Prove that the arithmetic progression topology on $\mathbb{Z}$ is Hausdorff. Prove that the arithmetic progression topology on $\mathbb{Z}$ is Hausdorff.
My thoughts on this:
Arithmetic progression topology is generated by the collection of basis elements: $\mathcal{A}=\{x+y\mathbb{Z}: x,y \in \mathbb{Z}, y\neq 0\}$. So fix a $y$ in this basis and suppose that $a,b$ are distinct integers. I claim that the disjoint neighborhoods in the topology are of the form:
$U=a+y\mathbb{Z}$ and $V=b+y\mathbb{Z}$. Showing this is pretty easy, but I'm wondering if ending the proof this way is sufficient to show that it is Hausdorff. Any guidance is appreciated. Thanks
 A: It's a bit more work than that claim: we have to start with distinct $a,b \in \Bbb Z$ and find two basic sets of the form $U(x,y):= x+y\Bbb Z$ with $y \neq 0$, such that $a \in U(x,y)$, $b \in U(x',y')$ and $U(x,y) \cap U(x',y') = \emptyset$.
We cannot just take $U(a,y)$ and $U(b,y)$ because if $b-a$ is a multiple of $y$ these sets are actually the same, and not disjoint; we do have that $a \in U(a,y), b \in U(b,y)$ so that's OK. We only have to ensure that we choose $y$ such that $b-a$ is not a multiple of $y$. We can choose $y$ with $|y| > (b-a)$ e.g. (we start with fixed but arbitrary $a,b$, so $b-a$ is some fixed number, so plenty such $y$ exist. We can choose it to be prime, if that suits us (that's a variant of this topology that I've also seen). But having such a $y$ we can see that $U(a,y) \cap U(b,y) = \emptyset$ and then we are done showing Hausdorffness.
(If $z \in U(a,y) \cap U(b,y)$ then $z = a + n_1 y = b + n_2$ for some $n_1,n_2 \in \Bbb Z$ so $(b-a)= (n_1- n_2)y$ and so $(b-a)$ is a multiple of $y$ which we ensured is not the case, so no intersection point exists, to be complete).
