product of non-empty set of non-empty sets is non-empty

In this entry, we show that the statement:

(*) the non-empty generalized cartesian product of non-empty sets is non-empty

is equivalent to the axiom of choice (AC).

Proposition 1.

AC implies (*).

Proof.

Suppose $C=\{A_j\mid j\in J\}$ is a set of non-empty sets, with $J\ne \mathrm{\varnothing }$. We want to show that

$$B:=\prod _{j\in J}{A}_j$$

is non-empty. Let $A=\bigcup C$. Then, by AC, there is a function $f:C\to A$ such that $f(X)\in X$ for every $X\in C$. Define $g:J\to A$ by $g(j):=f(A_j)$. Then $g\in B$ as a result, $B$ is non-empty.∎

Remark. The statement that if $J\ne \mathrm{\varnothing }$, then $B\ne \mathrm{\varnothing }$ implies $A_j\ne \mathrm{\varnothing }$ does not require AC: if $B$ is non-empty, then there is a function $g:J\to A$, and, as $J\ne \mathrm{\varnothing }$, $g\ne \mathrm{\varnothing }$, which means $A\ne \mathrm{\varnothing }$, or that $A_j\ne \mathrm{\varnothing }$ for some $j\in J$.

Proposition 2.

(*) implies AC.

Proof.

Suppose $C$ is a set of non-empty sets. If $C$ itself is empty, then the choice function is the empty set. So suppose that $C$ is non-empty. We want to find a (choice) function $f:C\to \bigcup C$, such that $f(x)\in x$ for every $x\in C$. Index elements of $C$ by $C$ itself: $A_x:=x$ for each $x\in C$. So $A_x\ne \mathrm{\varnothing }$ by assumption. Hence, by (*), the (non-empty) cartesian product $B$ of the $A_x$ is non-empty. But an element of $B$ is just a function $f$ whose domain is $C$ and whose codomain is the union of the $A_x$, or $\bigcup C$, such that $f(A_x)\in A_x$, which is precisely $f(x)\in x$.∎