TITLE: Prime numbers and Topology
QUESTION [5 upvotes]: I was doing this homework for my university in which I had to prove that the set of prime numbers was infinite, just like Harry Furstenburg did by considering the following topology:
Let $\mathcal{O} \subset \mathcal{P}(\mathbb{Z})$ be a topology defined as:

$\mathcal{O}:=\{\emptyset\}\cup\{A\subset\mathbb{Z}:A$ is an arbitrary union of arithmetic progressions$\}$

Remember that an arithmetic progression is a subset $S\subset\mathbb{Z}$ defined as $S:=\{r+nd : n \in \mathbb{Z}\}$, where $r,d\in\mathbb{Z}$ and $d\neq0$.
Everything went well until I got to the last question, which I unfortunately couldn't even start. It said: 

Is $(\mathbb{Z},\mathcal{O})$ a metrizable topology?

Thank you very much.

REPLY [2 votes]: There are only countably many arithmetic progressions, and the arithmetic progressions form a base $\mathscr{B}$ for $\mathcal{O}$, so the space is second countable. It’s clearly $T_1$, since for any distinct integers $m$ and $n$ the nbhd 
$$\left\{m+k\big(|m-n|+1\big):k\in\Bbb Z\right\}\in\mathscr{B}$$
of $m$ misses $n$. 
Let $B=\{r+nd:n\in\Bbb Z\}\in\mathscr{B}$. If $m\in\Bbb Z\setminus B$, $\{m+nd:n\in\Bbb Z\}$ is an open nbhd of $m$ disjoint from $B$, so $B$ is clopen. The space is therefore $T_3$ and second countable, so it’s metrizable by the Uryson metrization theorem. In fact, since it’s a countable metrizable space without isolated points, it’s homeomorphic to $\Bbb Q$ with the usual topology.