Cardinal Product Equinumerous to Ordinal Product

Theorem
Let $S$ and $T$ be sets that are equivalent to their cardinal numbers.

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

Let $\cdot$ denote ordinal multiplication and let $\times$ denote the Cartesian product.

Then:


 * $S \times T \sim \left|{ S }\right| \cdot \left|{ T }\right|$

Proof
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:


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

Suppose $F\left({x_1,y_1}\right) = F\left({x_2,y_2}\right)$.

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

Furthermore, take any $z \in \left|{ S }\right| \cdot \left|{ T }\right|$.

By the Division Theorem for Ordinals, it follows that:


 * $z = \left|{ S }\right| \cdot a + b$ for some unique $a$ and $b \in \left|{ S }\right|$.

Moreover, $a \in \left|{ T }\right|$ since otherwise, this would contradict the fact that $z \in \left|{ S }\right| \cdot \left|{ T }\right|$.

Therefore, $a = f\left({x}\right)$ for some $x \in S$ and $b = g\left({y}\right)$ for some $y \in T$.

Thus, $F : S \times T \to \left|{ S }\right| \cdot \left|{ T }\right|$ is a surjection.

Therefore, $F : S \times T \to \left|{ S }\right| \cdot \left|{ T }\right|$ is a bijection by the definition of bijection.

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