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I was reading The Foundations of Mathematics by Ian Stewart which discussed the existence of a universal set $\Omega$. It included the following:

If we select from the putative set $\Omega $ the subset comprising everything that is a set but does not belong to itself, we get:

$S = \{A \in \Omega \mid A \not \in A \}$

Now ask the key question: is $S \in S$?

- If $S \in S$, then, according to the defining predicate, $S \not \in S$.

- If $S \not \in S$, then $S$ satisfies the defining predicate, so $S \in S$.

Our flight of fancy in assuming the existence of a universe has led to a paradox. Therefore there cannot be a universal set.

How does this paradox imply that universal set $\Omega$ cannot exist?

Mageentta
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  • Because assuming its existence , we can derive the contradiction , so the existence of such a set makes the set theory inconsistent. The only way out is to forbid such a set (and not only that, in fact set theory became much weaker because of this surprising , not at all expected contradiction) – Peter Jan 14 '23 at 13:29
  • The duplicate is perhaps slightly different, but my answer there touches on exactly this point. – Asaf Karagila Jan 14 '23 at 13:52

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This comes from the axiom of specification. The existence of the set $\Omega$ allows you to construct $S = \lbrace A \in \Omega \mid A \notin A \rbrace$. Whithout $\Omega$, you cannot define a set using a proprety that's about sets (such as $\notin$ itself). The only assumption you made here is the existence of $\Omega$. Hence, it cannot exist.

Zag
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    I hope I understood your correctly. Axiom of specification works for any given set and any predicate (including the one describe above for the subset S). However, if the given set is a universal set, using that predicate leads to a paradox, so there cannot be a universal set. Correct me if I am wrong. Thanks. – Mageentta Jan 14 '23 at 21:49