# Prime numbers and Topology

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.

• It means there is a homeomorphism (bi-continuous, bijective) map from your topology onto a metric space. – BananaCats Category Theory App Mar 11 '15 at 17:44
• See arxiv.org/pdf/1008.0713.pdf. They define a norm. Then a metric is just $d(x,y) = | x - y|$ – BananaCats Category Theory App Mar 11 '15 at 17:52
• @EnjoysMath: I think it'd be better if you had written $\lVert x-y\rVert$. That is the notation of the paper, and far less confusing: $\lvert x-y\rvert$ has a commonly accepted meaning (for integers), which would make the topology discrete. – tomasz Mar 11 '15 at 18:00
• Thank you Enjoys Math for that incredible pdf and thank you tomasz as well, I totally agree with that. – GSF Mar 11 '15 at 19:15

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.