Axioms of ZFC in set theory

Question

While studying set theory, I confronted an exercise that encouraged me to prove that there does not exist a set $X$ such that $P(X) \subset X$. On the process of proving this exercise, I thought it is true that $X \in X$ cannot be occurred if $X$ is a set. The following is my proof.

pf) Suppose that there exists a set $X$ such that $X \in X$. Then $V=\{X : X \in X\}$ is a set. By the separation schema, there exists a set $S=\{x \in V : x \notin x\}$ becomes a set.

My question is this: Is $S$ the Russell's paradox? If not, can S become the Russell's paradox by changing the formula?

Answer 1

Your suggested proof is not going to work.

Firstly, since $X\notin X$ is true for all $X$, then $\{X\mid X\in X\}$ is empty and in particular it is a set. Since the empty class is a set.

Secondly, you are defining $S$ as the subset of $\{X\mid X\in X\}$ of those elements satisfying $X\notin X$. So an element of $S$ is both an element of itself and not. Since exactly one of these can be true, even if some $X$ was an element of itself, $S$ is always empty.

Finally, you're not really proving that $\mathcal P(X)$ is not a subet of $X$. You're going a step further. But that's actually a point of failure here. In $\sf ZFC$, which is the standard set theory we use, we can prove that $X\notin X$ is true for all $X$ by using the Axiom of Regularity (also known as Axiom of Foundation). But in a naive setting, we don't usually assume the Axiom of Regularity.

And indeed, it is consistent that all the axioms of $\sf ZF$ except the Axiom of Regularity hold, but there is a set that contains itself. In fact, it is consistent that the collection $\{X\mid X\in X\}$ is not even a set! So, does that mean that in those cases you could have $\mathcal P(X)\subseteq X$? Not quite. But it gives you a hint that using $X\notin X$ is probably not the correct approach here.

The idea is to relativise Russell's paradox to $X$. Namely, consider $B=\{x\in X\mid x\notin x\}$ and then prove that $B\notin X$. But since $B\subseteq X$ by definition, that shows that $\mathcal P(X)\nsubseteq X$ after all. (Note that if we assume $\sf ZFC$, then indeed $B=X$, and we get $X\notin X$.)

Let me add that some set theories restrict the formulas that define subsets, rather than requiring that a collection of sets is a set if it was "carved out a pre-existing set". For example, Quine's New Foundations; Positive set theory; and others. In those set theories, $x\notin x$ is not a valid formula for defining subsets, so $B$ is actually not a set.

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Further reading:

1. Why is “the set of all sets” a paradox, in Layman's terms?
2. If $S=\{x:x\in x\}$, is $S\in S$ knowable? (Naive set theory)
3. New Foundations
4. Positive set theory