## Topological proof of the infinitude of primes

Perhaps everyone knows how to prove “there exist infinitely many primes” by one or more ways. Here I just want to share a proof using topology and prime factorization theorem. It is also (perhaps highly) possible that most of the readers know this proof except me, but I just think that it is interesting so I want to go through it here. Of course, the prerequisite is some point set topology.

Theorem. There exist infinitely many primes.

Proof. (by H. Furstenberg, 1955) Define a topology on $\mathbb{Z}$ by setting the basic open sets to be sets in the form

$S_{a,b}=\{am+b:m\in \mathbb{Z}\}$

where $a$ is a nonzero integer and $b$ is an integer. This is a set of an arithmetic sequence. We call the collection $\mathcal{C}$. To show that this is a basis for some topology, we only need to show:

(1) $\displaystyle \mathbb{Z}=\bigcup_{a,b} S_{a,b}$;

(2) The intersection of finitely many members of $\mathcal{C}$ is in $\mathcal{C}$.

(1) is clear because $\mathbb{Z}=S_{1,0}$. For (2), suppose $S_{a_i,b_i}$ is a basic set for each $i=1,\cdots,n$. The intersection is either empty, or contains some integer $c$, and also

$\displaystyle \bigcap_{i=1}^n S_{a_i,b_i}=S_{k,c}$

where $k$ is the lowest common multiple of $a_1,\cdots,a_n$.  So (2) is justified and $\mathcal{C}$ is a basis of some topology.

Notice that each $S_{a,b}$ is closed because its complement

$\mathbb{Z}\setminus S_{a,b}=S_{a,b+1}\cup S_{a,b+2}\cup \cdots \cup S_{a,b+a-1}$

is an open set.

By the prime factorization theorem, every integer other than 1 or -1 appears in some $S_{p,0}$ where $p$ is a positive prime. Also $0\in S_{2,0}$. Thus

$\displaystyle \mathbb{Z}\setminus\{1,-1\}=\bigcup_{p \text{ prime }} S_{p,0}$,

which is a union of closed sets. If there are only finitely many primes, then the right hand side is a finite union of closed sets, which is closed. Then $\mathbb{Z}\setminus\{1,-1\}$ is closed so the set $\{1,-1\}$ is open, which is impossible, because every open set is infinite, as each $S_{a,b}$ is so. $\square$