Divisor topology In my topology course we defined the following set
$\mathcal{B}=\{D_n:n\in\mathbb{N}_2\}$,
Where for each $n\in\mathbb{N}_2$ we define $D_n:=\{x\in\mathbb{N}_2:x\mid n\}$ and $\mathbb{N}_2=\{2,3,4,5,...\}$. The topology generated by $\mathcal{B}$ is called the divisor topology on $\mathbb{N}_2$.
Given $n\in\mathbb{N}_2$, determine the interior and accumulation points of $\{kn:k\in\mathbb{N}\}$.
I've tried the division algorithm to find a basis element for every point in the set, but it's been useless, any help is appreciated <3
 A: Hint. I will solve the first question and it should help you enough to deal with the second one.
I claim that $\cal B$ is a base of the topology.
Since $n \in D_n$, the elements of $\cal B$ cover $\Bbb N_2$. Moreover, let $D_s$ and $D_t$ in $\cal B$ and $x \in D_s \cap D_t$. Then $x | s$ and $x | t$, whence $x | \gcd(s,t)$. Thus $x \in D_{\gcd(s,t)} \subseteq D_s \cap D_t$, which proves the claim. It follows that every open set is a union of elements of $\cal B$.
Let $M_n = \{kn \mid k \in \Bbb N\}$ and let $r$ be in the interior of $M_n$. Then there is an $s$ such that $r \in B_s$ and $B_s \subseteq M_n$. This means that $r$ divides $s$ and that every divisor of $s$ in $\Bbb N_2$ is a multiple of $n$. In particular, if $p$ is a prime divisor of $r$, then $p$ is also a divisor of $s$, and hence a multiple of $n$. This is only possible if $n = p$. Moreover, since every prime divisor of $s$ must be a multiple of $p$, $s$ has to be a power of $p$. Finally, since $r$ divides $s$, $r$ also has to be a power of $p$.
It follows that if $n$ is not prime, the interior of $M_n$ is empty. If $p$ is prime, then for every $k > 0$, $B_{p^k} \subseteq M_p$ and thus the interior of $M_p$ is the set of powers of $p$.
