Prove that the closure of any uncountable subset of $X$ is all of $X$.

Question

Consider the countable complement topology on a set $X$, defined as follows: $\mathcal{T}_{\infty}=\{U|X\setminus U\text{ is infinite or empty or all of }X\}$. Show that the closure of any uncountable subset of $X$ is all of $X$.

I really have no idea where to start here, and would appreciate guidance. I am in my first Topology class, working with the Munkres text.

Answer 1

Let $C$ be an uncountable subset of $X$ and $D$ its closure, $X-D$ is open, it implies that $X-D$ is $X-F$ where $F$ is countable or $X-D$ is empty. Suppose that $X-D=X-F$ where $F$ is countable. This implies that $C\cap (X-F)$ is not empty otherwise $C\subset F$, but $C$ is not countable. But since $C\subset D$, $C\cap (X-D)=C\cap (X-F)$ is empty. Contradiction.

Answer 2

The only closed sets are $X$ and sets that are at most countable. If $A$ is uncountable the only closed set containing $A$ can be $X$, so $A$ is dense. So sets are either closed (finite or countable) or dense (when uncountable).