Isn’t this very large set excluded by the axiom of regularity (foundation)

Question

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 ?

Answer 1

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.