Prove that the set of condensation points is closed I'm having a difficult time with this one. First off, $x$ is a condensation point of $A$ if every open nbhd $U$ of $x$ contains uncountably many points of $A$. Let us denote the set of condensation points as $A^c$ I have started a proof.
Proof. Let $X$ be a topological space and let $A \subset X$. We seek to show that $A^c$ is closed. To see this, we show that $X \setminus A^c$ is open.
Is this a good approach to the proof - to show closedness by showing openness of the complement? Also, I was wondering: does it follow that $X \setminus A^c$ contains only countably many points of $A$ or is that not the proper negation?
 A: Let $x_0\in\overline{S^c}$ and $x_o\in U\in\tau$.
Then $U\cap S^c \setminus \{x_o\}\neq \emptyset$
Let $x_1\in U\cap S^c \setminus \{x_o\}$
$x_1\in S^c$ and $x_1\in U$ implies $U\cap S$ contains an uncountable subset of $S$.
$x_0\in U\in\tau$ is arbitrary open set, hence $x_0\in S^c$
A: Let $p\in X\setminus A^c$. $p$ is not a condensation point of $A$, so there exists an open neighborhood $U$ of $p$ such that $U\cap A$ is countable. Let $q\in U$. It is clear that $U$ is an open neighborhood of $q$ such that $U\cap A$ is countable. This implies that $q\in X\setminus A^c$ and $U\subseteq X\setminus A^c$.
Finally, we can conclude that $X\setminus A^c$ is open, because we showed that all of its points are interior points.
A: We can show directly that $A^c$ is closed. Let $p\in\overline {A^c}.$ Let $U$ be any open nbhd of $p.$ We need to show that $U\cap A$ is uncountable and that therefore $p\in A^c.$ As follows:
There exists $q\in U\cap A^c$ because $p\in\overline {A^c}.$ Now U is also an open nbhd of $q,$ with $q\in A^c,$ so $U\cap A$ is uncountable.
