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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.

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a. any finite set is countable. b. what have you tried ? – Belgi Aug 16 '12 at 21:02
@Belgi: Many places mean "countably infinite" when saying "countable". To distinct they use "at most countable" for finite or countably infinite. – Asaf Karagila Aug 16 '12 at 21:03
I was trying to use the fact that Q is dense in R and that the intervals with rational endpoints is countable . By countable I mean countably infinite . – Ester Aug 16 '12 at 21:05
I also think there is a problem in the question: "Let $C$...s.t $S$" . should it be $C$ in the intersection ? – Belgi Aug 16 '12 at 21:08
@Belgi: no, that would make for a circular definition... for example, $C=\emptyset$ would satisfy it. – tomasz Aug 16 '12 at 21:09
up vote 1 down vote accepted

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.

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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 ...

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Yes , I was trying exactly that , but I cannot show the disjoint part . I feel that I am missing something very easy . – Ester Aug 16 '12 at 21:13
How is it possible that $x\in\emptyset$? Did you mean $x\in(p,q)$ and $(p,q)\cap C=\emptyset$, perhaps? – Cameron Buie Aug 16 '12 at 21:14
@Cameron: Yes; somehow I lost the middle characters. – Brian M. Scott Aug 16 '12 at 21:38

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