3
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ a. any finite set is countable. b. what have you tried ? $\endgroup$
    – Belgi
    Aug 16 '12 at 21:02
  • 1
    $\begingroup$ @Belgi: Many places mean "countably infinite" when saying "countable". To distinct they use "at most countable" for finite or countably infinite. $\endgroup$
    – Asaf Karagila
    Aug 16 '12 at 21:03
  • $\begingroup$ 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 . $\endgroup$
    – Ester
    Aug 16 '12 at 21:05
  • $\begingroup$ I also think there is a problem in the question: "Let $C$...s.t $S$" . should it be $C$ in the intersection ? $\endgroup$
    – Belgi
    Aug 16 '12 at 21:08
  • $\begingroup$ @Belgi: no, that would make for a circular definition... for example, $C=\emptyset$ would satisfy it. $\endgroup$
    – tomasz
    Aug 16 '12 at 21:09
2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 ...

$\endgroup$
3
  • $\begingroup$ Yes , I was trying exactly that , but I cannot show the disjoint part . I feel that I am missing something very easy . $\endgroup$
    – Ester
    Aug 16 '12 at 21:13
  • $\begingroup$ How is it possible that $x\in\emptyset$? Did you mean $x\in(p,q)$ and $(p,q)\cap C=\emptyset$, perhaps? $\endgroup$ Aug 16 '12 at 21:14
  • $\begingroup$ @Cameron: Yes; somehow I lost the middle characters. $\endgroup$ Aug 16 '12 at 21:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.