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?

  • $\begingroup$ 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

1 Answer 1


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 $M$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.