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

AC implies (*).

###### Proof.

Suppose $C=\{A_{j}\mid j\in J\}$ is a set of non-empty sets, with $J\neq\varnothing$. 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 $J\neq\varnothing$, then $B\neq\varnothing$ implies $A_{j}\neq\varnothing$ does not require AC: if $B$ is non-empty, then there is a function $g:J\to A$, and, as $J\neq\varnothing$, $g\neq\varnothing$, which means $A\neq\varnothing$, or that $A_{j}\neq\varnothing$ for some $j\in J$.

(*) 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 $A_{x}\neq\varnothing$ 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 $f(x)\in x$. ∎

Title product of non-empty set of non-empty sets is non-empty ProductOfNonemptySetOfNonemptySetsIsNonempty 2013-03-22 18:44:28 2013-03-22 18:44:28 CWoo (3771) CWoo (3771) 7 CWoo (3771) Derivation msc 03E20