# Condition for Cartesian Product Equivalent to Associated Cardinal Number

## Theorem

Let $S$ and $T$ be nonempty sets.

Let $\card S$ denote the cardinal number of $S$.

Then:

$S \times T \sim \card {S \times T} \iff S \sim \card S \land T \sim \card T$

where $S \times T$ denotes the cartesian product of $S$ and $T$.

## Proof

### Necessary Condition

If $S \times T \sim \card {S \times T}$, then there is a mapping $f$ such that:

$f : S \times T \to \card {S \times T}$ is a bijection.

Since $f$ is a bijection, it follows that:

$S$ is equivalent to the image of $S \times \set x$ under $f$ where $x \in T$.

This, in turn, is a subset of the ordinal $\card {S \times T}$.

$\card {S \times T}$ is an ordinal by Cardinal Number is Ordinal.

By Condition for Set Equivalent to Cardinal Number, it follows that $S \sim \card S$.

Similarly, $T \sim \card T$.

$\Box$

### Sufficient Condition

Suppose $f: S \to \card S$ is a bijection and $g: T \to \card T$ is a bijection.

Let $\cdot$ denote ordinal multiplication, while $\times$ shall denote the Cartesian product.

Define the function $F$ to be:

$\forall x \in S, y \in T: \map F {x, y} = \card S \cdot \map g y + \map f x$

It follows that $F: S \times T \to \card S \cdot \card T$ is a injection.

By Condition for Set Equivalent to Cardinal Number, it follows that $S \times T \sim \card {S \times T}$.

$\blacksquare$