Can there exist a model $M$ of $\sf ZFC$ and an external permutation $j$ on $M$ such that $j[(V_{\alpha+1})^M]=(V_\alpha)^M$ for each infinite $\alpha$?
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asked Sep 21 at 11:57
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2 Answers
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The answer is no.
If $j$ maps each $V_{\omega+n+1}^M$ onto $V_{\omega+n}^M$ for $n\in\omega^M$, then $j[V_{\omega+\omega}^M]=V_{\omega+\omega}^M$ and so it cannot be that $j[V_{\omega+\omega+1}^M]=V_{\omega+\omega}^M$ also, if $j$ is one-to-one, since all the elements in the target are already used.
answered Sep 21 at 14:19
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$\begingroup$ so this can be used if $M$ is a model of $\sf Z$. $\endgroup$ Commented Sep 21 at 17:43
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$\begingroup$ No, because even $V_\omega$ doesn't necessarily exist in Z, let alone $V_{\omega+\omega+1}$. $\endgroup$ Commented Sep 21 at 20:22
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$\begingroup$ Well the magic word in the question is "can", $V_{\omega+\omega}$ is a model of $\sf Z$, the point is that we don't want $V_{\omega+\omega+1}$ to be there, because of the exhaustive argument you depicted. $\endgroup$ Commented Sep 21 at 20:37
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$\begingroup$ Oh, I thought you meant the argument I gave can be used also with models of Z. $\endgroup$ Commented Sep 21 at 21:37
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$\begingroup$ @I see, I meant we can have $M$ if we state it in terms of $\sf Z$ instead of $\sf ZF$. Even if we have $V_\omega^M$ I think we can arrange for $j[V_\omega^M]$ to be some subset of $V_\omega^M$ such that the complement of $j[V_\omega^M]$ is infinite and thus has a room for images of extra-elements of $V_{\omega+1}$ under $j$. $\endgroup$ Commented Sep 22 at 11:51
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This is just an answer to a related condition that if we require $M$ to be a model of Zermelo set theory, then there does exist such a model.
Let $M$ be a countable transitive model of $\sf Z $ + $V_\omega$ exists + every set is an element of some $V_{\omega+n}$ for a natural $n$
Fix a well ordering $R$ on $\mathcal P(M)$.
Now we take the $R$-least bijection between $V_\omega^M$ and $\omega^M$, call it $j_0$, now let $j_1$ be the $R$-least bijection between $V_{\omega+1}^M\setminus V_\omega^M$ and $V_\omega^M \setminus \omega^M $, now $j_2$ is the $R$-least bijection from $V_{\omega+2}^M\setminus V_{\omega+1}^M$ to $V_{\omega+1}^M \setminus V_\omega^M$, and so on each $j_{n>2}$ is the $R$-least bijection from $V_{\omega+n}^M\setminus V_{\omega+n-1}^M$ to $V_{\omega+n-1}^M \setminus V_{\omega+n-2}^M$.
Define: $j=\bigcup_{n\in \omega} j_n$
answered Sep 23 at 19:07