Product of Finite Sets is Finite
Theorem
Let $S$ and $T$ be finite sets.
Then $S \times T$ is a finite set.
Proof 1
By the definition of Cartesian product:
- $S \times T = \set {\tuple {s, t}: s \in S, t \in T}$
Then by definition of set union:
- $S \times T = \ds \bigcup_{s \mathop \in S} \set s \times T$
Also, the mapping $g_s: \set s \times T \to T$ defined by:
- $\map {g_s} {s, t} = t$
is a bijection.
Therefore, since $T$ is finite, so is $\set s \times T$ for all $s \in S$.
Since $S$ is finite, the result follows from Finite Union of Finite Sets is Finite.
$\blacksquare$
Proof 2
Let $\card S$ denote the cardinal number of $S$.
Let $\cdot$ denote ordinal multiplication.
By Cardinal Product Equinumerous to Ordinal Product, it follows that $S \times T \sim \card S \cdot \card T$.
But then $\card S$ and $\card T$ are members of the minimally inductive set.
Therefore, $\card S \cdot \card T \in \omega$ by Natural Number Multiplication is Closed.
Since $S \times T$ is equinumerous to a member of the minimally inductive set, it follows that $S \times T$ is finite.
$\blacksquare$