Timeline for Can there exists a model of ZFC with permutation that sends successor infinite stages to their predecessors?
Current License: CC BY-SA 4.0
12 events
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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 15:03 | 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 |