Proof of Cantor's theorem by the diagonal argument

Proof

We use the diagonal argument.

1. Let $A$ be a set and let $f \colon A \to \mathcal{P}(A)$ be a function to the power set.
2. Let $D = \{a \in A : a \notin f(a)\}$. Then $D \subseteq A$, so $D \in \mathcal{P}(A)$.
3. Suppose that $D = f(d)$ for some $d \in A$.
4. If $d \in D$, then $d \notin f(d)$ by the definition of $D$. Since $f(d) = D$, this says $d \notin D$, a contradiction.
5. If $d \notin D$, then $d \notin f(d)$, since $f(d) = D$. By the definition of $D$, this means $d \in D$, a contradiction.
6. So no $d \in A$ has $f(d) = D$. Hence $f$ is not surjective.
7. The map $s \colon A \to \mathcal{P}(A)$, $s(a) = \{a\}$, is injective, since $\{a\} = \{a'\}$ implies $a = a'$. So $|A| \le |\mathcal{P}(A)|$ in the order of cardinals.
8. Suppose that $|A| = |\mathcal{P}(A)|$. By the definition of cardinality, $A$ and $\mathcal{P}(A)$ are equinumerous, so there is a bijection $A \to \mathcal{P}(A)$.
9. A bijection is surjective, which contradicts step 6. So $|A| \neq |\mathcal{P}(A)|$.
10. By steps 7 and 9, $|A| < |\mathcal{P}(A)|$.