The Core Model

Dodd, A.

doi:10.1017/cbo9780511600586

Cited by 80 publications

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“…Since a measurable cardinal is Keisler we may assume that there is no inner model with a measurable cardinal. We shall show that κ is Keisler in the Dodd-Jensen core model K (see [1]). Using our assumption ¬L It can be shown that every Keisler cardinal is inaccessible.…”

Section: Proof (2)mentioning

confidence: 86%

Cardinal elementary extensions

Donder

2000

Proc. Amer. Math. Soc.

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Abstract. We determine the consistency strength of some model theoretic extension properties for cardinals.Let M, N be two nonempty sets. We say that N extends M iff for every structureis an elementary extension of M. In this note we make some observations on this property for cardinals. More precisely we shall determine the consistency strength of the two properties "κ + extends κ" and "κ ++ extends κ". First note that if some τ > κ extends κ, then κ is weakly Π 1 1 -indescribable. Now let us look at the property "κ ++ extends κ" which is easy to analyze. Its consistency strength over ZFC is simply given by "for some regular τ > κ V τ extends V κ ". Let us call a κ with the latter property sublime (with witness τ ). Clearly, a witness τ for the sublimity of κ is inaccessible. So if we force with Col(κ + , < τ) we get a model where κ ++ extends κ. This gives one direction of our previous claim. For the other one just assume that some regular τ > κ extends κ. We shall show that κ is sublime in L with witness τ . To get this we only need to know that τ extends κ in L. For then L τ extends L κ which yields that κ is sublime in L. To see that τ extends κ in L we need a small but familiar argument. Since we shall need it once more we make it a lemma.

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Section: Proof (2)mentioning

confidence: 86%

Cardinal elementary extensions

Donder

2000

Proc. Amer. Math. Soc.

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“…Then D is a good set of indiscernibles of order type c for K λ , ∈ F . We now appeal to the Jensen Indiscernibles Lemma (see [2], 16.10), and use the uncountable cofinality of c, to claim that there is E ⊇ D with E ∈ K and elements of E good indiscernibles for the same structure. But this entails E being a homogeneous set for the function F .…”

Section: Theorem 26mentioning

confidence: 99%

On possible non-homeomorphic substructures of the real line

Welch

2002

Proc. Amer. Math. Soc.

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Abstract. We consider the problem, raised by Kunen and Tall, of whether the real continuum can have non-homeomorphic versions in different submodels of the universe of all sets. This requires large cardinals, and we obtain an exact consistency strength: Theorem 1. The following are equiconsistent: sufficiently elementary submodel of the universe of sets with R M not homeomorphic to R.The reverse direction is a corollary to:We further consider the large cardinal consequences of the existence of a topological space with a proper substructure homeomorphic to Baire space.

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“…j(a) = a for a < k, j(k) > k and M is closed under sequences of its elements of length < /(k). 1. The forcing notion.…”

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confidence: 99%

Regular cardinals in models of ${\rm ZF}$

Gitik¹

1985

Trans. Amer. Math. Soc.

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Abstract. We prove the following Theorem. Suppose M is a countable model of ZFC and k is an almost huge cardinal in M. Let A be a subset of k consisting of nonlimit ordinals. Then there is a model NA of ZF such that S0 is a regular cardinal in NA iff a e A for every a > 0. 0. Introduction. We consider the following question. What are the restrictions in ZF on the class of all regular cardinals? Clearly, S0 is always regular and Nu, Sw+l0, SuU and Sa, for a the least s.t. Sa = a, are singular. For a limit a, if Na is regular, then it is already quite large and its existence is unprovable in ZF. In ZFC, Na+1 is a regular cardinal for every a. Feferman and Levy [3] proved that it is not true in ZF alone. They built a model of ZF + "Sj is singular". Combining their method with Easton's work [2], it is possible to build a model of ZF + "for every a, Xa or Sa + 1 is singular". Now to make both S" and Sa+1 some additional assumptions are needed.Thus, the Dodd and Jensen Covering Lemma [1] implies that at least O1' is needed, and by Mitchell [11] it looks like at least a hypermeasurable cardinal is needed. On the other hand, a model of ZF + "Va > 0 Na is a singular cardinal" is constructed in [5] from the much stronger assumption Con(ZFC + "V«3k > a k is strongly compact"). The simplest question, which is unclear how to handle by methods of [5], is the following. Is there a model of ZF s.t. those regular alephs are only N0 and Hx! Or, for example, let A = {a + I20\a g On}. Is there a model of ZF s.t. those regular alephs are exactly {Sa\a G A U {0}}, i.e., N0,N120,N240,...?We shall prove the following Theorem. Suppose M is a countable model of ZFC and k is an almost huge cardinal in M. Let A be a subset of k consisting of nonlimit ordinals. Then there is a model NA of ZF s.t. those regular alephs are exactly {NJa G A U {0}}.As an immediate consequence we obtain Theorem. If ZFC + (3k) (k is an almost huge cardinal) is consistent, then there is a model M of ZFC s.t. for every class A of M consisting of nonlimit ordinals there exists a model NA ofZFs.t.those regular alephs are exactly {NJa E^U {0}}.Notice, that for A = 0 it gives a model with all uncountable Sa's singular, but the assumption here is stronger than those of [5].

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The Core Model

Cited by 80 publications

References 0 publications

Cardinal elementary extensions

Cardinal elementary extensions

On possible non-homeomorphic substructures of the real line

Regular cardinals in models of ${\rm ZF}$

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