Is this a $T_1$ topological space and what other properties (or name) does this topological space on the set of prime powers have?

Question

This question is motivated by the following answer: https://mathoverflow.net/a/432059/165920

Let $\hat{\mathbb{P}} := \{ p^k | k\ge 1 , p \text{ prime } \} = \{ 2, 3, 4, 5, 7, 9, 11, 13,16,\cdots \}$ be the set of prime powers and let

$$S_a := \{p^k:1\le k \le v_p(a),p \text{ prime }\}$$

We can define a topology on this set using the sets $S_a$ as the closed sets:

$$S_1 = \emptyset , S_0 = \hat{\mathbb{P}}$$

and if

$$I \subset \mathbb{N}_0$$

then

$$\cap_{i \in I} S_i = S_{\gcd(I)}$$

so the intersection of any collection of closed sets is closed. (By Bezout's theorem, the $\gcd$ of an infinte set of numbers can be defined.)

Then we have:

$$S_a \cup S_b = S_{ab/\gcd(a,b)}$$

$$S_a \cap S_b = S_{\gcd(a,b)}$$

Since

$$S_{a_1} \cup S_{a_2} = S_{\frac{a_1 a_2}{\gcd(a_1,a_2)}}$$

we conclude by induction on the number of closed sets in a finite collection, that their union is also a closed set.

The topology is not Hausdorff ($T_2$) but it is irreducible (https://en.wikipedia.org/wiki/Hyperconnected_space):

Suppose that $q_1 \neq q_2, q_1,q_2 \in \hat{\mathbb{P}}$ and there exist open sets

$q_1 \in U_1 = \hat{\mathbb{P}}-S_{a_1},q_2 \in U_2= \hat{\mathbb{P}}-S_{a_2}$ such that:

$$\emptyset = U_1 \cap U_2 = ( \hat{\mathbb{P}}-S_{a_1}) \cap ( \hat{\mathbb{P}}-S_{a_2}) = \hat{\mathbb{P}}-(S_{a_1} \cup S_{a_2})$$

$$=\hat{\mathbb{P}}-S_n \neq \emptyset$$

where $n = \frac{a_1 a_2}{\gcd(a_1,a_2)}$ and the last $\neq$ is because the set on the right is a difference between an infinite set and a finite subset, hence not empty.

The topology is $T_0$:

Let $q_1 \neq q_2, q_1,q_2 \in \hat{\mathbb{P}}$.

First case: $\gcd(q_1,q_2)=1$. Let $U := \hat{\mathbb{P}}-S_{q_2}$. Then $q_1 \in U$ and $q_2 \notin U$. Second case: $\gcd(q_1,q_2)>1$. Without loss of generality assume that $q_1 < q_2$. Then, since both $q_1,q_2$ are prime powers with a non-trivial $\gcd$, we must have $q_1|q_2$. Set $U := \hat{\mathbb{P}}-S_{q_1}$. Then $q_2 \in U$ and $q_1 \notin U$. Hence the topological space is $T_0$.

Questions:

1. Is this a $T_1$ space?

2. Does this topological space maybe have a name?

3. What other properties does it have? ( https://en.wikipedia.org/wiki/Topological_property )

Answer 1

$T_1$ spaces have closed singletons. However here given any prime number $p$, the subset $\{p^2\}$ is not closed.

For instance any $a$ such that $S_a$ contains $4$, has $\nu_2(a) \geq 2$ so $2 \in S_a$. Therefore $$\{4\} \neq \bigcap_{a} S_a$$ where the intersection is on all closed sets containing $4$.

---

About the topology, the set $\hat{\mathbb P}$ is a preorder for divisibility. The closed sets satisfy the "lower set" property : if $F$ is closed, $x \in F$ and $y$ divises $x$ then $y \in F$.

It looks like an instance of Alexandrov topology on preorders.