Theorem: there does not exist a set that contains all sets.

By contradiction, suppose that such a set exists and call it \(E\). Then it would include \(\mathcal{P}(E)\) (all subsets of \(E\)) i.e. there would be an injection \(g\) from \(\mathcal{P}(E)\) to \(E\). So there would also be a surjection \(f\) from \(E\) to \(\mathcal{P}(E)\) because every element \(x\) of \(\mathcal{P}(E)\) could have as (unique) antecedent \(g(x)\).

Define \(D = \{x \in E \mid x \notin f(x)\} \in \mathcal{P}(E)\). Since \(f\) is surjective, \(\exists y \in E; f(y) = D\). Now, we have two cases:

  • If \(y \in D\), then \(y \notin f(y) = D\)
  • If \(y \notin D\), then \(y \in f(y) = D\)

We have a contradiction in both cases. Therefore, the assumption that there exists a set containing all sets is false. Thanks to Cantor for this proof! :smile: