Isn't this very large set excluded by the axiom of regularity (foundation)? The axiom of regularity is
$$
\forall x:(x \neq \emptyset \to \exists z \in x:z \cap x = \emptyset)
$$
and the negation is
$$
\exists x:(x \neq \emptyset \wedge \forall z \in x:z \cap x \neq \emptyset)
$$
A 'set' $x$ constructed as follows
$$
x := \{z_1,z_2,z_3,\dots\}; \qquad
\begin{cases}
\emptyset \notin x, \\
z_1 = \{z_2,\dots\} = \{\{z_3,\dots\},\dots\} = \dots, \\
z_2 = \{z_3,\dots\} = \{\{z_4,\dots\},\dots\} = \dots, \\
\dots
\end{cases}
$$
should satisfy the negated statement of the axiom, since
$$
x \cap z_n = \{z_{n+1}\} \neq \emptyset, \qquad x \ni z_1 \ni z_2 \ni z_3 , \dots
$$
Now this is a large set that the axiom of reguarity rules out but the set
$$
x' = x \cup \{\emptyset\} = \{\emptyset,z_1,z_2,z_3,\dots\}
$$
is still larger with infinite descending $\ni$-series and yet not ruled out, since
$$
\exists z \in x':z \cap x' = \emptyset, \qquad \text{namely $z = \emptyset$}
$$
Is there a mistake in this reasoning ?
 A: The thing is that your theory is not "just" the axiom of regularity, as witnessed by your attempt to construct sets, which generally require additional axioms to justify why the construction results in a set.
If we look at $x\cup\{\varnothing\}$, it's true that it has an element which is disjoint from itself, but if you allow even the most basic Separation axioms, you can construct $x$ out of $x\cup\{\varnothing\}$ by simply look at $\{y\in x\cup\{\varnothing\}\mid\exists z(z\in y)\}$.
To summarise, it's true that $x\cup\{\varnothing\}$ doesn't immediately contradict the axiom of regularity. But since set theory is not just the axiom of regularity, but includes a lot of other axioms, it's easy to see that it does lead to a contradiction, it simply takes an ever so slightly longer proof.
A: There is no mistake in your reasoning! The axiom of regularity alone cannot rule out your larger set although it does contain an infinite descending membership chain. Everything depends on whether your theory prove the existence of a transitive closure for every set. If it doesn't (like Zermelo), then adding regularity as a single sentence like the one you wrote won't be sufficient to rule out infinite descending membership chains. But for the set you wrote, one can easily negate it in Zermelo, since by separation I can retrieve the set $x$ from $x'$ and this would negate Regularity. In order to get rid of having infinite descending membership chains in a theory like Zermelo (or any theory that doesn't prove transitive closures), you need to add Regularity as a schema! $$ \forall \vec{p} \, (\exists x.\phi(x) \to \exists z:\phi(z) \land \forall y (\phi(y) \to y \not \in z))$$; for every formula $\phi$.
The transitive closure of a set $x$ is a set that contains all elements of $x$, elements of elements of $x$, elements of elements of elements of $x$, etc... Formally that is:
$TC(x)=\{ y: \forall \ trs \ t( x \subseteq t \to y \in t) \}$
Where: $trs(t) \iff \forall z \in t (z \subseteq t)$
In nutshell: the axiom of Regularity alone by itself cannot negate existence of $x'$, but together with the other axioms of set theory, especially Separation $x'$ cannot exist. And if you want to get rid of infinite descending membership, then you need to add Regularity as a schema if your theory doesn't prove transitive closures, if it does, you can add it as a single axiom.
