Proving $\mathbb{Z × N}$ is countable.

Question

How would I prove that $\mathbb{Z × N}$ is countable? The hint given was to follow to indicated order. Thanks!

Prove $\mathbb{Z}$ is countable and then cartesian product of two countable set is countable. — Kushal Bhuyan, Dec 05 '15 at 07:54
The point is to prove cartesian product is contable. You can do it trough some kind of zig-zag bijective function between the pairs and $\Bbb N$. Pairs represent points in a plane... think a way to fill the plane with a function. You can start from one corner, in zig-zag, to the infinity, including jumps to negative values of $\Bbb Z$. Be creative my friend :) — Masacroso, Dec 05 '15 at 07:56
...or if you have already proved that $\Bbb Q\times \Bbb Q$ is countable, then use subset of a countable set is countable. — Jesse P Francis, Dec 05 '15 at 08:02
Possible duplicate of Bijecting a countably infinite set $S$ and its cartesian product $S \times S$ — Dietrich Burde, Dec 05 '15 at 09:36

copper.hat · Answer 1 · 2015-12-05T22:38:18.047

3

Let $p_0,p_1,p_2$ be three distinct positive primes. I am taking $\mathbb{N}=\{1,2,...\}$.

Define $\phi:\mathbb{N} \to \mathbb{Z} \times \mathbb{N}$ by $\phi(n) = \begin{cases} (n_1,n_2+1), & n = p_0^0p_1^{n_1}p_2^{n_2} \\ (-n_1,n_2+1), & n = p_0^1p_1^{n_1}p_2^{n_2} \\ (0,1), & \text{otherwise} \end{cases}$.

We see that $\phi(p_0^0p_1^{n_1}p_2^{n_2-1}) = (n_1,n_2)$, $\phi(p_0^1p_1^{n_1}p_2^{n_2-1}) = (-n_1,n_2)$, hence $\phi$ is surjective. It follows that $\mathbb{Z} \times \mathbb{N}$ is countable.

edited Dec 05 '15 at 22:38

answered Dec 05 '15 at 08:03

copper.hat

172,524

@user236182: Thanks for catching that. – copper.hat Dec 05 '15 at 14:20
@user236182: Thanks again. – copper.hat Dec 05 '15 at 22:39
@user236182: I don't get the 'you haven't edited...' part of your last comment, $n_2 \ge 0$ so $\phi$ is OK, and the part showing surjectivity seems fine? – copper.hat Dec 05 '15 at 22:48

score 3 · Answer 2 · answered Dec 05 '15 at 08:33

$$\begin{array}{c|ccccccc} \mathbb{Z}&\searrow&\vdots&\vdots&\vdots&\vdots&\vdots\\ &10&\searrow&\vdots&\vdots&\vdots&\vdots\\ &5&11&\searrow&\vdots&\vdots&\vdots\\ &2&6&12&\searrow&\vdots&\vdots\\ &1&3&7&13&\downarrow&\vdots&\hspace{1in}\mathbb{N}\\\hline &4&8&14&\swarrow&\vdots&\vdots\\ &9&15&\swarrow&\vdots&\vdots&\vdots\\ &16&\swarrow&\vdots&\vdots&\vdots&\vdots\\ &\swarrow&\vdots&\vdots&\vdots&\vdots&\vdots\\ \end{array}$$

Proving $\mathbb{Z × N}$ is countable.

2 Answers2