Exercise 4, § 3

Let E be a set and let $f$ be a mapping of $\mathfrak{P}(\mathrm{E}) - \{\emptyset\}$ into E such that for each non-empty subset X of E we have $f(\mathrm{X}) \in \mathrm{X}$ ("choice function").

(a) Let $\mathfrak{b}$ be a cardinal and let A be the set of all $x \in \mathrm{E}$ such that $\mathrm{Card}\,(\overset{-1}{f}(x)) \leqslant \mathfrak{b}$. Show that, if $\mathfrak{a} = \mathrm{Card}\,(\mathrm{A})$, then $2^{\mathfrak{a}} \leqslant 1 + \mathfrak{a}\mathfrak{b}$ (note that if $\mathrm{Y} \subset \mathrm{A}$ and $\mathrm{Y} \neq \emptyset$, then $f(\mathrm{Y}) \in \mathrm{A}$).

(b) Let B be the set of all $x \in \mathrm{E}$ such that, for each non-empty subset X of $\overset{-1}{f}(x)$, $\mathrm{Card}\,(\mathrm{X}) \leqslant \mathfrak{b}$. Show that $\mathrm{Card}\,(\mathrm{B}) \leqslant \mathfrak{b}$.