# Can we have a set that is hereditarily equinumerous to itself, of any size?

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 $$x$$.

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

## 1 Answer

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).

• any nonempty cardinality means any cardinal larger than the empty set. Dec 8, 2020 at 20:35
• But how are you formulating that statement in the language of set theory? Dec 8, 2020 at 20:36
• OK, I'll add the formulation. thanks Dec 8, 2020 at 20:37
• Ah, I didn't realize you meant simultaneously such a set exists for each cardinality. Dec 8, 2020 at 20:59
• 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. Dec 8, 2020 at 21:07