Cantor's theorem

Statement

Let $A$ be a set and let $f \colon A \to \mathcal{P}(A)$ be a function to the power set of $A$. Then $f$ is not surjective. So $|A| < |\mathcal{P}(A)|$.

Remarks

Proof: let $D = \{a \in A \mid a \notin f(a)\}$. If $D = f(d)$ for some $d \in A$, then $d \in D$ if and only if $d \notin D$, a contradiction. The map $a \mapsto \{a\}$ is an injection, so $|A| \le |\mathcal{P}(A)|$, and the inequality is strict. It follows that there is no largest cardinal. The proof is a form of the diagonal argument, and it is close to Russell's paradox.

Explanations

L1 human

For each set, its set of subsets is strictly larger. For a set with $3$ elements, this is $3 < 8$. For infinite sets, it gives an endless tower of larger and larger infinities.