Set of all Sets
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The validity of the material on this page is questionable.
This is circular: We need to demonstrate the contradiction and then demonstrate why ZF (specifically the AoF) prevents it happening.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by resolving the issues.
Theorem
Forming the set $\SS$ of all sets leads to a contradiction.
Proof
Let $\SS$ be the set of all sets.
Then $\SS$ must be an element of itself, in symbols, $\SS \owns \SS$.
Thus we have an infinite descending sequence of membership:
- $\SS \owns \SS \owns \SS \owns \cdots$
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.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by resolving the issues.
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