5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ It means there is a homeomorphism (bi-continuous, bijective) map from your topology onto a metric space. $\endgroup$ – BananaCats Category Theory App Mar 11 '15 at 17:44
  • $\begingroup$ See arxiv.org/pdf/1008.0713.pdf. They define a norm. Then a metric is just $d(x,y) = | x - y|$ $\endgroup$ – BananaCats Category Theory App Mar 11 '15 at 17:52
  • 1
    $\begingroup$ @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. $\endgroup$ – tomasz Mar 11 '15 at 18:00
  • $\begingroup$ Thank you Enjoys Math for that incredible pdf and thank you tomasz as well, I totally agree with that. $\endgroup$ – GSF Mar 11 '15 at 19:15
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.