Prime Integer Topology is $T_2$ but not $T_3$

According to this $$\pi$$-Base page, the "Prime Integer Topology" is an example of a topological space which is $$T_2$$ but not $$T_3$$. The space is defined as $$(\mathbb{Z}^+,\tau)$$ where $$\tau$$ is the topology generated by a basis consisting of sets of the form: $$U_p(b)=\{b+np:n\in\mathbb{Z}\}$$ Where $$p$$ is prime and $$b$$ is a positive integer. I have managed to prove this is indeed a $$T_2$$ space since given positive integers $$y>x$$, if we denote $$d=x-y$$, and we denote by $$p$$ the smallest integer greater than $$d$$, then $$U_p(x)$$ and $$U_p(y)$$ are disjoint open neighborhoods.

However, I was unable to prove this space isn't a $$T_3$$ space. I managed to prove that the set of prime numbers together with $$1$$ is closed in this topology, but it didn't help me, and I'm not sure that's the right direction. In addition, my number theory background is somewhat limited so I may not have the right tools to solve this problem. Any hints or suggestions would be appreciated.

The closure of $$U_a(b)$$ contains all multiples of $$a$$ and so if we have two open sets $$U_a(b)$$ and $$U_c(d)$$ then their closures have all multiples of the least common multiple of $$a$$ and $$c$$ in common. (This also for when $$a$$ and $$c$$ even need not be prime, i.e. the relatively prime integer topology.)
In a Hausdorff $$T_3$$ space we always have that two distinct points have open neighbourhoods with disjoint closures ($$T_{2\frac12}$$ this is often called) and this is thus disproved by the above observation. As we already know the space is Hausdorff it thus cannot be regular.
• Can you elaborate on what exactly is the closure of $U_a(b)$? It's not very intuitive – Dean Gurvitz Jan 30 at 20:08
• @DeanGurvitz Show that $U_a(b)$ intersects every basic neighbourhood of an integer of the form $ka$. – Henno Brandsma Jan 30 at 23:04