How do you prove that set of all 2-element sets does not exist basing on russell's paradox. Seems pretty obvious to me but no idea how to make a proper proof.

  • $\begingroup$ All two sets? If there were only two then life would be a lot simpler. $\endgroup$ – badjohn Oct 22 '19 at 16:21
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    $\begingroup$ The assertion stating this exists is not inconsistent as a single statement in the way that that asserting the existence of the Russell set is. In the set theory NFU, for example, there is such a set. So the question needs to be sharpened by situating it within a particular theory. In ZFC, for example, its existence can be disproved because of the axiom of union. $\endgroup$ – Malice Vidrine Oct 22 '19 at 20:21
  • $\begingroup$ What do you mean by "basing on"? $\endgroup$ – Andrés E. Caicedo Oct 24 '19 at 1:39

It seems as though you want to prove the non-existence of the set of all two-element sets as a corollary of Russell's paradox. But there's an important difference between the class of all pairs and the Russell class. Notice the theory consisting of the single sentence $$\exists y\forall x(x\in y\leftrightarrow x\notin x)$$ is inconsistent. But there are consistent set theories in which $\{z:\exists xy(z=\{x,y\})\}$ is actually a set (like $\mathsf{NFU})$. So you can't disprove the existence of such a set except with respect to a particular theory.

If you're thinking about something like Zermelo set theory, or one of its extensions, then you likely already know the argument that there can be no universal set; the separation scheme would let us show that the Russell class is a set, resulting in the usual contradiction.

To disprove the existence of a set of all two-element sets in Zermelo, we note that in the presence of the other axioms (particularly whichever axioms ensure that there's at least one thing, and also something else), Pairing implies that every set is a member of some two-element set. So suppose we have our set of all pairs; the axiom of Union says that if we have this set, we can form the set of all elements that are a member of some two-element set. And that's the set of all sets, something we already know leads to contradiction. So we can have no such set.


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