Exercise 19.9

Proof. For the following denote the collection ${\left\{A_{\alpha }\right\}}_{\alpha \in J}$ by $A$.

$\text{(⇒)}$ First suppose that the choice axiom is true. Then by Lemma 9.2 there exists a choice function

where $c(A)\in A$ for each $A\in A$, noting that this is true since $A$ is a collection of nonempty sets. Then consider, set $x_{\alpha }=c(A_{\alpha })$ for each $\alpha \in J$ so that $x_{\alpha }=c(A_{\alpha })\in A_{\alpha }$. Therefore clearly $x={(x_{\alpha })}_{\alpha \in J}\in \prod <!--\; FUNCTION\; APPLICATION\; -->A_{\alpha }$ so that $\prod <!--\; FUNCTION\; APPLICATION\; -->A_{\alpha }$ is not empty.

$\text{(⇐)}$ Now suppose that ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{\alpha \in J}{A}_{\alpha }$ is nonempty for any indexed family ${\left\{A_{\alpha }\right\}}_{\alpha \in J}$ of nonempty sets when $J\ne \varnothing$. Let $A$ be a collection of disjoint nonempty sets where $A\ne \varnothing$. Then the ${\left\{A\right\}}_{A\in A}$ is a nonempty family of nonempty sets. Hence ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{A\in A}A$ is nonempty so that there is an $x={(x_A)}_{A\in A}\in {\prod <!--\; FUNCTION\; APPLICATION\; -->}_{A\in A}A$, and thus $x_A\in A$ for every $A\in A$. Now let $C={\left\{x_A\right\}}_{A\in A}$ so that clearly $C\subset \bigcup <!--\; FUNCTION\; APPLICATION\; -->A$. Consider any $A\in A$ so that $x_A\in C$ and $x_A\in A$, and hence $x_A\in C\cap A$. Suppose that $y\in C\cap A$ so that $y\in C$ and hence there is a $B\in A$ where $y=x_B$. We also have that $x_B=y\in A$. If $B\ne A$ then $x_B\in B$ and $x_B\in A$, which is not possible since $B$ and $A$ are disjoint as they are distinct elements of $A$. So it must be that $B=A$ and hence $y=x_B=x_A$. Since $y$ was arbitrary, this shows that $C\cap A$ has only a single element $x_A$. This suffices to show the choice axiom. □