Russell's paradox with bounded comprehension

Question

Consider the set $S = \{A, \varnothing\}$ and define $A = \{x \in S|x \not\in x\}$; this is the same as Russell's paradox except with bounded comprehension, ie $A\in A\iff A\not\in A$.

I think the problem lies in the fact that the composition of $S$ depends on $A$, but the composition of $A$ depends on $S$. My question is: what about the Axiom of Separation (or the other axioms) precludes this construction?

NB: I haven't studied set theory except for reading a little bit in some introductory books in my own time.

Answer 1

Your definition is circular. You define $S$ using $A$, and you define $A$ using $S$.

And that's what precludes the definition in this case.

Answer 2

User wrote:

My question is: what about the Axiom of Separation (or the other axioms) precludes this construction?

It has nothing to do with any axioms of set theory (or circularity). This problem can be resolved with a purely logical argument.

Theorem

There do not exist sets $A$ and $S$ such that $A = \{x \in S|x \not\in x\}$ and $S = \{A, \varnothing\}$.

Proof

1. Suppose to the contrary that we do have sets $A$ and $S$ such that $\forall x:[x\in A \iff x \in S \land x\notin x]$ and $S = \{A, \varnothing\}$.

2. From (1), $A\in A \iff A\in S \land A\notin A$

3. From (1), $A\in S$

4. From (2) and (3), we can obtain the contradiction $A\in A$ and $A\notin A$

5. Therefore, $\neg \exists A,S: [\forall x:[x\in A \iff x \in S \land x\notin x]\land S = \{A, \varnothing\}]$

Note

Similarly, we can prove that there do not exist sets $A$ and $S$ such that $A = \{x \in S|x \not\in x\}$ and $A\in S$.