I am quite new to topology and I am particularly interested in gaining an intuitive understanding for the following proof:

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I am wondering if someone would be able to slow down the sequence of thoughts here so that I can put more of the puzzle together. For example:

  1. In which sense is the described topology "metrizable"

  2. How can arithmetic progressions in $(-\infty,\infty)$ be both open and closed (and I don't fully get why this is implied via the complement of the union). By consequence why does this imply finite progressions are closed?

  3. How does all this help to build the picture of the final conclusion.

I am "ok" with the basics of topology / measure theory / diff. geo - just in case you need to gauge how much you need to tailor the answer.

  • 1
    $\begingroup$ For (2) it might help to think of an example: the complement of $\{\ldots, -6, -3, 0, 3, 6, \ldots\}$ is the union of $\{\ldots, -5, -2, 1, 4, \ldots\}$ and $\{\ldots,-4, -1, 2, 5,\ldots\}$ and thus open, which means $\{\ldots, -6,-3,0,3,6,\ldots\}$ is itself closed. $\endgroup$ – Elliot G Mar 20 '20 at 5:49

Let $X=\Bbb Z$ where the set of all $B(a,n)=\{a+kn: k \in \Bbb Z\}$, where $a,n \in \Bbb Z$ with $n \neq 0$, forms a base for a topology on $\Bbb Z$. The $B(a,n)$ is an arithmetic progression with difference $n$.

Checking the two standard demands on a base: $X=B(0,1)$ so $X$ is covered by base elements (or every $a\in X$ is in $B(a,k)$ for any $k \neq 0$ can also be used).

If $a \in B(a_1, k_1) \cap B(a_2,k_2)$ then $$a \in B(a,\gcd(k_1,k_2)) \subseteq B(a_1, k_1) \cap B(a_2,k_2)$$

so that the second property (about intersections) is also satisfied. See here or Munkres e.g.

So the arithmetic progressions form a base for a unique topology, in which the $B(a,k)$ and their unions form the open sets.

If $a \neq b$ then $B(a,k) \cap B(b,k)$ form disjoint open neighbourhoods of $a$ and $b$ if $k > (b-a)$ e.g. So $X$ is Hausdorff in this topology.

$$X\setminus B(a,k) = \bigcup\{B(a+i,k): i=1,\ldots,|k|-1\}$$

so that the complement of each $B(a,k)$ is open and all $B(a,k)$ are open-and-closed. It follows that $X$ is Tychonoff and regular (even normal, as $X$ is countable and hence Lindelöf) and as it also has a countable base, Urysohn's metrisation theorem (explained here) tells us it's metrisable. But this is only "nice to know", not relevant to the proof itself. We only need that the $B(a,n)$ are also closed for the actual "prime proof".

Now observe that all open sets are infinite as unions of infinite sets $B(a,k)$.

If $P$ is the set of primes

$$X\setminus\{-1,1\} = \bigcup_{p \in P} B(0,p)$$

as each integer $\neq 1,-1$ has a prime divisor.

And if $P$ were finite, the union on the right hand set is closed (as a finite union of closed sets) and so $\{-1,1\}$ is open, contradiction as it's finite, so cannot be open.


  • $\begingroup$ As extra extra info: $\Bbb Z$ in this topology is homeomorphic to $\Bbb Q$ as it is a countable metric space without isolated points (a classic theorem due to Sierpiński). $\endgroup$ – Henno Brandsma Mar 20 '20 at 9:27

I will make things clearer.

Define $A(a,d) \subset S$ as $A(a,d) = \{a+kd : k \in S\}$, the arithmetic progression containing the element $a$ with common difference $d$. Now, form the set of subsets $\{A(a,d) : a , d \in S, \color{red}{d \neq 0}\}$. This is a set of subsets, the most important thing to note is that every $A(a,d)$ is an infinite set.

Now, we form a topology by using these as open sets. So an open set in $S$ is now defined as an arbitrary union of the $A(a,d)$. You can verify the second condition of the basis yourself.

So the $A(a,d)$ are open , obviously because they are in the basis forming the open sets. Why are they closed?

An example first : Why is $A(1,2)$ closed? Because $A(1,2) \cup A(0,2) = S$ (Every integer on division by $2$ leaves a remainder $0$ or $1$). So, $A(1,2) = A(0,2)^C$ , and because $A(0,2)$ is open, $A(1,2)$ is closed as the complement of an open set is closed.

Similarly, note that $\cup_{i=1}^d A(a+i,d) = S$ for any $a,d$. Thus, taking whatever open sets you want to the other side with complement, you can see that any $A(a,d)$ is closed, because it is the complement of a union of open sets, which is an open set itself.

Now, a finite union of closed sets is always closed. However, an infinite union need not be closed.

Also, note that every closed set is infinite, because it is the complement of an open set : and every open set that does not contain an element misses at least one arithmetic progression containing it.

This is not necessary, but I'll still cover it.

Why is this topology metrizable? We look at the Theorem of Urysohn in this regard :

Every regular space with a countable basis is metrizable.

So we need to look at why this space is regular. That it has a countable basis is clear from the fact that the chosen basis itself is countable (indexed by $S \times S$, for example).

But why is it regular? For this, simply note that it is Hausdorff : given $m,n \in \mathbb S$, take $t > |m-n|$, then $A(m,t)$ and $A(n,t)$ are disjoint open sets containing $m,n$ respectively.

For regularity, we note that Hausdorff spaces become regular under this condition : given $x$ and $x \in U$ open, there is $x \in V \subset U$ such that $\bar V \subset U$ and $V$ is open.

But this is obvious : if $x \in U$ then $U$ being a union of APs, contains some AP having $x$ as an element. That AP can be $V$, which is open and closed. Hence, from the Urysohn metrization theorem, it is clear that the space is metrizable.

As for the final flourish : $\{-1,1\}$ cannot be open. This is because of the comment regarding the infinitude of any open set. For the same reason, $\cup_p A(0,p)$ cannot be closed.

Therefore, the union $\cup_p A(0,p)$ must be infinite, because if it were finite,then the set would be closed, a contradiction to its complement not being open!

That concludes the proof : the union is infinite implies that there are infinitely many distinct primes.

This topology is called the Furstenberg topology.

Also read up on the related Golomb space.


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