TITLE: Topology on Integers such that set of all Primes is open
QUESTION [7 upvotes]: In my topology homework we are asked to describe a topology on the Integers such that:

set of all Primes is open.
for each $x\in\mathbb Z$, the set $\{x\}$ is not open.
$\forall x,y \in\mathbb Z$ 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:

$U = \emptyset$; or
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!

REPLY [3 votes]: A cheap way to go about it is the following:

Let $\mathbb{P}$ denote the set of prime numbers, and $\mathbb{P}^\prime = \mathbb{Z} \setminus \mathbb{P}$.
As both $\mathbb{P}$ and $\mathbb{P}^\prime$ are countably infinite, pick bijections $f : \mathbb{P} \to \mathbb{Q}$, $g : \mathbb{P}^\prime \to \mathbb{Q}$.
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).