# Furstenberg’s proof of the infinitude of the primes

Another very classical proof that’s just too beautiful to ignore. This time, the theorem is millennia old, but it’s really the proof that I’m interested in.

Lemma. There are infinitely many primes.

Proof. Define the sets

for with . Note that if ; say it contains some , then

Hence, the intersection of two sets of this form is again of this form (or empty). Thus, they form the basis for a topology on . Notice that the sets

are also open.

Now suppose that there were finitely many primes. Then the intersection

is a finite intersection of opens, hence open. On the other hand, it equals the set of integers divisible by no prime number, which is . But all basic open sets are infinite, so this set can never be open.

Remark. This proof is in essence the usual proof, where we consider and conclude that some prime has to divide it and this prime can be none of the . Indeed, when we showed that the intersection of basic opens is open, we went to a set whose period is (which for gives period ). Since 1 is in , so should be.

That said, it doesn’t seem possible to adopt this proof to reprove other theorems like ‘there are infinitely many primes congruent to 3 modulo 4’. The problem is that the above proof seems to rely on a global analysis of the set : it has to be an infinite set. So the method is too crude to prove theorems with congruence restrictions.