Define a topology on the integers , called the evenly spaced integer topology, by declaring a subset U ⊆ to be an open set if and only if it is a union of arithmetic sequences S(a, b) for a ≠ 0, or is empty (which can be seen as a nullary union (empty union) of arithmetic sequences), where
Equivalently, U is open if and only if for every x in U there is some non-zero integer a such that S(a, x) ⊆ U. The axioms for a topology are easily verified:
- ∅ is open by definition, and is just the sequence S(1, 0), and so is open as well.
- Any union of open sets is open: for any collection of open sets Ui and x in their union U, any of the numbers ai for which S(ai, x) ⊆ Ui also shows that S(ai, x) ⊆ U.
- The intersection of two (and hence finitely many) open sets is open: let U1 and U2 be open sets and let x ∈ U1 ∩ U2 (with numbers a1 and a2 establishing membership). Set a to be the least common multiple of a1 and a2. Then S(a, x) ⊆ S(ai, x) ⊆ Ui.
This topology has two notable properties:
- Since any non-empty open set contains an infinite sequence, a finite non-empty set cannot be open; put another way, the complement of a finite non-empty set cannot be a closed set.
- The basis sets S(a, b) are both open and closed: they are open by definition, and we can write S(a, b) as the complement of an open set as follows:
The only integers that are not integer multiples of prime numbers are −1 and +1, i.e.
Now, by the first topological property, the set on the left-hand side cannot be closed. On the other hand, by the second topological property, the sets S(p, 0) are closed. So, if there were only finitely many prime numbers, then the set on the right-hand side would be a finite union of closed sets, and hence closed. This would be a contradiction, so there must be infinitely many prime numbers.