)ℓ(푞)≤푖<휔 be a ≤-increasing sequence such 푞 * * * * * * that 훼푖 ∈ 푎푖 for all 푖. Let 푟 and 푠 be conditions such that 푟 , 푠 ≤ 푞, ℓ(푟 ) = ℓ(푠 ), 푟* 푞 푠* 푞 * ′ * * and 푓푘 (훼푘) = 푓푘 (훼푘) for ℓ(푞) ≤ 푘 < ℓ(푟 ). Then there exist conditions 푟 ≤ 푟 and 푠′ ≤* 푠* and 휌 : Q ↓ 푟′ → Q ↓ 푠′ such that: ∙ 휌 is an isomorphism. ′ ∙ If 퐺0 is Q-generic with 푟 ∈ 퐺0, and 퐺1 is the Q-generic filter generated ′ by 휌[퐺0 ∩ Q ↓ 푟 ], then 퐺0,⃗훼 = 퐺1,⃗훼. Proof. We choose direct extensions 푟′ ≤* 푟 and 푠′ ≤* 푠 such that: 푟′ 푠′ ∘ dom(푓푘 ) = dom(푓푘 ) for 푘 < ℓ(푟). 푟′ 푟′ 푠′ 푠′ 푟′ 푠′ 푟′ 푠′ ∘ dom(푓푘 ) ∪ 푎푘 = dom(푓푘 ) ∪ 푎푘 , 훼푘 = 훼푘 , 퐴푘 = 퐴푘 for 푘 ≥ ℓ(푟). 푟′ 푠′ To help readability let 훽푘 be the common value of 훼푘 and 훼푘 , let 퐵푘 be the 푟′ 푠′ 푟′ common value of 퐴푘 and 퐴푘 , and let 푑푘 be the common value of dom(푓푘 ) and 푠′ 푟′ 푟′ 푠′ 푠′ dom(푓푘 ) for 푘 < ℓ(푟) and the common value of dom(푓푘 ) ∪ 푎푘 and dom(푓푘 ) ∪ 푎푘 for ℓ(푟) ≤ 푘. ORDINAL DEFINABLE SUBSETS OF SINGULAR CARDINALS 13

We will now define maps 휌 : Q ↓ 푟′ → Q ↓ 푠′ and 휎 : Q ↓ 푠′ → Q ↓ 푟′ with 휌(푟′) = 푠′, and argue that they are order-preserving and mutually inverse. The ′′ reader may find the following slogan useful: “휌 alters 푟푘 on 푑푘 to make it look like the 푘-level of an extension 푠′′ of 푠′ with stem(푠′′, 푠′) = stem(푟′′, 푟′)”. Let 푟′′ ≤ 푟′. We define 휌(푟′′) to be the condition 푠′′ such that:

′ 푠′′ 푟′′ 푠′′ 푠′ 푠′ 푠′′ I For 푘 < ℓ(푟 ), dom(푓푘 ) = dom(푓푘 ), 푓푘 dom(푓푘 ) = 푓푘 , 푓푘 푟′′ 푠′ 푟′′ 푟′′ 푠′ dom(푓푘 ) ∖ dom(푓푘 ) = 푓푘 dom(푓푘 ) ∖ dom(푓푘 ). ′ ′′ 푠′′ 푟′′ 푠′′ 푠′ 푠′ I For ℓ(푟 ) ≤ 푘 < ℓ(푟 ), dom(푓푘 ) = dom(푓푘 ), 푓푘 dom(푓푘 ) = 푓푘 , 푠′′ 푟′′ 푠′ 푠′′ 푟′′ 푟′′ 푟′′ 푓푘 (휂) = 푓푘 (훽푘)∩휂 for 휂 ∈ 푎푘 , 푓푘 dom(푓푘 )∖푑푘 = 푓푘 dom(푓푘 )∖푑푘. ′′ I For ℓ(푟 ) ≤ 푘: 푠′′ 푠′ 푟′′ 푠′′ 푠′ 푠′ ∙ dom(푓푘 ) = dom(푓푘 ) ∪ (dom(푓푘 ) ∖ 푑푘), 푓푘 dom(푓푘 ) = 푓푘 and 푠′′ 푟′′ 푟′′ 푟′′ 푓푘 (dom(푓푘 ) ∖ 푑푘) = 푓 (dom(푓푘 ) ∖ 푑푘). 푠′′ 푠′ 푟′′ ∙ 푎푘 = 푎푘 ∪ (푎푘 ∖ 푑푘). 푠′′ 푟′′ ∙ 퐴푘 = 퐴푘 . 휎(푠′′) for 푠′′ ≤ 푠′ is deﬁned in an exactly similar way with the roles of 푟′, 푟′′ and 푠′, 푠′′ reversed. Note in particular that:

′′ ′ ′′ 푟′ 푠′ 푟′ 푠′ 훼(푟 ) 푟′′ ∙ For ℓ(푟 ) ≤ 푘 < ℓ(푟 ), 훽푘 = 훼푘 = 훼푘 ∈ 푎푘 ∩푎푘 , and 푓푘 (훽푘) = 푓푘 (훽푘). ′′ ∙ The deﬁnitions of 휌 and 휎 are “level by level” in the sense that 휌(푟 )푘 (resp ′′ ′′ ′′ 휎(푠 )푘) depends only on 푟푘 (resp 푠푘). ′′ ′′ 푟′′ 푟′′ 푠′′ ∙ On each level 푘 of 푟 (resp 푠 ) 휌 (resp 휎) only alters 푎푘 and 푓푘 (resp 푎푘 푠′′ and 푓푘 ) inside the set 푑푘. It is very easy to verify that 휌(푟′) = 푠′. We claim that 휌 and 휎 are order- preserving, by symmetry we only need verify this for 휌. From Lemmas8 and7 we see that it is enough to verify that 휌 is order-preserving for extensions of the form 푟′′ + ⟨푥⟩ ≤ 푟′′ and 푟′′′ ≤* 푟′′.

Claim 16.1. Let 푟′′ ≤ 푟′ and let 푥 be such that 푟′′ +⟨푥⟩ is defined. then 휌(푟′′)+⟨푥⟩ is well-defined and 휌(푟′′ + ⟨푥⟩) = 휌(푟′′) + ⟨푥⟩.

′′ 휌(푟 ) 푟′′ ′′ ′′ Proof. We note that 퐴푘 = 퐴푘 for all 푘 ≥ ℓ(푟 ), so that 휌(푟 ) + ⟨푥⟩ is well- deﬁned. Now we do a case analysis:

′′ ′ 휌(푟 ) 푟′′ 푠′ I 푘 < ℓ(푟 ): 푓푘 is obtained by altering the values of 푓푘 on dom(푓푘 ) to ′′ ′′ ′′ 푠′ 휌(푟 )+⟨푥⟩ 휌(푟 ) 푟 +⟨푥⟩ 푟′′ agree with 푓푘 , and by definition 푓푘 = 푓푘 . Also 푓푘 = 푓푘 ′′ 휌(푟 +⟨푥⟩) 푠′ and then 푓푘 is obtained by altering its values on dom(푓푘 ) to agree ′′ ′′ 푠′ 휌(푟 +⟨푥⟩) 휌(푟 )+⟨푥⟩ with 푓푘 , so easily 푓푘 = 푓푘 . ′′ ′ ′′ 푟′′ 휌(푟 ) I ℓ(푟 ) ≤ 푘 < ℓ(푟 ): Note that 푑푘 ⊆ dom(푓푘 ). By definition, 푓푘 is 푟′′ 푠′ obtained by altering the values of 푓푘 on 푑푘, so as to agree with 푓푘 푠′ 푟′′ 푠′ on dom(푓푘 ) and to agree with 휂 ↦→ 푓푘 (훽푘) ∩ 휂 on 푎푘 . By definition ′′ ′′ ′′ ′′ 휌(푟 )+⟨푥⟩ 휌(푟 ) 푟 +⟨푥⟩ 푟′′ 휌(푟 +⟨푥⟩) 푓푘 = 푓푘 . Also 푓푘 = 푓푘 , and 푓푘 is obtained from it 휌(푟′′+⟨푥⟩) 휌(푟′′)+⟨푥⟩ by the scheme of alteration described above, so 푓푘 = 푓푘 . ′′ I 푘 = ℓ(푟 ): This is the most complicated case. Start by recalling that ′′ 푟′ 푟′ 푠′ 푠′ 푟′′ 푟′′ 푟 +⟨푥⟩ 푑푘 = dom(푓푘 ) ∪ 푎푘 = dom(푓푘 ) ∪ 푎푘 ⊆ dom(푓푘 ) ∪ 푎푘 = dom(푓푘 ), and also that 푟′′ ≤ 푟′. 14 J. CUMMINGS, S.-D. FRIEDMAN, M. MAGIDOR, A. RINOT, AND D. SINAPOVA

푟′′+⟨푥⟩ 푟′′ 푟′′ 푟′′+⟨푥⟩ By definition 푓 (휂) = 푓푘 (휂) for 휂 ∈ dom(푓푘 ) and 푓 (휂) = ′′ 푟′′ 휌(푟 +⟨푥⟩) 푥 ∩ 휂 for 휂 ∈ dom(푎푘 ). When we alter values to obtain 푓푘 , we obtain: 휌(푟′′+⟨푥⟩) 푟′′ 푟′′ ∙ 푓 (휂) = 푓푘 (휂) for 휂 ∈ dom(푓푘 ) ∖ 푑푘. 휌(푟′′+⟨푥⟩) 푟′′ ∙ 푓 (휂) = 푥 ∩ 휂 for 휂 ∈ 푎푘 ∖ 푑푘. 휌(푟′′+⟨푥⟩) 푠′ 푠′ ∙ 푓 (휂) = 푓푘 (휂) for 휂 ∈ dom(푓푘 ). ′′ 휌(푟′′+⟨푥⟩) 푟 +⟨푥⟩ 푠′ ∙ 푓 (휂) = 푓푘 (훽푘) ∩ 휂 for 휂 ∈ 푎푘 . In the last of these cases we note that 푟′′ ≤ 푟′ and ℓ(푟′′) = 푘, so that ′′ 푟′ 푟′′ 푟 +⟨푥⟩ 휌(푟′′+⟨푥⟩) 훽푘 = 훼푘 ∈ 푎푘 and so 푓푘 (훽푘) = 푥∩훽푘. It follows that 푓 (휂) = 푠′ 푥 ∩ 휂 for 휂 ∈ 푎푘 , so we conclude that: 휌(푟′′+⟨푥⟩) 푟′′ 푟′′ ∙ 푓 (휂) = 푓푘 (휂) for 휂 ∈ dom(푓푘 ) ∖ 푑푘. 휌(푟′′+⟨푥⟩) 푠′ 푟′′ ∙ 푓 (휂) = 푥 ∩ 휂 for 휂 ∈ 푎푘 ∪ (푎푘 ∖ 푑푘). 휌(푟′′+⟨푥⟩) 푠′ 푠′ ∙ 푓 (휂) = 푓푘 (휂) for 휂 ∈ dom(푓푘 ). ′′ ′′ 휌(푟 ) 푠′ 푟′′ 휌(푟 ) By definition again we have 푎푘 = 푎푘 ∪ (푎푘 ∖ 푑푘), dom(푓푘 ) = 푠′ 푟′′ dom(푓푘 ) ∪ (dom(푓푘 ) ∖ 푑푘) and: ′′ 휌(푟 ) 푟′′ 푟′′ ∙ 푓푘 (휂) = 푓푘 (휂) for 휂 ∈ dom(푓푘 ) ∖ 푑푘. ′′ 휌(푟 ) 푠′ 푠′ ∙ 푓푘 (휂) = 푓푘 (휂) for 휂 ∈ dom(푓푘 ) ∖ 푑푘. 휌(푟′′)+⟨푥⟩ 휌(푟′′) 휌(푟′′)+⟨푥⟩ 휌(푟′′) Since 푓푘 ⊇ 푓푘 and 푓푘 (휂) = 푥 ∩ 휂 for 휂 ∈ 푎푘 , we see 휌(푟′′+⟨푥⟩) 휌(푟′′)+⟨푥⟩ that 푓푘 = 푓푘 . ′′ ′′ ′′ 푟 +⟨푥⟩ 푟′′ 푟 +⟨푥⟩ 푟′′ I 푘 > ℓ(푟 ): By definition 푎푘 = 푎푘 and 푓푘 = 푓푘 , so that by the 휌(푟′′+⟨푥⟩) 휌(푟′′) 휌(푟′′+⟨푥⟩) definition of the operation 휌 we have 푎푘 = 푎푘 and 푓푘 = 휌(푟′′) 휌(푟′′)+⟨푥⟩ 휌(푟′′) 휌(푟′′)+⟨푥⟩ 휌(푟′′) 푓푘 . Also 푎푘 = 푎푘 and 푓푘 = 푓푘 , from which it 휌(푟′′)+⟨푥⟩ 휌(푟′′+⟨푥⟩) 휌(푟′′)+⟨푥⟩ 휌(푟′′+⟨푥⟩) follows that 푎푘 = 푎푘 and 푓푘 = 푓푘 . We also have 휌(푟′′+⟨푥⟩) 푟′′+⟨푥⟩ 푟′′ 휌(푟′′) 휌(푟′′)+⟨푥⟩ 퐴푘 = 퐴푘 = {푦 ∈ 퐴푘 : 푥 ⊆ 푦} = {푦 ∈ 퐴푘 : 푥 ⊆ 푦} = 퐴푘 . Claim 16.2. Let 푟′′′ ≤* 푟′′ ≤ 푟′, then 휌(푟′′′) ≤* 휌(푟′′).

′′ 푟′′ 푟′′′ ′′ Proof. For 푘 < ℓ(푟 ) we have 푑푘 ⊆ dom(푓푘 ) ⊆ 푓푘 , while for 푘 ≥ 푙ℎ(푟 ) we have ′′ ′′ ′′′ ′′′ 푟 푟푘 푟 푟푘 ′′ 푑푘 ⊆ dom(푓푘 ) ∪ 푎 ⊆ dom(푓푘 ) ∪ 푎 . The deﬁnition of 휌(푟 )푘 involves altering 푟′′ ′′ 푟′′ 푟′′ ′′ the values of 푓푘 (for 푘 < ℓ(푟 )) or 푓푘 and 푎푘 (for 푘 ≥ ℓ(푟 )) in a uniform manner 푟′ ′ ′′ inside 푑푘, using only the values of 푓푘 and (for 푙ℎ(푟 ) ≤ 푘 < ℓ(푟 )) the values of 푟′′ 푟′ ′′′ 푟′′ 푟′ 푟′′′ 푟′ 푓푘 (훼푘 ), and similarly for 휌(푟 ). Since 푓푘 (훼푘 ) = 푓푘 (훼푘 ), it is now routine to ′′′ ′′ verify that 휌(푟 ) ≤ 휌(푟 ). Claim 16.3. 휌 and 휎 are mutually inverse. Proof. Let 휌(푟′′) = 푠′′. To verify that 휎(푠′′) = 푟′′, the key points are that 푟′′ ≤ 푟′ ′ ′′ 푠′′ 푟′′ and that for 푙ℎ(푟 ) ≤ 푘 < ℓ(푟 ) we have 푓푘 (훽푘) = 푓푘 (훽푘). It follows that for such ′′ 푟′ 푠′ 휎(푠 ) 푠′′ 푟′′ values of 푘 and for all 휂 ∈ dom(푓푘 )∩푎푘 we have 푓푘 (휂) = 푓푘 (훽푘)∩휂 = 푓푘 (훽푘)∩ 푟′′ ′′ ′′ 휂 = 푓푘 (휂), which is the only diﬃcult step in the proof that 휎(푠 ) = 푟 .

′ Claim 16.4. If 퐺0 is Q-generic with 푟 ∈ 퐺0, and 퐺1 is the Q-generic filter ′ generated by 휌[퐺0 ∩ Q ↓ 푟 ], then 퐺0,⃗훼 = 퐺1,⃗훼. ORDINAL DEFINABLE SUBSETS OF SINGULAR CARDINALS 15

′′ ′′ ′ ′ ′ 휌(푟 ) 푟′′ Proof. Let 푟 ≤ 푟 and recall that 푟 , 푠 ≤ 푞. we claim that 푓푘 (훼푘) = 푓푘 (훼푘) for all 푘 with ℓ(푞) ≤ 푘 < ℓ(푟′′). ′′ ′ 휌(푟 ) 푠′ 푟′ 푟′′ I If 푘 < ℓ(푟 ) then 푓푘 (훼푘) = 푓푘 (훼푘) = 푓푘 (훼푘) = 푓푘 (훼푘). ′ ′′ 푞 푠′ ′′ ′ I If ℓ(푟 ) ≤ 푘 < ℓ(푟 ) then since 훼푘 ∈ 푎푘 ⊆ 푎푘 and 푟 ≤ 푟 , we have ′′ 휌(푟 ) 푟′′ 푟′ 푓푘 (훼푘) = 푓푘 (훽푘) ∩ 훼푘 = 푓푘 (훼푘).

This concludes the proof of Lemma 16. We are now ready to complete the proof of Lemma 15.

Proof of Lemma 15, Part Two. Recall that we need to derive a contradiction from the assumption that there are 푟* and 푠* satisfying the hypotheses of Lemma 16 * * such that 푟 휓(훾, 훿, 푥˙) and 푠 ¬휓(훾, 훿, 푥˙), where (crucially)푥 ˙ is a Q⃗훼-name. * * Let us force with Q below 푟 to obtain a generic object 퐺0 containing 푟 , and * then use 휌 to obtain a generic object 퐺1 such that 퐺1 contains 푠 . Since 휌 ∈ 푉 ,

푉 [퐺0] = 푉 [퐺1] and since 퐺0,⃗훼 = 퐺1,⃗훼, 푖퐺0 (푥 ˙) = 푖퐺1 (푥 ˙) = 푦 say. This is an immediate contradiction. Combining the results we have obtained, we have proved the Main Theorem. Theorem. Suppose that 휅 < 휆 where cf(휅) = 휔, 휆 is inaccessible and 휅 is a limit of 휆-supercompact cardinals. There is a forcing poset Q such that if 퐺 is Q-generic then: ∙ The models 푉 and 푉 [퐺] have the same bounded subsets of 휅. ∙ Every infinite cardinal 휇 with 휇 ≤ 휅 or 휇 ≥ 휆 is preserved in 푉 [퐺]. ∙ 휆 = (휅+)푉 [퐺]. ∙ For every 푥 ⊆ 휅 with 푥 ∈ 푉 [퐺], (휅+)HOD푥 < 휆. As we mentioned in the Introduction, we can strengthen the conclusion of the Main Theorem. We start with (휅푛)푛<휔 and 휆 as before, but we now assume that the cardinals 휅푛 and 휆 are supercompact. By doing a suitable preparatory class forcing we may assume in addition that for every set-generic extension 푊 of 푉 , 푉 ⊆ HOD푊 . The idea of the preparation, which is essentially due to McAloon [9], is that we will make the 휅푛 and 휆 Laver-indestructible, and then do 휆-directed closed forcing above 휆 to arrange that every set of ordinals is coded unboundedly often into the values of the continuum function. We claim that if 퐺 is Q-generic then in 푉 [퐺] we have that 휆 is supercompact in 푉 [퐺] HOD푥 for all 푥 ⊆ 휅. To see this ﬁnd ⃗훼 as before such that HOD푥 ⊆ 푉 [퐺⃗훼], so that we have a chain of inclusions 푉 [퐺] 푉 [퐺] 푉 ⊆ HOD ⊆ HOD푥 ⊆ 푉 [퐺⃗훼]. 푉 [퐺] By the Intermediate Models Theorem the model HOD푥 is a set-generic extension of 푉 via some complete Boolean subalgebra B of r. o.(Q⃗훼). Since |B| < 휆, it follows 푉 [퐺] by the Levy-Solovay theorem that 휆 is supercompact in HOD푥 . Remark 6. Gitik [3] informed us that the same eﬀect should be possible using Me- rimovich’s supercompact extender-based forcing [10]. This weakens the hypotheses of the Main Theorem to one supercompact cardinal with an inaccessible cardinal above it. The details then appeared in [6]. 16 J. CUMMINGS, S.-D. FRIEDMAN, M. MAGIDOR, A. RINOT, AND D. SINAPOVA

Remark 7. Woodin [13] pointed out that it should also be possible to prove the Main Theorem using methods of Kafkoulis [7]. The idea is to start with a supercompact cardinal with an inaccessible cardinal above, construct a choiceless “Solovay-like” model 푉 * generated by subsets of 휅 arising in initial segments of a supercompact Prikry extension, and then force over 푉 * to restore choice.

References [1] M. Foreman and W. H. Woodin, The Generalized Continuum Hypothesis can fail everywhere, Annals of Mathematics 133 (1991), 1–35. [2] M. Gitik, Prikry-type forcings, in Handbook of set theory (ed. M. Foreman and A. Kanamori), 1351–1147, Springer, Dordrecht, 2010. [3] M. Gitik, private communication, February 2016. [4] M. Gitik and M. Magidor, Extender based forcings, Journal of Symbolic Logic 59 (1994), 445–460. [5] M. Gitik and M. Magidor, The singular cardinal hypothesis revisited, in Set theory of the continuum, 243–279, Springer, New York, 1992. [6] M. Gitik and C. Merimovich, Some applications of supercompact extender based forcings to HOD, arXiv:1608.00356, August 2016. [7] G. Kafkoulis, The consistency strength of an infinitary Ramsey property, Journal of Symbolic Logic 59 (1994), 1158–1195. [8] A. Mathias, On sequences generic in the sense of Prikry, Journal of the Australian Mathe- matical Society 15 (1973), 409–414. [9] K. McAloon, Consistency results about ordinal definability, Annals of Mathematical Logic 2 (1971), 449–467. [10] C. Merimovich, Supercompact extender based Prikry forcing, Archive for Mathematical Logic 50 (2011), 591–602 [11] J. Silver, On the singular cardinals problem, Proceedings of the International Congress of Mathematicians (Vancouver, B. C., 1974), Vol. 1 (1975), 265–268 [12] S. Shelah, Set Theory without choice: not everything on cofinality is possible, Archive for Mathematical Logic 36 (1997), 81–125 [13] W. H. Woodin, private communication, February 2017.

Department of Mathematical Sciences, Carnegie Mellon University, Pittsburgh PA 15213-3890, USA E-mail address: [email protected]

Kurt Godel¨ Research Center for Mathematical Logic, University of Vienna, A-1090 Vienna, Austria E-mail address: [email protected]

Einstein Institute of Mathematics, The Hebrew University of Jerusalem, Jerusalem, 91904, Israel E-mail address: [email protected]

Department of Mathematics, Bar-Ilan University, Ramat-Gan, 5290002, Israel E-mail address: [email protected]

Department of Mathematics, Statistics, and Computer Science, University of Illinois at Chicago, Chicago, IL 60607-7045, USA E-mail address: [email protected]