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:

$S = \O$ if and only if $S_i = \O$ for some $i \in I$

Proof

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 $i \in I$

$\Box$

Sufficient Condition

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

By the definition of the Cartesian product:definition 2:

$\ds \prod_{i \mathop \in I} S_i := \set {f: \paren {f: I \to \bigcup_{i \mathop \in I} S_i} \land \paren {\forall i \in I: \paren {\map f i \in S_i} } }$

For any $f: I \to \bigcup_{i \mathop \in I} S_i$ we have:

$\map f j \notin S_j$

Thus:

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

It follows that:

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

$\blacksquare$