All uncountable cardinals can be singular

Gitik, Moti

doi:10.1007/bf02760939

Cited by 65 publications

(93 citation statements)

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“…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].…”

mentioning

confidence: 85%

“…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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confidence: 96%

“…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.…”

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confidence: 99%

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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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“…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].…”

mentioning

confidence: 85%

mentioning

confidence: 96%

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

Gitik¹

1985

Trans. Amer. Math. Soc.

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“…The first result in this direction was from van den Berg (see [12] 1 ) who proved that WISC implies the existence of a proper class of regular cardinals, and so WISC must fail in Gitik's model of ZF [4]. This model is constructed assuming the existence of a proper class of certain large cardinals, and it has no regular cardinals bigger than ℵ 0 .…”

Section: M Robertsmentioning

confidence: 99%

The Weak Choice Principle WISC may Fail in the Category of Sets

Roberts

2015

Stud Logica

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Abstract.The set-theoretic axiom WISC states that for every set there is a set of surjections to it cofinal in all such surjections. By constructing an unbounded topos over the category of sets and using an extension of the internal logic of a topos due to Shulman, we show that WISC is independent of the rest of the axioms of the set theory given by a well-pointed topos. This also gives an example of a topos that is not a predicative topos as defined by van den Berg.

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“…Gitik's work exhibits a steady engagement with central and difficult issues of set theory and a masterful virtuosity in the application of sophisticated techniques over a broad range. Gitik [1980] had established through an iterated Prikry forcing the conspicuous singularization result that: If there is a proper class of strongly compact cardinals, then in a ZF inner model of a class forcing extension every infinite cardinal has cofinality ω. Mentioned earlier was the mid-1970s result that that NS ω1 is precipitous is equi-consistent with having a measurable cardinal.…”

Section: Into the 1990smentioning

confidence: 99%

Large Cardinals with Forcing

Kanamori¹

2012

Handbook of the History of Logic

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All uncountable cardinals can be singular

Cited by 65 publications

References 5 publications

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

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

The Weak Choice Principle WISC may Fail in the Category of Sets

Large Cardinals with Forcing

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