Talk:Rank-into-rank

Is &lambda; or &kappa; the rank-into-rank cardinal?
Regarding the current page contents, would it be safe to say that a cardinal &lambda; 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 &kappa; 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. &lambda; is larger than &kappa;, but it is not as strong a limit, in fact, it has cofinality &omega;. JRSpriggs 05:58, 6 May 2006 (UTC)

Elementary Embedding Of Vλ For Non-Inaccessible λ?
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? -- 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. V&lambda; satisfies ZFC except for instances of the axiom of replacement where the image would have rank &lambda;. In I2, M like V is a model of ZFC. V&lambda;+1 satisfies ZFC except for instances of replacement, pairing, or powerset where one of the given sets has rank &lambda;. In I0, L(V&lambda;+1) satisfies ZF (without choice). I hope that helps. JRSpriggs (talk) 09:21, 29 January 2010 (UTC)

Is I0 inconsistent?
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 &lambda; is the smallest fixed point of j above the critical point &kappa; of j, then M does not contain the set j "&lambda; (the image of j restricted to &lambda;).". However according to this article, rank-into-rank axiom I0 says "There is a nontrivial elementary embedding of L(V&lambda;+1) into itself with the critical point below &lambda;.". Now, j "&lambda; is a subset of &lambda; and thus an element of V&lambda;+1. If we apply Kunen's result to the submodel V'  = M'  = L(V&lambda;+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(V&lambda;+1) may not satisfy the axiom of choice. JRSpriggs (talk) 18:21, 26 January 2010 (UTC)

I1 and $$\mathrm{HOD}$$
The article asserts `Axiom I1 implies that $$V_{\lambda+1}$$ (equivalently, $$H(\lambda^+)$$) 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, $$V_{\lambda+1}$$ can never satisfy "V=HOD", for any infinite $$\lambda$$, just because there are only $$\lambda$$-many ordinal parameters available, and $$V_{\lambda+1}$$ has cardinality strictly bigger than $$\lambda^{<\omega}$$. On the other hand, $$H(\lambda^+)$$ _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)


 * 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 $$V_{\lambda+1}$$ doesn't model ZF.
 * It doesn't matter whether the argument can be carried out inside $$V_{\lambda+1}$$ or not. The point is that if for every $$x\in V_{\lambda+1}$$, there is an ordinal $$\eta\leq\lambda$$ such that $$x$$ is definable over $$V_{\lambda+1}$$ from the parameter $$\eta$$, then we get a surjection $$\pi:\omega\times(\lambda+1)\to V_{\lambda+1}$$ (set $$\pi(\varphi,\eta)=$$ the unique $$x\in V_{\lambda+1}$$ such that $$V_{\lambda+1}$$ satisfies $$\varphi(\eta,x)$$, if there is such a unique $$x$$, and $$\pi(\varphi,\eta)=\emptyset$$ otherwise). (Remark: This function is not definable over $$V_{\lambda+1}$$, since $$\varphi$$ has arbitrary complexity, but it is definable in $$V$$, which is all that matters.) But in $$V$$, 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)