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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.

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The argument as given in Counterexamples in Topology (examples 60/61, p. 82/83):

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.

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  • $\begingroup$ Can you elaborate on what exactly is the closure of $U_a(b)$? It's not very intuitive $\endgroup$ – Dean Gurvitz Jan 30 at 20:08
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    $\begingroup$ @DeanGurvitz Show that $U_a(b)$ intersects every basic neighbourhood of an integer of the form $ka$. $\endgroup$ – Henno Brandsma Jan 30 at 23:04

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