A Countable Set of Real Numbers Let $S$ be a subset of $\mathbb{R}$. Let $C$ be the set of points $x$ in $\mathbb{R}$ with the property that $S \cap (x - r, x + r)$ is uncountable for every $r > 0$. Show that $S - C$ is finite or countable.
Thanks for any help.
 A: HINT: You’re on the right track. Suppose that $x\in S\setminus C$; show that there is an open interval $(p,q)$ with rational endpoints such that $x\in(p,q)$ and $(p,q)\cap C=\varnothing$. There are only countably many such intervals, so ...
A: Instead of considering arbitrary neighborhood $(x - r, x + r)$ for $x \in \mathbb{R}$ and $r > 0$, you can consider just those open intervals where $x \in \mathbb{Q}$ and $r \in \mathbb{Q}$. These form a countable basis for the topology on $\mathbb{R}$. Let $(U_n)_{n \in \mathbb{N}}$ denote a countable enumeration of these open intervals. Then you have that $C$ is the set of all $x \in \mathbb{R}$ such that $S \cap U_n$ is uncountable for all $n$ such that $x \in U_n$. 
Hence $S - C$ is the union of over all $n \in \mathbb{N}$ of $S \cap U_n$ such that $S \cap U_n$ is countable, i.e.
$S - C = \bigcup_{n \text{ st } |S \cap U_n| = \aleph_0} S \cap U_n$
A countable union of countable sets is countable. You are done. 

By the way, the points in $C$ are usually called condensation points. If $S$ happens to be closed, the result above is a step in proving the Cantor Bendixson Theorem. This theorem states that every closed set is the union of a perfect set and a countable set. Perfects sets have cardinality $2^{\aleph_0}$. Hence the Cantor Bendixson Theorem states that no closed subset of $\mathbb{R}$ is a counterexample for the continuum hypothesis. 
