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Sep 22 at 11:51 comment added Zuhair Al-Johar @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 $j$.
Sep 21 at 21:37 comment added Joel David Hamkins Oh, I thought you meant the argument I gave can be used also with models of Z.
Sep 21 at 20:37 comment added Zuhair Al-Johar 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.
Sep 21 at 20:22 comment added Joel David Hamkins No, because even $V_\omega$ doesn't necessarily exist in Z, let alone $V_{\omega+\omega+1}$.
Sep 21 at 17:43 comment added Zuhair Al-Johar so this can be used if $M$ is a model of $\sf Z$.
Sep 21 at 17:40 vote accept Zuhair Al-Johar
Sep 21 at 15:09 history undeleted Joel David Hamkins
Sep 21 at 15:09 history edited Joel David Hamkins CC BY-SA 4.0
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Sep 21 at 15:02 history deleted Joel David Hamkins via Vote
Sep 21 at 14:38 history edited Joel David Hamkins CC BY-SA 4.0
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Sep 21 at 14:19 history answered Joel David Hamkins CC BY-SA 4.0