Theorem 2 (Cantor)

— For each cardinal $\mathfrak{a}$, we have $2^\mathfrak{a} > \mathfrak{a}$.

We have $\mathrm{Card}(\mathfrak{P}(\mathfrak{a})) = 2^\mathfrak{a}$ (no. 5, Proposition 12). The mapping $x \to \{x\}$ $(x \in \mathfrak{a})$ is an injection of $\mathfrak{a}$ into $\mathfrak{P}(\mathfrak{a})$, whence $\mathfrak{a} \leqslant 2^\mathfrak{a}$. Hence it is enough to show that $\mathfrak{a} \neq 2^\mathfrak{a}$, i.e., that for every mapping $f$ of $\mathfrak{a}$ into $\mathfrak{P}(\mathfrak{a})$, the image $f(\mathfrak{a})$ is distinct from $\mathfrak{P}(\mathfrak{a})$. Let X be the set of all $x \in \mathfrak{a}$ such that $x \notin f(x)$. If $x \in X$, we have $x \notin f(x)$, whence $f(x) \neq X$; if $x \in \mathfrak{a} - X$, we have $x \in f(x)$ and $x \notin X$, whence $f(x) = X$. This shows that $X \notin f(\mathfrak{a})$ and proves the theorem.