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!
Best Answer
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.