# Set of all Sets

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## 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.

$\blacksquare$