This post is entirely unrelated to what I’ve been talking about involving sheaves and ringed spaces. But I saw it today in class and thought it was so fitting with the name and purpose of this blog that I’d present it anyway, because I love it so much. It’s a proof that there are infinitely many prime numbers (due to Furstenberg) – but instead of Euclid’s more direct approach, it uses topology!

Let’s create a basis for a topology on $\mathbb{Z}$; for integers $m, d$ with $d > 0$, denote by $A(m,d) = \left\{ m + d n : n\in\mathbb{Z}\right\}$ the infinite arithmetic progression of all integers congruent to $m$ modulo $d$. I claim that $\mathcal{B} = \left\{ A(m,d) : m\in\mathbb{Z}, d\in\mathbb{N}\right\}$ is a basis for a topology on $\mathbb{Z}$. Firstly, note that we can write $\mathbb{Z} = A(0,1)$, and we can agree that the empty set is equal to an empty union of arithmetic progressions. As we’re defining a basis our open sets are just unions of the $A(m,d)$ so we don’t need to check anything here. But we do need to check whether intersections of two basic open sets $A(m_1,d_1), A(m_2, d_2)$ can be expressed as unions of other basic opens of the form $A(m,d)$.

Firstly, if $A(m_1, d_1)\cap A(m_2, d_2) = \varnothing$ (i.e. $d_1 = d_2$ and $m_1 \neq m_2$) then we can vacuously write the intersection as an empty union, as agreed. So instead suppose $A(m_1, d_1)$ and $A(m_2, d_2)$ have nonempty intersection. Then without loss of generality we may assume they intersect at some point $m$. “Re-centering” each arithmetic progression on $m$ (as they are only defined modulo their respective differences $d_1$ and $d_2$ anyway), we now need to compute $A(m,d_1)\cap A(m,d_2)$. But it is not hard to see that this is just $A(m,\text{lcm}(d_1, d_2))$. But this is just another basic open set in $\mathcal{B}$ – it’s not even a union of more than one open set! So the whole construction works very well – we can take arbitrary unions of these sets, take finite intersections, and we know that we can express the whole space $\mathbb{Z}$ and the empty set in terms of the basic opens $A(m,d)$. So we really have a basis for a topology on $\mathbb{Z}$, whose open sets consist of unions of infinite arithmetic progressions.

Now the sets $A(m,d)$ are open, but they are also closed! To see this, note that $A(m,d)^c = \bigcup_{m'\not\equiv m \pmod d} A(m',d)$ is just a union of all those $d-1$ arithmetic progressions of the same step-size $d$ but shifted throughout the whole interval $\left\{m+1, \dots, m+d-1\right\}$. So the complement of a basic open set is a (finite, even) union of basic open sets, and thus is open too. Hence the basic open sets $A(m,d)$ are both closed and open.

Now consider the union $\bigcup_{p \text{ prime}} A(0,p)$. Since every integer except $1$ and $-1$ is divisible by at least one prime (by the fundamental theorem of arithmetic), the above union gives $\bigcup_{p \text{ prime}} A(0,p) = \mathbb{Z} \backslash \left\{1,-1\right\}$. The complement of this set, namely $\left\{1,-1\right\}$, is finite – hence there is no way that it can be expressed as a union of arithmetic progressions of the form $A(m,d)$, which are all infinite sets. But therefore as these are the basic open sets in the topology, the set $\left\{1,-1\right\}$ is not open. Hence its complement, $\mathbb{Z}\backslash \left\{1,-1\right\} =\bigcup_{p \text{ prime}} A(0,p)$, is not closed.

But $\bigcup_{p \text{ prime}} A(0,p)$ is a union of closed sets (as I showed, each arithmetic progression is both open and closed). If this was a union of a finite number of closed sets – and therefore if it were indexed by finitely many primes $p$ – then it would itself be closed, because in a topology finite unions of closed sets are always closed. And we definitely have a base for a topology, as shown above. Therefore, the union must involve an infinite number of closed sets $A(0,p)$, indexed by infinitely many primes $p$. And therefore, there are infinitely many primes!!

Do you prefer Furstenberg’s topological proof or Euclid’s streamlined proof? The latter has the advantage that it’s a lot more direct, quicker, and probably seems like it’s more related to prime numbers. Furstenberg’s proof, on the other hand, takes a long time to set up and it feels very removed from anything to do with primes, using a lot more abstract machinery and a proof by contradiction, while Euclid’s proof is direct. However, the contradiction is very easy to understand, while many people (myself included, until I was wisely advised to the contrary) believe that Euclid’s proof uses a contradiction. I guess this subtlety in Euclid’s proof – depending on how it’s presented – can make it slightly confusing. The topological proof seems more simple in each individual step it takes, and for this reason I prefer it. But both proofs rest on the fundamental fact that every one of the infinitely many integers (except the units) are divisible by at least one prime.

Edit: This post has generated a lot of comments with great advice and helpful pointers, for which I’m really grateful! I have made several changes to the post accordingly.