# Cartesian Product is Empty iff Factor is Empty

## Theorem

$S \times T = \O \iff S = \O \lor T = \O$

Thus:

$S \times \O = \O = \O \times T$

### Family of Sets

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

 $\ds S \times T$ $\ne$ $\ds \O$ $\ds \leadstoandfrom \ \$ $\ds \exists \tuple {s, t}$ $\in$ $\ds S \times T$ Definition of Empty Set $\ds \leadstoandfrom \ \$ $\ds \exists s \in S$ $\land$ $\ds \exists t \in T$ Definition of Cartesian Product $\ds \leadstoandfrom \ \$ $\ds S \ne \O$ $\land$ $\ds T \ne \O$ Definition of Empty Set $\ds \leadstoandfrom \ \$ $\ds \neg \leftparen {S = \O}$ $\lor$ $\ds \rightparen {T = \O}$ De Morgan's Laws: Conjunction of Negations

So by the Rule of Transposition:

$S = \O \lor T = \O \iff S \times T = \O$

$\blacksquare$