# A bit of topology

May 5, 2012 2 Comments

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

Let’s define a topology on . 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 .

Another fun fact about this topology is that each is also closed. Indeed

.

This is probably poor notation, but I mean that runs from to , but skips . So is the complement of an open set and hence closed. All I will really need is that is closed

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

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 of integers is periodic if for some positive integer , we have iff . Then the union of a period set with a period set is a set of period (here periods do not have to be minimal). Complements of periodic sets are also periodic (if the period is then the complement will be the union of the “missing” congruence classes mod ). If there were only finitely many primes, then is periodic, contradiction. But, if we unravel this argument, really we are just looking at so it’s Euclid’s argument all over again.

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