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It's hard not to be amused and perhaps even amazed when first encountering Furstenberg'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.

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  • $\begingroup$ 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 $\endgroup$
    – Gerhard Paseman
    Commented Oct 18, 2010 at 6:42
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    $\begingroup$ As BCnrd pointed out in his comments to another MO question, the topology is the one inherited from $\hat{\mathbf Z}$. $\endgroup$
    – Chandan Singh Dalawat
    Commented Oct 18, 2010 at 7:07
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    $\begingroup$ 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. $\endgroup$
    – Christian Blatter
    Commented Oct 18, 2010 at 7:43
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    $\begingroup$ A standard and more accurate name is "profinite topology". $\endgroup$
    – YCor
    Commented Aug 31, 2021 at 13:27

2 Answers 2

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The answer to your question is yes, but it is a stretch to claim that the topology is due to Furstenberg. There is an extended discussion on Furstenberg'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 Furstenberg.

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.

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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.

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  • $\begingroup$ 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 $\neq1$). $\endgroup$
    – Chandan Singh Dalawat
    Commented Oct 22, 2010 at 10:18
  • $\begingroup$ 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. $\endgroup$
    – Greg Kuperberg
    Commented Oct 22, 2010 at 12:52
  • $\begingroup$ 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 ! $\endgroup$
    – Chandan Singh Dalawat
    Commented Oct 23, 2010 at 4:56

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