Understanding limit points given the topology $X = \mathbb R$ and $\mathcal T = \{ U\subset X | \> X - U \>\> \text{is X or is countable}\}$ This specific topology was already discussed in this problem, but I am having difficulty thinking about the limit points for a given subset under this topology ($X = \mathbb R$ and $\mathcal T = \{ U\subset X | \> X - U \>\> \text{is X or is countable}\}$). For example, if we consider an arbitrary interval $(a,b)$, what would its limit points be? 
Here is how I was thinking: given a set $A$, $\overline{A} = A \cup \text{limit points of } A$. It seems to me that every open set is either $\emptyset$ or uncountable. I wanted to consider the closure because that would give me a glimpse into its limit points. However, given the topology above, with $A = (a,b)$, I don't know what the closure "looks" like/what the set is. Any help would be appreciated.
 A: Note that by construction, the closed sets of $X$ are precisely and the countable subsets of $X$ and $X$ itself. On the other hand, the closure of a subset $A \subset X$ can be characterized as
$$
\overline{A} = \bigcap_{F \ \supset \ A\\ F \text{ closed}} F.
$$ 
In particular if $|A| > \aleph_0$, then $A$ is not contained in any countable set. Thus the only closed set which contains $A$ is $X$, and so $\overline{A} = X$. 
(Note that this is a very general fact, we haven't used that $X = \mathbb{R}$ anywhere).
You can think of it this way: since closed sets are countable, the open sets of $X$ will be all of $X$ except a countable amount of points. When $X$ is uncountable (as in your case), this tells us that the open sets of $X$ are "almost all the space" since closed sets are "small".  
Another way of thinking about this: take any subset $A \subset \mathbb{R}$. A point $p \in X$ will be a limit point of $A$ if for any open set $U \ni p$ we have $A \cap U \neq \emptyset$. In other terms, $p$ will be a limit point of $A$ if for each $C \subset X$ countable we have
$$
(X \setminus C) \cap A \neq \emptyset.
$$
Thus, if $p$ were  not to be a limit point, there should exist a countable set $C \subset X$ such that $A \subset C$. Once again, this is impossible if $A$ is uncountable, thus any point of $X$ is a limit point of $A$ in that case. If $A$ is countable, then it is closed, and so the only limit points are its own points.
