Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

It's hard not to be amused and perhaps even amazed when first encountering Fürstenberg's clever "topological" proof that there are infinitely many primes. Closer inspection, however, reveals the disappointing truth that there really isn't anything topological going on there, as pointed out by BCnrd in a comment to this answer.

Nevertheless, the topology on $\mathbb{Z}$ introduced in the proof, where an open set is defined as any union of arithmetic sequences, does seem both natural and interesting.

My question is this: Can anything useful be done with this topology? Useful would include a new theorem, a simplification to a proof of a known result, or even fresh insight into standard material.

share|improve this question
1  
This probably should be a Community Wiki. –  Dick Palais Oct 18 '10 at 6:37
    
For the particular topology on the integers, it seems unlikely, especially as most (non-topological) results about the integers assume little or nothing about that topology. I can imagine that an infinite product or some topological modification (something other than compactification) might produce interesting results. Gerhard "Ask Me About System Design" Paseman, 2010.10.17 –  Gerhard Paseman Oct 18 '10 at 6:42
14  
As BCnrd pointed out in his comments to another MO question, the topology is the one inherited from $\hat{\mathbf Z}$. –  Chandan Singh Dalawat Oct 18 '10 at 7:07
12  
Here is a version of Fuerstenberg's proof that does not mention topology: We argue about periodic subsets of $\mathbb Z$. The set of all numbers prime to a given $p$ is periodic and the intersection of two periodic sets is periodic. If there were only finitely many primes the set $\{-1,1\}$ would be periodic. –  Christian Blatter Oct 18 '10 at 7:43
    
Wiki-hammered.$ $ –  S. Carnahan Oct 18 '10 at 8:04

2 Answers 2

The answer to your question is yes, but it is a stretch to claim that the topology is due to Furstenburg. There is an extended discussion on Furstenburg's proof in the comments to this answer. The short version is as Chandan Singh Dalawat said in the comments above: this topology on the integers is the profinite topology, and people had been studying profinite topologies long before Furstenburg.

The topology is useful in the sense that profinite completions are useful. In particular, you may argue that it is a natural topology on the fundamental group of a circle (or the punctured complex affine line), since its profinite completion is the geometric fundamental group of the multiplicative group $\mathbb{G}_m$. It also appears in some form whenever one uses the ring of adeles $\mathbb{A}_\mathbb{Q}$, which you may encounter when studying Tate's thesis or automorphic representations.

share|improve this answer
2  
Since Prüfer, Neue Begründung der algebraischen Zahlentheorie, Math. Annalen 94 (1925), 198-243 (gdz.sub.uni-goettingen.de/dms/load/img/?PPN=PPN235181684_0094). –  Chandan Singh Dalawat Oct 18 '10 at 8:06

A more rigorous version of Scott's answer: If a topology on a group $G$ is translation-invariant, then it also defines a uniformity on $G$, by definition a distinguished set of neighborhoods of the diagonal $G \times G$ that is analogous to a metric. Actually, in the present example with $G = \mathbb{Z}$, the uniformity comes from a metric. Like the metric spaces that they generalize, uniform spaces have completions. The completion of $\mathbb{Z}$ with respect to the uniformity cited by Furstenberg is exactly the adelic profinite completion of $\mathbb{Z}$. Or if $G$ is any group, there is a similar topology generated by finite-index subgroups, and a uniformity, and the completion is the profinite completion.

share|improve this answer
    
For $G$ to embed into its profinite completion $\hat G$, it should be residually finite (for every $x\in G$, there should be a finite quotient $H_x$ of $G$ in which the image of $x$ is $\neq1$). –  Chandan Singh Dalawat Oct 22 '10 at 10:18
    
That's true. If $X$ is a uniform space, then it does not necessarily embed into its completion either; rather it has a canonical Hausdorff quotient which embeds. –  Greg Kuperberg Oct 22 '10 at 12:52
    
Sorry, I meant to say "For every $x\neq1$ in $G$" instead of "For every $x\in G$". But this lapse doesn't seem to have led to any confusion ! –  Chandan Singh Dalawat Oct 23 '10 at 4:56

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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