Infinitude of Primes: Golomb's Topological Proof

Let $\mathbb{Z}^{+}$ denote the positive integers, $\mathbb{N}=\mathbb{Z}\cup\{0\}.$ Let $a,b\in\mathbb{Z}^{+};$ we use the notation $N_{an+b}$ to refer to the one-sided arithmetic progression $\{an+b: n\in \mathbb{N}\}.$

Now we define a basis for a topology $\tau$ on $\mathbb{Z}^{+}$ to be the set of arithmetic progressions $N_{an+b}$ where $a$ and $b$ are relatively prime. That is, $\tau$ is the set of all such arithmetic progressions taken as sets, countable unions of such sets, complements, and finite intersections, these were more described at a greater length in the page devoted to Harry Fürstenberg's proof that there are infinitely many primes.

For example, since $a=3$ and $b=7$ are relatively prime the arithmetic progression $N_{3n+7}=\{7,10,13,16,19,\ldots\}$ is member of the basis of topology $\tau.$ $N_{2n+11}=\{11,13,15,17,19,\ldots\}$ is another example, implying that $N_{an+b}\cup N_{2n+11}=\{7,10,11,13,15,16,17,19,\ldots\}$ and $N_{3n+7}\cap N_{2n+11}=\{13,19,25\ldots\}$ are both in $\tau.$


There are infinitely many prime numbers.


Consider the topological space $(\mathbb{N},\tau).$ Let $p$ be a prime number. Recall that closed sets are the complements of open sets. We claim that $\{np\}=N_{p,0}$ is closed. We can directly compute $\{np\}=\{0,p,2p,3p,\ldots\}$ and so

$\begin{align} \{np\}^{c}&=\mathbb{N}\setminus\{np\}\\ &=\{0,1,\ldots,p-1,p+1,\ldots,2p-1,2p+1,\ldots\}\\ &=N_{p,1}\cup N_{p,2}\cup \ldots \cup N_{p,p-1}. \end{align}$

but since each set $N_{an+b},$ with $a$ prime and $b\in[1,a-1],$ is a basic open set, $\{np\}^{c}$ is the union of open sets, and therefore is open. Thus $(\{np\}^{c})^{c}=\{np\}$ is closed.

Define $\displaystyle X=\cup_{p\space\text{prime}}\{np\}$ and assume that the number of primes is finite. This would make $X$ as a finite union of closed sets closed. However, since every integer different from $0$ or $1$, is of the form $np$ for some $n$ and some $p,$ we see that $X^{c}=\{0,1\}.$ Since $X^{c}$ is finite and nonempty, it cannot be an element of the topology $\tau.$ This is because every open set, unless empty, contains an arithmetic progression - an element of the topological basis. Therefore $X^{c}$ is not open. Therefore $X$ cannot be closed. Thus the number of primes cannot be finite!

It is interesting to compare topologies used to proving the infinitude of primes. Superficially, both Fürstenberg and Golomb use the bases of arithmetic progressions. But, say, while in Fürstenberg's topology arithmetic sequence are disconnected, some of them are connected in Golomb's topology Szczuka. Moreover, Fürstenberg's topology is neither connected, not locally connected; Golomb's is connected but not locally connected.


  1. Du Alaud, There's Infinitely Many Primes, III: A Second Topological Proof, 9 December 2013
  2. Solomon W. Golomb, A Connected Topology for the Integers, The American Mathematical Monthly, Vol. 66, No. 8 (Oct., 1959), pp. 663-665
  3. Paulina Szczuka, The Connectedness of Arithmetic Progressions in Furstenberg's, Golomb's, and Kirch's Topologies, Demonstratio Matemetica, Vol. XLIII No 4 2010
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