Closure in the Co-Countable Topology on $\Bbb R$ I'm working on this question and was wondering if it's correct. I am working with the co-countable topology on $\mathbb{R}$. Consider $A \subset \mathbb{R}$ such that $A  = \mathbb{R} \setminus \{p\}$. I ned to find $\operatorname{cl}(A)$.
My answer/work:
Since $\{p\}$ is a countable set, by definition of a co-countable topology, $A$, being the complement of $\{p\}$, must be open. Hence, $\operatorname{int}(A) = \mathbb{R} \setminus \{p\}$. But what about the closure of $A$? Since $A$ is itself uncountable, it can only be 'contained' or closed in $\mathbb{R}$. Does that mean that $\operatorname{cl}(A) = \mathbb{R}$? What about $\{p\}$ then? Or is $\operatorname{cl}(A) = \mathbb{R} \cup \{p\}$? But isn't $\mathbb{R} \cup \{p\} = \mathbb{R}$?
 A: Yes, $\overline A=\Bbb R$, since $A$ is not closed, from which it follows that the only closed subset of $\Bbb R$ which contains $A$ is $\Bbb R$ itself.
A: Yes, it is the case that $\operatorname{cl}(A)=\mathbb{R}$. Here is an argument that works in more general contexts: by definition, a subset $X\subseteq\mathbb{R}$ is closed if and only if either $X$ is countable or $X=\mathbb{R}$. Now, $\operatorname{cl}(A)$ is a closed set containing $A$; since $A$ is uncountable, this leaves only one option for what $\operatorname{cl}(A)$ can be: namely, $\mathbb{R}$ itself. More generally, this same argument shows that $\operatorname{cl}(B)=\mathbb{R}$ for any uncountable subset $B\subseteq\mathbb{R}$.
A: The closure has to contain $A$, so there are only two options here: $\mathbb{R}$ or $\mathbb{R}\setminus\{p\}$. Since $\mathbb{R}\setminus\{p\}$ is not a closed set (its complement is clearly not open in this topology), it can't be the closure. So it has to be $\mathbb{R}$.
A: The closed sets are all sets that are at most countable, or $\Bbb R$. So the only closed superset of an uncountable set $A$ is $\Bbb R$ so
$$A \text{ uncountable } \implies \overline{A}=\Bbb R$$
in this topology....
