Non-existence claims making use of the axiom of regularity

Question

We have that an intersection with the empty set is empty, $$\forall x.\, \varnothing \cap x = \varnothing$$

and so with the witness $\varnothing$ for $y$, we get

$$\forall x.\, \varnothing\in x \rightarrow \exists y.\, (y\in x \land y \cap x = \varnothing)$$

and so in particular

$$\forall x.\, \varnothing\in x \rightarrow \left(x \neq \varnothing \rightarrow \exists y.\, (y\in x \land y \cap x = \varnothing)\right)$$

So every set holding the empty set is regular in the sense given by the axiom of regularity.

Now if we have a theory allowing us to form what would naturally be denoted as

$$x=\{x, \varnothing\},$$

(e.g. via allowing for comprehensions over a predicate $z=\varnothing\ \lor z \in x$, in which however the definiendum $x$ would notably depend on $x$ itself, calling for the appropriate axioms)

then we got an infinite descending chain, $x\in x$.

As such, is the active non-existence claim I saw occasionally here on SE as well as in the third sentence of the axiom of regularity Wikipedia article wrong?

The axiom implies that no set is an element of itself (...)

Answer 1

No, the set $x=\{x,\varnothing\}$ will still violate the Axiom of Regularity. Consider the set $X=\{x\}$ with $x=\{x,\varnothing\}$, then $X$ is nonempty, yet its only element $x$ does not have the property that $x\cap X$ is empty, since $x\cap X=\{x\}$.