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Is it consistent to add to ZFC-Reg. the existence of a nonempty set $\mathcal H_\mathcal H$ that is hereditarily equinumerous to itself?

If that is consistent, then is it consistent that we can have $\mathcal H_\mathcal H$ being of any nonempty cardinality? Formally that is:

$\forall x (x \neq \emptyset \implies \exists y \forall z [z \in TC(\{y\}) \Rightarrow z \sim x])$

Where $``TC(x)"$ stands for "transitive closure of $x$" defined in the usual manner, and $`` z \sim x"$ stands for existence of a bijection between $z$ and $x$.

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  • $\begingroup$ A set satisfying $x=\{x\}$, also called a quine atom, shows that the first question has a positive answer. $\endgroup$ Dec 8, 2020 at 15:22
  • $\begingroup$ @tomasz, the second condition of mine is a flagrant violation of Aczel's anti-foundation axiom! $\endgroup$
    – Zuhair
    Dec 8, 2020 at 20:32

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Yes, and the proof is nearly identical to the argument in this answer: just start with an $M_0$ that consists of a set which is hereditarily of some size, and then formally construct a cumulative hierarchy on it. Or, if you want to simultaneously get such hereditarily equinumerous sets of all cardinalities, take $M_0$ to be a class-sized structure which consists of such sets for all cardinalities.

More generally, that construction shows that given any structure $M_0$ in the language of set theory which satisfies extensionality, there is a class model of ZFC without regularity which contains a transitive set that is isomorphic to $M_0$. This works even if $M_0$ is a proper class (take a direct limit of the construction over all set-sized substructures of $M_0$), as long as each element of $M_0$ has only a set of elements (this assumption is needed for Power Set to hold).

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  • $\begingroup$ any nonempty cardinality means any cardinal larger than the empty set. $\endgroup$
    – Zuhair
    Dec 8, 2020 at 20:35
  • $\begingroup$ But how are you formulating that statement in the language of set theory? $\endgroup$ Dec 8, 2020 at 20:36
  • $\begingroup$ OK, I'll add the formulation. thanks $\endgroup$
    – Zuhair
    Dec 8, 2020 at 20:37
  • $\begingroup$ Ah, I didn't realize you meant simultaneously such a set exists for each cardinality. $\endgroup$ Dec 8, 2020 at 20:59
  • $\begingroup$ Yes, I want it for all cardinalities. But regarding your answer, you simply took $M_0$ to be a set that is hereditarily of size $\kappa$, but my question is about the very existence of such sets. What is the proof that for example for $\aleph_1$ there can exist a set that is hereditarily of size $\aleph_1$ in the first place? I mean I know that if such a set can exist then one can iterate powers over it thereby building a hierarchy over it that would satisfy ZFC and its existence, but what's the proof that it can exist in the first place. $\endgroup$
    – Zuhair
    Dec 8, 2020 at 21:07

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