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Work in $ZF$. Let $j:V\to V$ be a non-trivial elementary embedding which is iterable, so that we can iterate it and form models $M_\alpha, \alpha\in ON,$ with $M_0=V,$ and elementary embeddings $j_{\alpha, \beta},$ for $\alpha\leq \beta.$

Let $M=\bigcap_\alpha M_\alpha.$

What can we say about $M$?

It seems it is not difficult to show that $M$ is also a model of $ZF$, since it suffices to show that it is closed under Godel operations.

Which large cardinals of $V$ are preserved in $M$?

Of course the answer seems to be trivial for some large cardinals, in particular for those below the critical point of $j$, where their existence require information in $V_{crit(j)}$ So I am particularly interested in those large cardinals whose definitions require a proper class of information, like supercompact cardinals, ...

What interesting properties $M$ can have? In particular what can we say about the relation between $V$ and $M$?

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Remark. As it is stated in the comments by Hamkins, any such embedding is iterable, so we can remove the extra assumption of iterability of $j.$

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  • $\begingroup$ Is $j\colon V\to V$? If so, $M=V$ since $M_\alpha$ is $V$ for all $\alpha$. No? Am I missing something? $\endgroup$
    – Asaf Karagila
    Commented Oct 24, 2014 at 5:46
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    $\begingroup$ @AsafKaragila Why is $M_\omega=V$? $M_\omega$ is defined as a direct limit, and the maps are not the identity. $\endgroup$ Commented Oct 24, 2014 at 6:01
  • $\begingroup$ @Andres: Ah, yes. That's what I was missing. I had a hunch it might be in the limit steps. Those can be sneaky bastards sometimes. Thanks! (In fact, it's even easy to see why $M_\omega\neq V$, because $\kappa_\omega$ must be regular in $M_\omega$ but singular in $V$...) $\endgroup$
    – Asaf Karagila
    Commented Oct 24, 2014 at 6:02
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    $\begingroup$ Yair, we may apply $j$ to any $j\upharpoonright V_\alpha$, and take the union to form $j_{1,2}=j(j)=\bigcup_\alpha j(j\upharpoonright V_\alpha)$. This is presumably what Mohammad means by iterating. $\endgroup$ Commented Oct 24, 2014 at 10:57
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    $\begingroup$ Andres, why doesn't the usual argument show that all iterates are well-founded? Let $\xi$ be least such that $j_{0,\lambda}(\xi)$ is in the ill-founded part of some $M_\lambda$, with $\lambda$ a limit. But in some $M_\alpha$ with $0<\alpha<\lambda$, there will be born a smaller $\xi'<\xi$ with $j_{\alpha,\lambda}(\xi')$ also in the ill-founded part. This argument seems to use only ZF in the language with j, in order that we may refer in any model to the corresponding iterates of $j$. $\endgroup$ Commented Oct 24, 2014 at 13:44

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Note that $M=\bigcap_{\alpha}M_\alpha$ is just the $\mathrm{OR}^{\mathrm{th}}$ iterate $M_{\mathrm{OR}}$ cut off at height $\mathrm{OR}$, so we have for example $V_\kappa\preceq V_\lambda\preceq M$ where $\kappa=\mathrm{crit}(j)$ and $\lambda=\kappa_\omega(j)$, where $\kappa_0(j)=\mathrm{crit}(j)$ and $\kappa_{n+1}(j)=j(\kappa_n(j))$ and $\kappa_\omega(j)=\sup_{n<\omega}\kappa_n(j)$. So in terms of large cardinals with a first-order description, $M$ has the same kinds as does $V_\kappa$. If there is an elementary $k:V\to V$ with $\kappa_\omega(k)<\kappa$ then arguing like in A weak reflection of Reinhardt by super Reinhardt cardinals, we get an embedding $\ell:M\to M$ witnessing that $\kappa_n(k)$ is Reinhardt for $M$, for some $n<\omega$. (That is, for some $n<\omega$, we get $k_n(\kappa)=\kappa$, and then we can just stretch out the structure $(V_\kappa,{k_n\upharpoonright V_\kappa})$ to produce such an $\ell$.)

There are some facts established in Reinhardt cardinals and iterates of V on the relationship between the iterates $M_\alpha$ and $V$. In particular, it is shown that for each $\alpha\geq\omega$, every set is set-generic over $M_\alpha$, but $M_\alpha$ is not a set-ground of $V$. And $\mathrm{HOD}\subseteq M_\omega$ but $\mathrm{HOD}\not\subseteq M$.

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