# Is a set of prime numbers open in Furstenberg's topology?

I was reading Furstenberg's Proof of Infinitude of Primes and I wonder if a set of prime numbers is open in this topology.

Thanks!

• Better to make your question self-contained by including the definition of the topology. – Gerry Myerson May 25 '13 at 8:53

As each sequence $S(a,b) = a \mathbb{Z} + b$ contains non-primes, i.e. $b | ab + b$, this cannot be true.

A set in any topological space is open iff for every point in it there exists an open set containing it and contained in the set.

Now, can you find an open set contained in "a set" of primes? In fact, can you find a basic open set $\,a\Bbb Z+b\subset\Bbb Z\,$ contained in a set (any set) of primes?

In order for $\Bbb P$, the set of prime numbers, to be open in the Furstenberg topology on $\bf N$, it must be the union of the basis sets, namely a union of infinite arithmetic sequences. Since $\Bbb P$ is nonempty, this means it would have to contain at least one such sequence. However this is not possible, since any infinite arithmetic sequence contains a composite number.

Let $\{an+d:n\in{\bf Z}\}\cap{\bf N}$ be any such arithmetic sequence. Then $(an+d)(am+1)$ is composite for sufficiently large choices of $n,m$, and it can be written as $a(anm+n+dm)+d$ which is in the arithmetic sequence. Thus, every such arithmetic sequence contains a composite number.

If you work over $\bf Z$ instead, then a much simpler argument works: every infinite arithmetic sequence contains negative integers, but the set of primes contains no negative integers.

These same arguments apply to any subset of $\Bbb P$, i.e. any nonempty set of prime numbers.

Every non-empty open set $U$ of $\Bbb{Z}$ contains a basis $a + b \Bbb{Z}$ for some $b > 0$. In particular,

$$\liminf_{n\to\infty} \frac{\# (U \cap [-n, n])}{2n+1} \geq \frac{1}{b} > 0.$$

But if $\Bbb{P}$ denotes the set of primes, it is not hard to prove that

$$\liminf_{n\to\infty} \frac{\# (\Bbb{P} \cap [-n, n])}{2n+1} = \prod_{p \in \Bbb{P}} \left( 1 - \frac{1}{p} \right) = 0.$$

Therefore $\Bbb{P}$ is not open in $\Bbb{Z}$.