Proof: Let $A_1, A_2, \ldots, A_n$ be countable sets and let $S = A_1 \times A_2 \times \cdots \times A_n$ . Since each $A_i$ is countable there exists a one-to-one function $f_i\colon A_i \to \mathbb{N}$ . The function $h\colon S \to \mathbb{N}$ defined by $$ h(a_1, a_2, \ldots a_n) = \prod_{i=1}^n p_i^{f_i(a_i)} $$ where $p_i$ is the $i$ th prime is, by the fundamental theorem of arithmetic, a bijection between $S$ and a subset of $\mathbb{N}$ and therefore $S$ is also countable.

Note that this result does not (in general) extend to the Cartesian product of a countably infinite collection of countable sets. If such a collection contains more than a finite number of sets with at least two elements, then Cantor's diagonal argument can be used to show that the product is not countable.

For example, given $B = \{0, 1\}$ , the set $F = B \times B \times \cdots$ consists of all countably infinite sequences of zeros and ones. Each element of $F$ can be viewed as a binary fraction and can therefore be mapped to a unique real number in $[0, 1)$ and $[0, 1)$ is, of course, not countable.