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WikiProject Mathematics (Rated Stub-class, Low-priority)
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 Field:  Foundations, logic, and set theory

Is λ or κ the rank-into-rank cardinal?[edit]

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)

Elementary Embedding Of Vλ For Non-Inaccessible λ?[edit]

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? -- (talk · contribs) 10:39, 6 August 2007 (UTC)

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)
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)

Is I0 inconsistent?[edit]

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)

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)