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Gitik Used a Cardinal TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 353, Number 12, Pages 4863{4897 S 0002-9947(01)02853-7 Article electronically published on July 17, 2001 A MEASURABLE CARDINAL WITH A CLOSED UNBOUNDED SET OF INACCESSIBLES FROM o(κ)=κ WILLIAM MITCHELL Abstract. We prove that o(κ)=κ is sufficient to construct a model V [C]in which κ is measurable and C is a closed and unbounded subset of κ containing only inaccessible cardinals of V . Gitik proved that o(κ)=κ is necessary. We also calculate the consistency strength of the existence of such a set C together with the assumption that κ is Mahlo, weakly compact, or Ramsey. In addition we consider the possibility of having the set C generate the closed unbounded ultrafilter of V while κ remains measurable, and show that Radin forcing, which requires a weak repeat point, cannot be improved on. 1. Introduction In [2], Gitik used a cardinal κ with o(κ)=κ + 1 to give a forcing construction of amodelV [C]inwhichκ is measurable, while the set C is closed and unbounded in κ and contains only inaccessibles of V . Gitik also showed that no such construction is possible unless o(κ) ≥ κ in K. The main result of this paper is that the lower bound o(κ)=κ is correct: Theorem 1.1. The following are equiconsistent: 1. There is a cardinal κ with oK (κ)=κ. 2. There is a measurable cardinal κ andaclosedunboundedsubsetC of κ such that each member of C is inaccessible in K. We begin with a proof of Gitik's result that 1.1(2) implies that o(κ)=κ in the core model K. This proof is essentially the same as that of Gitik, but is reorganized to feature the Prikry generic sequence more prominently. Afterward we will reexamine the proof to gain some further insight into the general problem. The proof depends on the following lemma: Lemma 1.2. Suppose that ~κ =(κn : n<!) is K-generic for Prikry forcing by a measure U 2 K on κ, and suppose that each κn is inaccessible and has a closed and K unbounded subset Cn containing only cardinals inaccessible in K.Theno (κ) ≥ κ. In order to see that the lemma implies the direction (2)=)(1) of theorem 1.1, let κ and C be as in clause 1.1(2). Using the measure on κ, force to add a Prikry sequence ( κn : n<!) cofinal in κ. For all but finitely many n 2 !,thesetCn = C \ κn is a closed and unbounded subset of κn containing only inaccessibles of K, and hence the lemma implies that oK (κ) ≥ κ. Received by the editors March 30, 2001. 2000 Mathematics Subject Classification. Primary 03E35, 03E45, 03E55. This work was partially supported by grant number DMS-962-6143 from the National Science Foundation. c 2001 American Mathematical Society 4863 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 4864 WILLIAM MITCHELL Proof of lemma 1.2. Assume, towards a contradiction, that δ = oK (κ) <κ. In order to allow a generalization of the proof, we will not use the assumption that ~κ is a Prikry sequence until the end of the proof. We begin by applying the covering lemma [3]. Let X be a precovering set such that ( κn : n<!) 2 X and ( Cn : n 2 ! ) 2 X and such that X contains all of its limit ordinals of cofinality at most δ. The assertion that X is a precovering set means that jXj <κ,thatX ≺ Vτ for some τ>κ=supκn,andthatX satisfies a certain closure condition. The covering lemma asserts that such a precovering set exists, and that if X is any such precovering set, then there is a function h 2 K,an X ordinal ρ<κ,andasystemSC of indiscernibles for K with the following properties, wherewewriteCX (λ)for CX(λ, β). β<oK (λ) 1. domain(C)= (α, β) 2 X : β<oK (α) ,andC(α, β) ⊂{ν 2 α \X : oK (ν)= β g. 2. If ν 2 X nρ, then either ν 2 h\ν or else ν 2CX(λ)whereλ =inf(h\ν nν) ≤ κ. 3. For any λ 2 X, CX (λ)isclosedinX \ λ. K 4.S If β0 <o (λ)andν is a ordinal of cofinality ! which is a limit point of CX (λ, β), then ν 2CX(λ, β)forsomeβ>β. β≥β0 0 5. If ν 2CX (λ) [fλg and CX (λ) is bounded in X \ ν, then there is η 2 X \ ν such that h\η is cofinal in X \ ν. We can assume, without loss of generality, that ρ<κ0.Foreachn<!,let λn be the least member of X \ h\κn.Thenκn ≤ λn ≤ κ,andifλn >κn,then X κn 2C (λn). The proof of lemma 1.2 relies on two observations: X Claim 1.3. 1. For each n<!,thesetC (λn) is cofinal in κn \ X. K X 2. If X contains all of its limit points of cofinality at most o (λn),thenC (λn) is bounded in κn \ X. X Proof of claim. For clause 1, suppose to the contrary that C (λn) is bounded in X \ κn. It follows by property 5 of the precovering set that there is η 2 X \ κn so that h\η is cofinal in X \ κn.LetA be the set of limit points of X \ h\η.Then 0 \ A is closed and unbounded in κn = sup(X κn), which has uncountable cofinality since X contains all of its limit points of cofinality !. Hence A \ Cn is closed and 0 th \ unbounded in κn; but this is impossible since the ! member τ of A Cn is singular in K,sinceh\η is cofinal in τ, contradicting the assumption that every member of Cn is inaccessible in K. K To prove clause 2, let τ = o (λn) and define a sequence of ordinals ηξ for ξ ≤ τ·!, as follows: 8 >0ifξ =0; <> supξ0<ξ ηξ0 if ξ is a limit ordinal; ηξ = X X >inf C (λ ,ι) n η − if ξ = τ · k + ι +1andC (λ ,ι) 6⊂ η ; :> k ξ 1 k ξ ηξ−1 otherwise: Since X contains all of its limit points of cofinality at most τ, each of the ordinals ηξ is in X for ξ ≤ τ · !. Furthermore, all of the ordinals ηξ are less than κn:The ordinals η0 and ηξ+1 are less than κn by definition, while if ξ is a limit ordinal then 0 either ηξ = ηξ0 =6 κn for some ξ <ξ,orelseηξ is singular while κn is regular. K Thus η = ητ·! is in C(λn,β)forsomeβ<o (λn). For each k<!it follows · C n 6 ? 2C that if ξk = τ k + β then (λn,β) ηξk = and hence ηξk <ηξk +1 (λn,β). License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use A MEASURABLE CARDINAL 4865 It follows that ητ·! is a limit point of C(λn,β), but this is impossible, as it would 0 0 follow by (3) that η 2C(λn,β )forsomeβ >β. We can now complete the proof of lemma 1.2 by observing that the assumption that ~κ is a Prikry sequence implies that λn = κ for all but finitely many n<!. K Thus o (λn)=o(κ)=δ, but this is impossible by claim 1.3 since the precovering set X was chosen to contain all of its limit points of cofinality at most δ. Hence oK (κ) ≥ κ. 1.1. A warm up. Before proving the other direction of theorem 1.1 we state the similar but simpler theorem 1.5. One direction of the proof is given by a second look at the proof of lemma 1.2, while other direction will be used to introduce some of the ideas of the proof of the main theorem in an easier context. Definition 1.4. We write E for the class of inaccessible cardinals of K,andD for the class of cardinals λ 2 E such that f α<λ: oK (α)=β g is stationary in λ for all β<λ. Gitik [2, theorem 1.1] proves that κ 2 D is equiconsistent with the existence of a closed, unbounded set C ⊂ E \ κ,whereκ is inaccessible. The following theorem shows that this construction cannot be directly iterated to obtain infinitely many such cardinals κ: Theorem 1.5. The following are equiconsistent: 1. There is an infinite increasing sequence ~κ =(κn : n<!) of inaccessible car- dinals such that for each n<!there is a closed unbounded set Cn ⊂ κn \ E. 2. There is an infinite increasing sequence ~λ of cardinals λn 2 D with κ = K sup~λ = limn<! o (λn). K Furthermore, the model in clause (1) can be constructed so that o (κn)=0for all n 2 !. Proof of (1)=)(2). Assume that ~κ is a sequence satisfying condition (1) and, as in the proof of lemma 1.2, pick a precovering set X and use it to define the sequence K ~λ. We will show that δ = limn o (λn) ≥ κ. Suppose to the contrary that δ<κ, and pick a second precovering set X0 ⊃ X such that X0 contains all of its limit points of cofinality at most δ. We now need to apply one further consequence of 0 0 the covering lemma, which states that if the ordinals λn are defined from X in the 0 same way as the ordinals λn were defined from X,thenλn = λn for all but finitely many n<!.
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