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Is λ or κ the rank-into-rank cardinal?

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Regarding the current page contents, would it be safe to say that a cardinal λ is (a) rank-into-rank iff it satisfies one of the four axioms? (which?) -- Schnee 11:54, 18 Dec 2003 (UTC)

Since I think that large cardinal properties which involve elementary embeddings belong to their critical points, I would call κ a rank-into-rank cardinal iff it is the critical point of any of the elementary embeddings mentioned in the definitions of I3, I2, I1, or I0. λ is larger than κ, but it is not as strong a limit, in fact, it has cofinality ω. JRSpriggs 05:58, 6 May 2006 (UTC)[reply]

Elementary Embedding Of Vλ For Non-Inaccessible λ?

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The article mentions elementary embeddings of Vλ but also says that λ cannot be inaccessible (assuming choice). But an elementary embedding is an isomorphism between models, so if λ isn't inaccessible, what exactly is Vλ being considered as a model of? -- 79.72.38.186 (talk · contribs) 10:39, 6 August 2007 (UTC)[reply]

A model of the theory of itself. Models are not required to be models "of" something given in advance. JRSpriggs (talk) 21:37, 24 January 2010 (UTC)[reply]
Perhaps my previous answer was not sufficiently responsive. Vλ satisfies ZFC except for instances of the axiom of replacement where the image would have rank λ. In I2, M like V is a model of ZFC. Vλ+1 satisfies ZFC except for instances of replacement, pairing, or powerset where one of the given sets has rank λ. In I0, L(Vλ+1) satisfies ZF (without choice). I hope that helps. JRSpriggs (talk) 09:21, 29 January 2010 (UTC)[reply]

Is I0 inconsistent?

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According to the article on Kunen's inconsistency theorem, one of its consequences says "If j is an elementary embedding of the universe V into an inner model M, and λ is the smallest fixed point of j above the critical point κ of j, then M does not contain the set j "λ (the image of j restricted to λ).". However according to this article, rank-into-rank axiom I0 says "There is a nontrivial elementary embedding of L(Vλ+1) into itself with the critical point below λ.". Now, j "λ is a subset of λ and thus an element of Vλ+1. If we apply Kunen's result to the submodel V' = M' = L(Vλ+1), does this not result in a contradiction since M' contains the forbidden element? JRSpriggs (talk) 21:37, 24 January 2010 (UTC)[reply]

I see now that I was forgetting that L(Vλ+1) may not satisfy the axiom of choice. JRSpriggs (talk) 18:21, 26 January 2010 (UTC)[reply]

I1 and

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The article asserts `Axiom I1 implies that (equivalently, ) does not satisfy "V=HOD"'. What is "V=HOD" supposed to mean here? The most obvious, which is the usual one in the ZF context, is "every set is definable from ordinal parameters". But with this interpretation, can never satisfy "V=HOD", for any infinite , just because there are only -many ordinal parameters available, and has cardinality strictly bigger than . On the other hand, _can_ satisfy "V=HOD" in this sense. So this needs to be clarified.

See Ordinal definable set for the meaning of V=HOD.
Your argument "because there are only λ-many ordinal parameters available, and Vλ + 1 has cardinality strictly bigger than λ< ω." cannot be carried out inside Vλ + 1. JRSpriggs (talk) 05:18, 1 July 2020 (UTC)[reply]
Regarding the meaning of "V=HOD", I believe the discussion in the article Ordinal definable set assumes we are working in a model of ZF, but doesn't model ZF.
It doesn't matter whether the argument can be carried out inside or not. The point is that if for every , there is an ordinal such that is definable over from the parameter , then we get a surjection (set the unique such that satisfies , if there is such a unique , and otherwise). (Remark: This function is not definable over , since has arbitrary complexity, but it is definable in , which is all that matters.) But in , there can be no such surjection for the usual reasons. So this interpretation of "V=HOD" is not tenable. But this is the most obvious interpretation. So the article should indicate how "V=HOD" is supposed to be interpreted. --Bezian (talk) 22:38, 2 July 2020 (UTC)[reply]