Cartesian Product of Countable Sets is Countable

Theorem
The cartesian product of two countable sets is countable.

Corollary
Let $$k > 1$$.

Then the cartesian product of $$k$$ countable sets is countable.

Informal Proof
Let $$S = \left\{{s_0, s_1, s_2, \ldots}\right\}$$ and $$T = \left\{{t_0, t_1, t_2, \ldots}\right\}$$ be countable sets.

If both $$S$$ and $$T$$ are finite, the result follows immediately.

Suppose either of $$S$$ or $$T$$ (or both) is countably infinite.

We can write the elements of $$S \times T$$ in the form of an infinite table:

$$\begin{array} {*{4}c} {\left({s_0, t_0}\right)} & {\left({s_0, t_1}\right)} & {\left({s_0, t_2}\right)} & \cdots \\ {\left({s_1, t_0}\right)} & {\left({s_1, t_1}\right)} & {\left({s_1, t_2}\right)} & \cdots \\ {\left({s_2, t_0}\right)} & {\left({s_2, t_1}\right)} & {\left({s_2, t_2}\right)} & \cdots \\ \vdots & \vdots  & \vdots & \ddots \\ \end{array} $$

This table clearly contains all the elements of $$S \times T$$.

Now we can count the elements of $$S \times T$$ by processing the table diagonally. First we pick $$\left({s_0, t_0}\right)$$. Then we pick $$\left({s_0, t_1}\right), \left({s_1, t_0}\right)$$. Then we pick $$\left({s_0, t_2}\right), \left({s_1, t_1}\right), \left({s_2, t_0}\right)$$.

We can see that all the elements of $$S \times T$$ will (eventually) be listed, and there is a specific number (element of $$\N$$) to index each of its elements with.

Thus we have the required one-to-one correspondence between $$S \times T$$ and $$\mathbb{N}$$, and our assertion is proved.

Formal 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.

Proof of Corollary
Let $$S_1, S_2, \ldots, s_k$$ be countable sets.

By the same argument, there exists an injection $$g: S_1 \times S_2 \times \cdots \times S_k \to \N^k$$.

Now let $$p_1, p_2, \ldots, p_k$$ be the first $$k$$ prime numbers.

From the Fundamental Theorem of Arithmetic, $$f: \N^k \to \N$$ defined as:
 * $$f \left({n_1, n_2, \ldots, n_k}\right) = p_1^{n_1} p_2^{n_2} \cdots p_k^{n_k}$$

is an injection.

The result follows from Composite of Injections is an Injection and Injection from Infinite to Countably Infinite Set, as above.