Meaning of the Axiom of regularity (foundation)

Question

Do I understand correctly that without the Axiom of regularity (aka Axiom of foundation): $$\forall x\left(x=\varnothing\ {\Large\lor}\ \exists y\left(y \in x\ {\Large\land}\ y \cap x = \varnothing\right)\right)$$ there would be a possibility that there are different sets $p\ne q$ such that $p=\{p\}$ and $q=\{q\}$, seemingly having identical structure $\{\{\{...\}\}\}$ but still not ruled out by the Axiom of extensionality, because they have different elements $p\ne q$?

Answer 1

Yes. This is consistent. Such sets ($x=\{x\}$) are sometimes called Quine Atoms. And indeed if $x,y$ are two different Quine atoms, then $x\neq y$ because $x\notin y$ and $y\in y$ (and vice versa) witness that.

Back in the 1960s you can find works related to the axiom of choice which used those instead of urelements (atoms which are not sets) for building permutation models.

Namely we start with a set of a particular size, whose elements are all different Quine atoms, and we construct a von Neumann-like hierarchy from that set (reiterating power sets), so the model we get is a model of $\sf ZFC-Fnd$, and with the usual urelements constructions we can construct counterexamples to the axiom of choice.

Notable users of this technique are Specker and Lauchli.