Talk:Rank-into-rank

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

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 V_λ For Non-Inaccessible λ?

The article mentions elementary embeddings of V_λ 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 V_λ being considered as a model of? -- 79.72.38.186 (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. V_λ 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. V_{λ 1} satisfies ZFC except for instances of replacement, pairing, or powerset where one of the given sets has rank λ. In I0, L(V_{λ 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 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(V_{λ 1}) into itself with the critical point below λ.". Now, j "λ is a subset of λ and thus an element of V_{λ 1}. If we apply Kunen's result to the submodel V' = M' = L(V_{λ 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_{λ 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 V_{λ 1} has cardinality strictly bigger than λ^{< ω}." cannot be carried out inside V_{λ 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)