# Russell's paradox but set with sets of 2-element sets. [closed]

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.

• All two sets? If there were only two then life would be a lot simpler. – badjohn Oct 22 '19 at 16:21
• 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. – Malice Vidrine Oct 22 '19 at 20:21
• What do you mean by "basing on"? – 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.