# Cartesian Product of Countable Sets is Countable/Corollary

## Contents

## Corollary to Cartesian Product of Countable Sets is Countable

Let $k$ be an integer such that $k > 1$.

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

## Proof 1

Let $S_1, S_2, \ldots, S_k$ be countable sets. Equivalently, there exists an injection $f_i: S_i \to \N$ for each $i \in \{ 1, \dots, k \}$.

Therefore, the mapping $g: S_1 \times S_2 \times \cdots \times S_k \to \N^k$ defined by (for $x_i \in S_i$):

- $\map g {x_1, x_2, \dotsc, x_k} = \tuple {\map {f_1} {x_1}, \map {f_2} {x_2}, \dotsc, \map {f_k} {x_k} }$

is also injective.

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:

- $\map f {n_1, n_2, \ldots, n_k} = p_1^{n_1} p_2^{n_2} \cdots p_k^{n_k}$

is an injection.

From Composite of Injections is Injection, $f \circ g: S_1 \times S_2 \times \cdots \times S_k \to \N$ is injective.

$\blacksquare$

## Proof 2

Proof by induction:

### Basis for the Induction

When $k = 2$, the case is the same as Cartesian Product of Countable Sets is Countable.

So shown for basis for the induction.

### Induction Hypothesis

This is our induction hypothesis:

- $\exists f_k: S_1 \times S_2 \times \cdots \times S_k \to \N$

where $f_k$ is an injection.

Now we need to show that for $n = k + 1$:

- $\exists f_{k+1}: S_1 \times S_2 \times \cdots \times S_k \times S_{k+1} \to \N$

where $f_{k+1}$ is an injection.

### Induction Step

This is our induction step:

By the induction hypothesis:

- $\exists f_k: S_1 \times S_2 \times \cdots \times S_k \to \N$

where $f_k$ is an injection.

Thus by definition, $S_1 \times S_2 \times \cdots \times S_k$ is countable.

By hypothesis $S_{k + 1}$ is countable.

So by the basis for the induction:

- $\exists g: \left({S_1 \times S_2 \times \cdots \times S_k}\right) \times S_{k+1} \to \N \times \N$

where $g$ is an injection.

By Cartesian Product of Countable Sets is Countable,

- $\exists r: \N \times \N \to \N$

where $r$ is an injection.

Therefore, by Composite of Injections is Injection:

- $f_{k+1} = r \circ g: S_1 \times S_2 \times \cdots \times S_k \times S_{k+1} \to \N$

is an injection.

The result follows by induction.

$\blacksquare$

## Sources

- 2008: Paul Halmos and Steven Givant:
*Introduction to Boolean Algebras*... (previous) ... (next): Appendix $\text{A}$: Set Theory: Countable Sets - 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Exercise $18.16$