Cardinality of Cartesian Product of Finite Sets

Theorem
Let $$S \times T$$ be the cartesian product of two sets $$S$$ and $$T$$.

Then:
 * $$\left|{S \times T}\right| = \left|{S}\right| \times \left|{T}\right|$$

where $$\left|{S}\right|$$ denotes cardinality.

This is convenient, given the symbology.

Proof
Let $$\left|{S}\right| = n$$ and $$\left|{T}\right| = m$$.


 * If either $$n = 0$$ or $$m = 0$$, then from Cartesian Product Null:
 * $$S \times T = \varnothing$$

and the result holds, as $$n m = 0 = \left|{\varnothing}\right|$$ from Cardinality of Empty Set.


 * So, we assume that $$n > 0$$ and $$m > 0$$.

For each $$a \in S$$, we define the mapping $$g_a: T \to \left\{{a}\right\} \times T$$ such that:


 * $$\forall y \in T: g_a \left({y}\right) = \left({a, y}\right)$$

The mapping $$g_a$$ is a bijection, so $$\left| {\left\{{a}\right\} \times T}\right| = m$$.

Now let:
 * $$\mathbb T = \left\{{\left\{{a}\right\} \times T: a \in S}\right\}$$

We define the mapping $$h: S \to \mathbb T$$:


 * $$\forall a \in S: h \left({a}\right) = \left\{{a}\right\} \times T$$

The mapping $$h$$ is a bijection, so $$\left|{\mathbb T}\right| = n$$.

Thus $$\mathbb T$$ is a partition of $$S \times T$$ containing $$n$$ sets.

Hence from Number of Elements in Partition:
 * $$\left|{S \times T}\right| = n m$$

and the result follows.