Cartesian Product of Countable Sets is Countable/Formal Proof 1

Theorem
The cartesian product of two countable sets is countable.

Proof
Let $S, T$ be countable sets.

From the definition of countable, there exists a injection from $S$ to $\N$, and similarly one from $T$ to $\N$.

Hence there exists an injection $g$ from $S \times T$ to $\N^2$.

Now let us investigate the cardinality of $\N^2$.

From the Fundamental Theorem of Arithmetic, every natural number greater than $1$ has a unique prime decomposition.

Thus, if a number can be written as $2^n 3^m$, it can be done thus in only one way.

So, consider the function $f: \N^2 \to \N$ defined by:
 * $f \left({n, m}\right) = 2^n 3^m$.

Now suppose $\exists m, n, r, s \in \N$ such that $f \left({n, m}\right) = f \left({r, s}\right)$.

Then $2^n 3^m = 2^r 3^s$ so that $n = r$ and $m = s$.

Thus $f$ is an injection.

Since $f: \N^2 \to \N$ is an injection and $\N$ is countably infinite, it follows from Injection from Infinite to Countably Infinite Set that $\N^2$ is countably infinite.

Now we see that as $g$ and $f$ are injective, it follows from Composite of Injections is an Injection that $f \circ g: S \times T \to \N$ is also injective.

Hence the result.