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Is the following inconsistent:

By "intersectional" set I mean a set having the intersectional set of every nonempty subset of it, being an element of it.

$\forall S \subset M: S\neq \varnothing \to \bigcap S \in M$

Where: $\bigcap S = \{x \mid \forall s \in S \, (x \in s) \}$

Sets $(\varnothing, \{\varnothing\}, V_\omega, H_x, \operatorname {Fin}(\omega.2 \setminus \omega))$ are examples; where $H_x$ is the set of all sets hereditarily strictly subnumerous to $x$; and $\operatorname {Fin}(x)$ is the set of all finite subsets of $x$.

Let the ambient theory of models be $\sf ZF$-$\sf Reg.$. Can we have a model $M$ of say Mac Lane set theory (with or without Regularity, and without Choice) that admits a non-trivial external automorphism and at the same time have $M$ be an intersectional set?

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  • $\begingroup$ Do you intend that the membership relation of $M$ is the ambient membership relation $\in$? $\endgroup$ Commented May 10 at 11:25
  • $\begingroup$ Also, I guess you intend that the automorphism is nontrivial. $\endgroup$ Commented May 10 at 11:25
  • $\begingroup$ @JoelDavidHamkins, Yes! For both, but the ambient theory of models is ZF-Reg. Also for Mac Lane set theory I take the version that doesn't have Regularity among its axioms. I've edited. Thanks! $\endgroup$ Commented May 10 at 11:29
  • $\begingroup$ Ah, sorry, I missed the Reg part. Why use such a weird ambient theory? With reg, no standard model has any nontrivial automorphism at all, and intersectionality is irrelevant. $\endgroup$ Commented May 10 at 11:49
  • $\begingroup$ @JoelDavidHamkins, I know! That's why you cannot have the ambient theory be ZF, since I need $\in$ to be the one of the ambient theory. I also made the typo of writing it first as $\sf ZF$, but it should be $\sf ZF-Reg$. $\endgroup$ Commented May 10 at 12:22

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One can easily make a model of ZF-Reg with numerous Quine atoms. Simply begin with a model of ZFCU, with numerous urelements, and then turn the urlements into Quine atoms, which are singleton sets $a=\{a\}$, and the result is a model of ZF-Reg. The atoms can be permuted, and these permutations extend to automorphisms of the whole universe.

But meanwhile, there can be supertransitive models of MacLane set theory or even ZFC-Reg inside the original model, with multiple Quine atoms. Such a model will be intersectional, but admit automorphisms.

For example, consider $M=V_{\omega_1}[A]$ in the original universe, with a set $A$ of urelements. In the new model, these turn into Quine atoms, and this model $M$ is supertransitive, hence intersectional, and a model of MacLane set theory (without Reg) and much more. But permutations of $A$ extend to automorphisms of $M$.

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  • $\begingroup$ I edited to explain. I removed the Boffa angle, which is unimportant. $\endgroup$ Commented May 10 at 13:16
  • $\begingroup$ One important issue, can one of these automorphisms be rank shifting? That is, it can shift some non-standard ordinal $\alpha$ say inwardly? That is, we have $j(\alpha)< \alpha$, for an automorphism $j$? $\endgroup$ Commented May 10 at 13:21
  • $\begingroup$ These models have only standard ordinals, which are fixed by every automorphism. Without Reg, in the general case, there is no satisfactory theory of "rank". That is why I find your ambient theory unsatisfactory as a foundational theory of sets. $\endgroup$ Commented May 10 at 13:26
  • $\begingroup$ I see. I wanted the model to be intersectional, admit external automorphism but I need that to be rank shifting, The rank is seen from inside the model as so, but from outside it is not so, its index is in reality a non-standard ordinal, i.e. a transitive set of transitive sets that has an infinitely descending membership subset. $\endgroup$ Commented May 10 at 13:41
  • $\begingroup$ I mean suppose the inside model satisfy a theory with regularity. So, it has a rank notion inside it, but the model is non-well founded externally, what I need is for this model to admit rank shifting automorphism and be intersectional at the same time. Can this be cooked inside Boffa set theory? $\endgroup$ Commented May 10 at 13:58

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