# Infinitude of Primes: A Topological Proof

Although topology made away with metric properties of shapes, it was helped very much by algebra in classification of knots. Following is a wonderful example (due to Harry Fürstenberg of the Hebrew University of Jerusalem, Israel) of a returned favor (albeit on a smaller scale): Euclid's theorem - one of the basic statements of arithmetic - is proven by very simple topological means! (Fürstenberg's proof has been recast elsewhere with the verbiage that avoids topological terms.)

We call a set closed if it contains all its near points. A set is open if its complement is closed. There are always two sets (the empty set and the space itself) that are simultaneously open and closed. Open and closed sets are also characterized by complementary properties:

$\begin{array}{ll} \text{Any union of open sets is open.} & &\text{Any intersection of closed sets is closed.}\\ \text{A finite intersection of open sets is open.} & & \text{A finite union of closed sets is closed.} \end{array}$

Let's prove the right two. The left ones will follow from de Morgan's laws,

$(\cup A)^{c} = \cap A^{c},$ and $(\cap A)^{c} = \cup A^{c},$

### Statement 1

Any intersection of closed sets is closed.

### Proof

Let point p be near $\cap F_{t},$ where $F_{t}$'s are closed sets for all values of parameter $t$ from some set $T.$ This means that every neighborhood of $p$ has a nonempty intersection with $\cap F_{t}$, and, therefore, with every $F_{t}.$ Since the latter are closed sets, $p\in F_{t}$ for all $t.$ Thus $p\in \cap F_{t}$.

### Statement 2

A finite union of closed sets is closed.

### Proof

Let $p$ be near $\cup F_{t},$ where $t\in T$ a finite set. $p$ must be near at least one of $F_{t}\mbox{'s}.$ For, if it were not the case, for every $t$ there would exist a neighborhood of $p$ that did not intersect $F_{t}.$ From the way the neighborhoods were defined, there would exist a neighborhood (take the smallest of the aforementioned neighborhoods) of $p$ that did not intersect any of $F_{t}\mbox{'s},$ nor would it intersect the union $\cup F_{t}$ in contradiction with nearness of $p$ to $\cup F_{t}.$ Hence $p$ is near one of $F_{t}\mbox{'s}.$ Therefore, $p$ belongs to that $F_{t},$ and, finally, $p\in \cup F_{t}.$ There are various ways to define a topology on a given space. One may start with neighborhoods and deduce from their properties properties of open and closed sets or those of the operation of closure. It's also possible to start with Statements 1 and 2 and the stipulation that the empty set and the whole space are closed, and define neighborhoods and open sets with all the expected properties of the latter. Of course, one can start with open sets as well. And, finally, neighborhoods must not be defined as balls in a metric space. The statements below capture the most important properties of the neighborhoods:

 (N) Each neighborhood of a point p contains p. The whole space is a neighborhood of all its points. A superset of a neighborhood of a point is a neighborhood of that point. The intersection of two neighborhoods of a point, is a neighborhood of that point. Each neighborhood of a point p contains a neighborhood of p which is also a neighborhood of each of its points.

Any family of sets that satisfy (N1-N4) can be used to define closed and open sets and other attributes of a topology. We now look at Fürstenberg's example.

Let $\mathbb{Z}$ be the set of all integers - positive, negative, and $0.$ For $a, b\in \mathbb{Z}$, $b > 0$ let

$N_{a, b} = \{ a + nb: n \in \mathbb{Z} \}.$

Each $N_{a, b}$ is a two-sided arithmetic progression. Note also that $a\in N_{a, b}$ for every b. $\mbox{"}N\mbox{"}$ in the definition is supposed to remind us of neighborhoods. Indeed, think of $N_{a, b}$ as those basic neighborhoods that are neighborhoods of each of their points (N4). Add to the collection of neighborhoods all supersets of $N_{a, b}\mbox{'s}$. With this definition we only need to verify property (N3). But note that $N_{a, b}\cap N_{a, c} = N_{a, \text{lcm}(b, c)},$ where $\mbox{lcm}(n,m)$ is the least common multiple of $n$ and $m$. From this N3 follows easily.

Call a set $U$ open if, for every $a\in U,$ there exists $b \gt 0,$ such that $N_{a, b}\subset U.$ We can check that Statements 1 and 2 hold as are their analogues for the open sets. Two facts are important:

 (O) Any non-empty open set is infinite. Besides being open, any set $N_{a, b}$ is also closed!

The first of these follows from the definition, the second from $N_{a, b} = \mathbb{Z}\space\setminus\cup N_{a+i, b},$ where the union is taken over $i = 1, 2, ..., b-1.$ $N_{a, b}$ is then closed as a complement of a finite union of open sets.

Now the punch line. Except for $\pm 1$ and $0,$ all integers have prime factors. Therefore each is contained in one or more $N_{0, p},$ where $p$ is prime. We thus arrive at the identity:

$\mathbb{Z}\space\setminus\{-1, 1\} = \cup N_{0, p},$

where union is taken over the set $\{p\}= \mathbf{P}$ of all primes. If the latter were finite, the right hand side would be closed as the union of a finite number of closed sets (O2). The set $\{-1, 1\}$ would then be open as a complement of a closed set. This would contradict (O1).

## Reference

1. M. Aigner, G. Ziegler, Proofs from THE BOOK, Springer, 2000
2. H. Fürstenberg, On the Infinitude of Primes, Amer. Math. Monthly 62 (1955), 353
3. K. Janich, Topology, UTM, Springer-Verlag, 1984  