# Topology on Integers such that set of all Primes is open

In my topology homework we are asked to describe a topology on the Integers such that:

1. set of all Primes is open.
2. for each $x\in\mathbb Z$, the set $\{x\}$ is not open.
3. $\forall x,y \in\mathbb Z$ distinct, there is an open $U\ni x$ and an open $V\ni y$ such that $U\cap V=\emptyset$

i was looking at Furstenberg's topology as in this proof: For $m, b \in\mathbb Z$ with $m > 0$ define $N(m,b):=\{mx + b : x ∈ Z\}$, an arithmetic progression streching towards infinity in both directions. A set $U$ is open if either:

1. $U = \emptyset$; or
2. For each $b\in U$ there is an $m>0$ such that $N(m,b)\subseteq U$.

but as I understand the set of Primes is not open in this topology. Now I'm not sure what should I do: is there a way to modify this topology to make set of Primes open or should I think of something completely different.

Any hints are appreciated! thanks!

• An idea for you to try: define a base (or perhaps subbase...?) for the wanted topology with sets that contain an infinite number of primes. If this possibly generates a topology then the set of all primes would be open, but I can't say off the top of my head whether it'd fulfill the other requirements – DonAntonio May 25 '13 at 9:25
• The above won't work: the intersection of two such sets could have a finite number of primes (zero primes is fine and can fit within the definition if we add it, say for the empty set). Perhaps the topology resulting from requiring that the complement of a set contains a finite number of primes would work? – DonAntonio May 25 '13 at 9:36
• This won't work either, because the complement of $\{p\}$ for any prime $p$ would have to be open, so that $\{p\}$ is open itself. – user79202 May 25 '13 at 9:50

1. Let $\mathbb{P}$ denote the set of prime numbers, and $\mathbb{P}^\prime = \mathbb{Z} \setminus \mathbb{P}$.
2. As both $\mathbb{P}$ and $\mathbb{P}^\prime$ are countably infinite, pick bijections $f : \mathbb{P} \to \mathbb{Q}$, $g : \mathbb{P}^\prime \to \mathbb{Q}$.
3. Define $U \subseteq \mathbb{Z}$ to be open iff $f [ U \cap \mathbb{P} ]$ and $g [ U \cap \mathbb{P}^\prime ]$ are both open subsets of $\mathbb{Q}$ (under the subspace topology).
• @Torben: $\mathbb{Q}$ is given the topology it inherits as a subspace of $\mathbb{R}$: $U \subseteq \mathbb{Q}$ is open iff there is an open $W \subseteq \mathbb{R}$ such that $U = W \cap \mathbb{Q}$. – user642796 May 26 '13 at 3:34