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Working in $\sf ZF-Reg.$, can we have a transitive model $M$ of $\sf ZF$ such that there exists a proper class (i.e. a subset of $M$ that is not an element of $M$) of infinitely descending infinite ordinals in $M$.

Now $\alpha$ is to be called infinitely descending infinite ordinal in $M$ if and only if $\alpha$ is a set that $M$ sees as an infinite successor von Neumann ordinal but externally it has a subset of it that is an infinitely descending membership chain beginning with $\alpha-1$, i.e. we have $\{\alpha-1, \alpha -2, \alpha-3, ...\} \subset \alpha$, where each $\alpha-n = \alpha -n-1 \cup \{\alpha -n-1\}$ for all $n=0,1,2,...$

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  • $\begingroup$ Ordinals are absolute between transitive models, so such a subset would have to be a set of "real" ordinals, and that can't happen. $\endgroup$
    – Wojowu
    Commented Oct 3, 2022 at 19:37
  • $\begingroup$ @Wojowu, the ambient theory of models here is $\sf ZF−Reg. $ and not $\sf ZF$, and a transitive model here is in this ambience, so it can have non-standard ordinals. In other words, the absolutness argument of ordinals in transitive models (where the ambient theory of models is taken to be $\sf ZFC$), doesn't apply to the case here. $\endgroup$
    – Zuhair Al-Johar
    Commented Oct 3, 2022 at 19:51
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    $\begingroup$ @Wojowu When regularity is dropped from the metatheory, weird things can happen in terms of transitive models - see e.g. this old answer of mine for an extreme example! (Of course, that also means one has to be extremely careful with one's definitions - in particular, if "von Neuman ordinal" means "hereditarily transitive set," then $\mathsf{ZF-Reg}$ can't even prove that the vNos are linearly ordered by $\in$ since Quine atoms are technically vNos by this definition). $\endgroup$
    – Noah Schweber
    Commented Oct 3, 2022 at 20:55
  • $\begingroup$ @NoahSchweber You're right we have to be careful with that we even mean with "ordinals". I believe one characterization of (well-founded) von Neumann ordinals valid in ZF-Reg is as transitive sets (strictly) totally ordered by $\in$. Isn't this enough to show they are absolute between transitive models? $\endgroup$
    – Wojowu
    Commented Oct 3, 2022 at 21:27
  • $\begingroup$ @Wojowu, per your last definition those can be non-well founded since a strict total order is not necessarily a well-order. Now call those as *strict ordinals" to discriminate them from *well ordinals" which are the von Neumanns. Now perhaps the notion of strict ordinals is absolute beween transitive models of $\sf ZF$ even over non-well founded transitive models, but this won't mean that the notion of "well ordinal" is absolute over those, i.e. you can have strict ordinals that some of these models see as well ordinals, but others see them as strict ordinals that are not well ordinals. $\endgroup$
    – Zuhair Al-Johar
    Commented Oct 4, 2022 at 6:20

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