A bit of topology

This proof is due to Hillel Fürstenberg, and is topological.

Let’s define a topology on \mathbb Z. We’ll declare any infinite arithmetic sequence to be open. That is, each set


is open. These are the basic open sets, so any open set can be obtained by taking unions and finite intersections of these. It’s not hard to see that a finite intersection of these is empty, or infinite. That is, there are no finite open sets other than \varnothing.

Another fun fact about this topology is that each U_{a,b} is also closed. Indeed

U_{a,b}=\mathbb Z\setminus\displaystyle\bigcup_{i\ne b}U_{a,i}.

This is probably poor notation, but I mean that i runs from 0 to a-1, but skips b. So U_{a,b} is the complement of an open set and hence closed. All I will really need is that U_{a,0} is closed

Now, if there were only finitely many primes, then \bigcup_{p}U_{p,0} would be a closed set. This set consists of any number which is a multiple of some prime. It’s complement is \{-1,1\} which has to be open. This is a contradiction, because finite sets cannot be open. Cool, huh?


2 Responses to A bit of topology

  1. fusionball says:

    There was a huge thread on mathoverflow a while back arguing about whether this is really just Euclid’s proof couched in the language of topology. The argument could also be phrased this way. Say that a set S of integers is periodic if for some positive integer n, we have x \in S iff x + n \in S. Then the union of a period p set with a period q set is a set of period pq (here periods do not have to be minimal). Complements of periodic sets are also periodic (if the period is p then the complement will be the union of the “missing” congruence classes mod p). If there were only finitely many primes, then \{1,-1\} is periodic, contradiction. But, if we unravel this argument, really we are just looking at p_1 p_2 \ldots p_n \pm 1 so it’s Euclid’s argument all over again.

  2. Pingback: And one more similar one… « Andy Soffer

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