## 29 March 2008

### Furstenberg's topological proof of the infinitude of primes

I just learned about Furstenberg's proof of the infinitude of primes (link goes to Wikipedia article); it was published in 1955 in the American Mathematical Monthly. (Link goes to JSTOR. Full citation: The American Mathematical Monthly, Vol. 62, No. 5. (May, 1955), p. 353. If you have to do any work to get the article from JSTOR -- for example, logging in through a proxy server because you're at home instead of on campus -- or you don't have JSTOR access -- don't worry, you're not missing much. The article's a third of a page, and the Wikipedia article is probably longer than Furstenberg's original.) Surprisingly, I hadn't seen this before.

Anyway, here's my version of the proof: topologize the integers by taking the (doubly infinite) arithmetic sequences as a basis. (The only thing that needs checking here, to show these form a basis, is that the nonempty intersection of two basis elements contains a basis element; this is true since the intersection of two arithmetic sequences is another arithmetic sequence.) Consider the set which consists of all the integers except -1 and 1; this is the union of the sequences pZ over all primes. Now, there are no nonempty finite open sets in this topology, so there are no cofinite closed sets except Z itself. Thus the union of the pZ isn't closed. So it can't be an union of finitely many closed sets, since finite unions of closed sets are closed. So there are infinitely many sets pZ, and thus infinitely many primes!

(Thanks to an anonymous commentator for pointing out some errors.)

Anonymous said...

Anonymous said...

Wow, even I (almost) understood that. Makes me want to read more. Thanks!

Anonymous said...

This is one of six proofs in "Proofs from the Book". Since you had not heard of this proof before and it is given on page 5, I presume you don't own it. I think it would be very much your kind of book!

Aaron said...

Ahh, topology! That reminds me of a homework problem that one of my friends got a while ago: to prove that a continuous function that is zero at every rational number must be zero everywhere. The topological proof is about four sentences long. Unfortunately for my friend, the homework was for an analysis class. :)

Anonymous said...

Why is that unfortunate? What was topology first developed for by analysis? The analytic proof should more or less be a special case of the topological one, written out with the specific definitions of the topology on the real line.

Anonymous said...

You write: "Now, there are no finite open sets in this topology, so there are no cofinite closed sets. Thus the union of the pZ isn't closed. So it can't be an intersection of finitely many closed sets, since finite intersections of closed sets are closed."

Three points: "no finite open sets" should read "no finite nonempty open sets"; "intersection of finitely many closed" should read "union of finitely many closed"; and "finite intersections of closed sets" should read "finite unions of closed sets."