Proposition 12

Let X be a set and let $\mathfrak{a}$ be its cardinal. Then the cardinal of the set $\mathfrak{P}(\mathrm{X})$ of all subsets of X is $2^{\mathfrak{a}}$.

Let $\alpha$ and $\beta$ be the elements of the cardinal 2. For each subset Y of X let $f_{\mathrm{Y}}$ be the mapping of X into 2 defined by $f_{\mathrm{Y}}(x) = \alpha$ if $x \in \mathrm{Y}$ and $f_{\mathrm{Y}}(x) = \beta$ if $x \in \mathrm{X} - \mathrm{Y}$. Let $u$ be the mapping $\mathrm{Y} \to f_{\mathrm{Y}}$ of $\mathfrak{P}(\mathrm{X})$ into $2^{\mathrm{X}}$. Conversely, with each mapping $g$ of X into 2 let us associate the subset $\overset{-1}{g}(\alpha)$ of X, and let $v$ be the mapping $g \to \overset{-1}{g}(\alpha)$ of $2^{\mathrm{X}}$ into $\mathfrak{P}(\mathrm{X})$. It is clear that $u \circ v$ and $v \circ u$ are the identity mappings of $2^{\mathrm{X}}$ and $\mathfrak{P}(\mathrm{X})$; hence $u$ and $v$ are bijections (Chapter II, § 3, no. 8, Proposition 8, Corollary) and therefore Card $(\mathfrak{P}(\mathrm{X})) = 2^{\mathfrak{a}}$.