Why this works for proving ZF axioms system is free of Russell's paradox?

Question

In a book on mathematical logic, the author explains why ZF axioms avoid universal set like this:

We may also now show that no universal set exists. Suppose $u$ is a set containing all sets. By comprehension, we may form the set $\{x\in u\,:\,x\notin x\}$, which yields Russell's paradox, a contradiction. Thus, not every collection of sets is itself a set.

The author is using reductio ad absurdum. If a set of axiom $\Sigma$ plus a proposition $\alpha$ produce a paradox, then we know that $\alpha$ is inconsistent with $\Sigma$. In our case, $\Sigma$ is the ZF, while $\alpha$ is "universal set exists".

Here comes my confusion, if we don't assume the consistence of ZF, how do we know that the paradox comes from $\alpha$ rather than the dubious ZF? But we can't presume the consistence of ZF, because what we are doing is exactly to prove it free from Russell's paradox.

Answer 1

This doesn't show that ZF is free from Russel paradox, or consistency of ZF (it's imaginable, though extremely unlikely, that ZF is in fact inconsistent).

This shows that ZF proves that no universal set exists. Reductio ad absurdum works with all theories, even inconsistent: if by adding hypothesis $H$ to theory $T$ we can get inconsistency, then $T$ can prove $\neg H$. Of course it's not interesting if $T$ is itself inconsistent (because then it can prove anything), but it's still correct.

Answer 2

Russell's Paradox arose from an inconsistency in an early attempt to axiomatize set theory by G. Frege. In his system, it was possible to both prove and disprove:

$~~~~\exists x: \forall y: [y\in x \iff y\notin y]$

In ZFC theory, it seems to be impossible to derive this result. You can, of course, derive its negation in ZFC theory.