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If we violate the partition principle and add to $\sf ZF$ the axiom that there exists a set $X$ that has a partition on it that is greater in cardinality than the set of singleton subsets of $X$.

Can $X$ be a standard stage $V_\alpha$ of the cumulative hierarchy?

Can $X$ be a non-standard stage $V_{\alpha'}$ of the cumulative hierarchy?

About the second question what I mean is there is a non-well founded model of $\sf ZF$ such that $\alpha'$ is seen as an ordinal internally but externally it is a transitive set of transitive sets but has subsets of it that are infinite descending membership chains.

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  • $\begingroup$ Partitioning a set into "more" is actually violating the weak partition principle. It is a stronger claim. $\endgroup$
    – Asaf Karagila ♦
    Commented Jun 10, 2022 at 9:58
  • $\begingroup$ You can't have a transitive set of transitive sets which has a descending membership chain. If you are a transitive set, you are well-founded. Full stop. Non-standard ordinals/models/etc. will always have a membership relationship that is not the real $\in$. $\endgroup$
    – Asaf Karagila ♦
    Commented Jun 10, 2022 at 10:20
  • $\begingroup$ Assuming ZF, a transitive set is always well-founded. All sets are. Non-standard models have a different membership relation, which means it is irrelevant what is the underlying set. $\endgroup$
    – Asaf Karagila ♦
    Commented Jun 10, 2022 at 10:31
  • $\begingroup$ I'm speaking about non-well founded models, the background theory for models here is ZF-Reg. So here we can have those transitive and not well founded. $\endgroup$
    – Zuhair Al-Johar
    Commented Jun 10, 2022 at 10:33
  • $\begingroup$ We had discussed this over some of your previous questions. There is no reasonable notion for "non-standard ordinals" the way you thinking about it; there is even less of a way to define $V_\alpha$ in those cases. $\endgroup$
    – Asaf Karagila ♦
    Commented Jun 10, 2022 at 10:35

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If you are asking whether or not a $V_\alpha$ could violate the partition principle, the answer is easily yes.

As we all know, it is always the case that $\Bbb R$ can be partitioned into $\aleph_1$ parts; but it is consistent that there is no injection from $\omega_1$ into the reals. Next, observe that $V_{\omega+1}$ is exactly the same cardinality as $\Bbb R$.

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  • $\begingroup$ What about the non-standard stages? $\endgroup$
    – Zuhair Al-Johar
    Commented Jun 10, 2022 at 10:17
  • $\begingroup$ That's an internal statement. I'm not sure what standard/non-standard have to do with it. $\endgroup$
    – Asaf Karagila ♦
    Commented Jun 10, 2022 at 10:18
  • $\begingroup$ I mean can we consistently say that we have a non-well founded model of $\sf ZF$, where there is a set $V_{\alpha'}$ in that model where $\alpha'$ is externally a non-standard ordinal (as described above) and such that this model satisfies that $V_{\alpha'}$ violates the weak PP $\endgroup$
    – Zuhair Al-Johar
    Commented Jun 10, 2022 at 10:58
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    $\begingroup$ @ZuhairAl-Johar Take $\alpha'$ to be the $\omega+1$ of any model of "ZF + PP fails at $V_{\omega+1}$" which is not an $\omega$-model. $\endgroup$
    – Elliot Glazer
    Commented Jun 11, 2022 at 14:43

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