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.

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    $\begingroup$ We can only prove that Russel's paradox does not occur. The consistency of ZFC remains an open question. $\endgroup$
    – Peter
    Commented Dec 28, 2021 at 13:02

2 Answers 2


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.

  • $\begingroup$ Why do you think is it extremely likely for ZF to be consistent? Also, do you think it is likely or unlikely for ZF to be arithmetically sound? $\endgroup$
    – user21820
    Commented Jan 7, 2022 at 14:16
  • $\begingroup$ There are a lot of open problems that can be formulated in ZF, and they seem hard. As proving something in consistent theory is harder than in inconsistent one, it is evidence in favor of consistency of ZF. I think it's likely that ZF is arithmetically sound, but I don't actually trust my intuition about it. $\endgroup$
    – mihaild
    Commented Jan 7, 2022 at 15:12
  • $\begingroup$ I agree that proving things in a consistent theory is harder, but most people are unable to actually think beyond a certain strength level, roughly around that of bounded Replacement. Though I do think ZF is consistent. Would you like to have a more in-depth discussion in chat? I think it might be a very interesting one. $\endgroup$
    – user21820
    Commented Jan 7, 2022 at 15:32

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.


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