Exercise 10.5

Proof. Suppose that $A$ is a collection of nonempty sets. Then, by the well-ordering theorem, there is a well-ordering $<$ of $\bigcup <!--\; FUNCTION\; APPLICATION\; -->A$. We construct a choice function $c:A\to \bigcup <!--\; FUNCTION\; APPLICATION\; -->A$. Consider any set $A\in A$. Since clearly, $A$ is then a nonempty subset of $\bigcup <!--\; FUNCTION\; APPLICATION\; -->A$, it follows that it has a unique smallest element $a$ according to $<$ since $\bigcup <!--\; FUNCTION\; APPLICATION\; -->A$ is well-ordered by $<$. So simply set $c(A)=a$ so that clearly then $c(A)=a\in A$. This shows that $c$ is in fact a choice function on $A$. □