# Condition for Cartesian Product Equivalent to Associated Cardinal Number

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## Theorem

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

Let $\left|{S}\right|$ denote the cardinal number of $S$.

Then:

$S \times T \sim \left|{S \times T}\right| \iff S \sim \left|{S}\right| \land T \sim \left|{T}\right|$

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

## Proof

### Necessary Condition

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

$f : S \times T \to \left|{S \times T}\right|$ is a bijection.

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

$S$ is equivalent to the image of $S \times \left\{{x}\right\}$ under $f$ where $x \in T$.

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

$\left|{S \times T}\right|$ is an ordinal by Cardinal Number is Ordinal.

By Condition for Set Equivalent to Cardinal Number, it follows that $S \sim \left|{S}\right|$.

Similarly, $T \sim \left|{T}\right|$.

$\Box$

### Sufficient Condition

Suppose $f: S \to \left|{S}\right|$ is a bijection and $g: T \to \left|{T}\right|$ is a bijection.

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

Define the function $F$ to be:

$\forall x \in S, y \in T: F \left({x, y}\right) = \left|{S}\right| \cdot g \left({y}\right) + f \left({x}\right)$

It follows that $F: S \times T \to \left|{S}\right| \cdot \left|{T}\right|$ is a injection.

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

$\blacksquare$