For the set of integers $ \geq 2 $, define topology to be $ U_k=\{n: n|k, n\in\mathbb{Z}_{\geq 2} \} $ for every $k \in \mathbb{Z}_{\geq 2} $ and their unions. Then what is the closure of $ \{n\} $?
My attempt: In order to find the closure of $ \{ n \} $, we need to find all the limit points of $ \{ n \} $, i.e. $$ x\in\overline{\{ n \}}\longleftrightarrow \text{$ \forall $ open set $ U $ containing $ x $}, U\cap\{ n \}\neq\emptyset. $$
The set $ \{ sn: s\in\mathbb{Z}_{\geq 1} \} $ satisfies the condition;
For every point $ y \in \{ sn:s\in\mathbb{Z}_{\geq 1} \}^{C} $, we can always find an open set $ U_{y} $ not containging $ n $.
So we conclude that $ \overline{\{ n \}}=\{ sn: s\in\mathbb{Z}_{\geq 1} \} $.
Question: Is my solution right? I am also looking for different approaches. Thanks~
You want to show that
$$\overline{\{n\}} = \{ln: l \in \mathbb{Z}_{\ge 1}\}$$
The right to left is clear: if we have $ln$ in some open set $U_k$ then $ln$ divides $k$, so $n$ divides $k$ and so $U_k \cap \{n\} \neq \emptyset$. As this holds for all open sets $U_k$, $ln \in \overline{\{n\}}$. Reversely, if $m$ is not a multiple of $n$, then $U_m$ is a neighbourhood of $m$ ($m | m$), that does not contain $n$ as $n$ does not divide $m$ (and $U_m$ contains only the divisors of $m$). So $m \notin \overline{\{n\}}$. This shows the equality.