Avoiding Russell's paradox with ZF axioms

Question

I recently learned about the axioms of Zermelo–Fraenkel and I am trying to understand how those axioms avoid paradoxes such Russell's paradox. Recall that the "set" from Russell's paradox is:

$$R=\{x:x\notin x\}$$ Which led to the obvious contradiction $R\in R\iff R\notin R$.

I think that I understand it, my interpretation is:

Under $\mathsf{ZF}$ axiomatic system, the only way we can construct such a 'set' is from existing sets. So, using the axiom of separation we must have some set $B$ such that:

$$R=\{x\in B:x\notin x\}$$

Moreover, by the axiom of regularity it's not hard to prove that every set $A$ is not an element of itself, That is, $A\notin A$. Therefore, the formula $\varphi(x):=[x\notin x]$ is always true. Therefore:

$$R=\{x\in B:\overbrace{x\notin x}^{\text{always true}}\}=B$$

So no new set was created and no contradiction has been occurred.

Is my interpretation valid? Is that how $\mathsf{ZF}$ axioms 'avoid' such axioms?

Feedbacks will be appreciated!

Answer 1

Not quite; you must still consider why the original contradiction $R \in R \iff R \not \in R$ no longer applies. That is because once you limit to $B$, you can no longer conclude $R \in R \iff R \not \in R$ anymore; you can only conclude that under the additional assumption that $R \in B$. So the conclusion is simply that $R \not \in B$, and there is no contradiction.

Note that this holds even if you remove the axiom of regularity.

Answer 2

Defining $R:=\{x\in B|x\notin x\}$, $R\in R\iff R\in B\land R\notin R$, which reduces by the tatutology $(p\iff q\land\neg p)\equiv(\neg p\land\neg q)$ to $R\notin R\land R\notin B$, rather than a contradiction. Regularity tells us further that $R=B$.

One more point: in theories such as ZF, we can do more than just prove certain sets exist. We can also prove a certain set doesn't exist by deriving a contradiction, including of course a hypothetical set whose elements are precisely those $x$ with $x\notin x$, by the usual argument. So we don't only fail to prove such a set does exist; we can also prove it doesn't. (A related concept is to say that the class of all sets $x$ with $x\notin x$ is not a set, but rather a proper class.)