Cartesian Product is Empty iff Factor is Empty/Family of Sets

Theorem
Let $I$ be an indexing set.

Let $\family {S_i}_{i \mathop \in I}$ be a family of sets indexed by $I$.

Let $\ds S = \prod_{i \mathop \in I} S_i$ be the Cartesian product of $\family {S_i}_{i \mathop \in I}$.

Then:
 * $S = \O$  $S_i = \O$ for some $i \in I$

Necessary Condition
By the axiom of choice, the contrapositive statement holds:
 * if $S_i \ne \O$ for all $i \in I$ then $S \ne \O$

By the Rule of Transposition, the converse holds:
 * if $S = \O$ then $S_i = \O$ for some $i \in I$

Sufficient Condition
Let $S_j = \O$ for some $j \in I$.

By the definition of the Cartesian product:
 * $\ds \prod_{i \mathop \in I} S_i := \set {f: \paren {f: I \to \bigcup_{i \mathop \in I} S_i} \land \paren {\forall i \in I: \paren {\map f i \in S_i} } }$

For any $\ds f: I \to \bigcup_{i \mathop \in I} S_i$ we have:
 * $\map f j \notin S_j$

Thus:
 * $f \notin \ds \prod_{i \mathop \in I} S_i$

It follows that:


 * $\ds \prod_{i \mathop \in I} S_i = \O$