Better understanding of topology in ://math.stackexchange.com/q/3195705/506847. Ler $ (\mathbb{N^{*}}, T) $ be topologic space from Prove that there exists a unique topology $T$ for which $ P $ is its subbasis and that topological space $ ( T, \mathbb{N ^{*}})$ is metrisable ..
Find the interior, closure, accumulation points and borders of:
$$ A = \left\{\infty\right\}$$
$$ B = \left\{2, \infty\right\} $$
$$C= \mathbb{N} $$
$$D= \left\{2n \colon n \in \mathbb{N}\right\} $$
I am having trouble understanding the topology so any hints would help. All I managed to deduce is that sets $\left\{ n \right\} $ are open and that 
$\left\{\infty\right\}$  is not because there is always a lot more elements in the basis elements that contain it. 
 A: The topology is often known as the one-point compactification of $\Bbb N.$ Given a set $U\subseteq\Bbb N^*,$ we have that $U$ is open if and only if one of the following holds:


*

*$U\subseteq\Bbb N$

*$\Bbb N^*\setminus U$ is a closed compact (that is, a finite) subset of $\Bbb N$
Thus, given a set $K\subseteq\Bbb N^*,$ we have that $K$ is closed if and only if one of the following holds:


*

*$\infty\in K$

*$K$ is a finite subset of $\Bbb N$
Consequently, two of your sets are open (and not closed), and two of them are closed (and not open). The interiors and closures (and thus the borders) are fairly straightforward to find from there. As for accumulation points, the fact that $\{n\}$ is open for all $n\in\Bbb N$ means that no element of $\Bbb N$ can be a point of accumulation of any subset of $\Bbb N^*.$ However, $\infty$ will be a point of accumulation for two of your sets.
Let me know if you're struggling to figure out any of the particulars, or if you simply want to confirm your conclusions.
