Set of all Sets
From ProofWiki
Theorem
Forming the set $\mathcal S$ of all sets leads to a contradiction.
Proof
Let $\mathcal S$ be the set of all sets.
Then $\mathcal S$ must be an element of itself, in symbols, $\mathcal S \owns \mathcal S$.
Thus we have an infinite descending sequence of membership:
- $\mathcal S \owns \mathcal S \owns \mathcal S \owns \cdots$
But by the axiom of foundation, no such sequence exists, a contradiction.
$\blacksquare$
This is circular: We need to demonstrate the contradiction and then demonstrate why ZF (specifically the AoF) prevents it happening.
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