In my recent paper with Makoto Kikuchi,

we proved the following theorem.

Theorem. All countable models $\langle M,\in^M\rangle\models\text{ZFC}$, whether well-founded or not, have the same inclusion relation $\langle M,\subseteq^M\rangle$, up to isomorphism.

And the same is true for much weaker theories, such as KP and even finite set theory, provided that one excludes the $\omega$-standard models without any infinite sets and also the $\omega$-standard models with amorphous sets.

The proof proceeds by proving that for most of the models of set theory $\langle M,\in^M\rangle$, the corresponding inclusion relation $\langle M,\subseteq\rangle$ is an $\omega$-saturated model of what we have called set-theoretic mereology, which is the theory of an unbounded atomic relatively complemented distributed lattice. Those are the basic facts of $\subseteq$ in set theory and that is a complete, finitely axiomatizable, decidable theory. Since the theory is complete, and the resulting models are $\omega$-saturated, it follows by the back-and-forth method that all countable saturated models are isomorphic.

But the possibility of amorphous sets throws a wrench in the works. For most models of set theory, we prove that the inclusion relation is $\omega$-saturated; but when the model is $\omega$-standard and has an amorphous set, then it is not $\omega$-saturated, and so this is an irrirating obstacle to the general phenomenon. What we want to know is whether the ZF models provide just one or many different isomorphism types for the inclusion relation.

Question. Do all countable $\omega$-standard models of ZF with an amorphous set have isomorphic inclusion relations?

Kikuchi and I state in the paper that we believe that the answer to this question will come from an understanding of the Tarski/Ersov invariants combined with a knowledge of the models of ZF with amorphous sets.

  • $\begingroup$ Joel, what happens when you add a proper class of "pairwise-generic" amorphous sets compared to adding just a single one? $\endgroup$ – Asaf Karagila Nov 20 '17 at 10:52
  • $\begingroup$ That is a good idea, and I don't know what happens to the subset relation when you do this. It is a good candidate model for another isomorphism type. $\endgroup$ – Joel David Hamkins Nov 20 '17 at 12:13
  • $\begingroup$ Well, what happens when you have the one amorphous set? $\endgroup$ – Asaf Karagila Nov 21 '17 at 9:40
  • $\begingroup$ If you add an amorphous set to an $\omega$-standard model of ZFC, then the subset relation is definitely not isomorphic to the original model, since the property that all subsets are either finite or co-finite is revealed by the subset relation. $\endgroup$ – Joel David Hamkins Nov 21 '17 at 11:12
  • $\begingroup$ Right, right. But start with a standard model (just to omit this $\omega$ from the notation everywhere). And add a single amorphous set to it. That gives you an "almost minimal" infinite cardinal, which I suspect is what you mean by this fact. That you get an $(\omega,\omega)$-gap in the cardinals with the lower chain being $\omega$ itself. But, I mean, what else do you get there which "bothers" the containment structure? $\endgroup$ – Asaf Karagila Nov 21 '17 at 11:20

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