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.