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I am referring to a question (and answer) given in Berkeley's Problems which I haven't understood properly.

$S$ is a subset of $\mathbb{R}$. $C$ is the collection of points $x$ in $R$ such that for every $\delta > 0$, ($x-\delta, x+\delta$) contains uncountably many points of $S$. Prove that $S \backslash C$ is countable (I call finite also as countable).

My attempt:

I only got so far as proving that $C$ is contained in $S^{\prime}$ as every basic open set around any point $x$ of $C$ contains infinitely many points of $S$ other than $x$.

My interpretation of Berkeley Answer:(may not be correct as I have not been able to understand it fully.)

In the answer, they say, suppose $y$ is a point of $S\backslash C$, then clearly $\exists$ $\delta > 0$ such that ($x-\delta, x+\delta$) (disjoint from $C$) contains only countably many points of $S$. By density of $Q$ in $R$, there is such an interval with rational endpoints. Why is this true? I believe that they should have written that such intervals can have only rational end points.

Then, they proceed by saying that since only countably many intervals exist with rational end points, $S\backslash C$ can contain only countably many points of $R$ as there are only a countable union of countably many points of $S$. Please help.

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    $\begingroup$ I'm not sure the statement is even correct (maybe I'm misreading); here you could take $S = \mathbb{R}$, $C = \{0\}$, and the desired conditions would be satisfied, but $S\setminus C = \mathbb{R}\setminus \{0\}$ which is not countable. $\endgroup$ – catfish Nov 11 '15 at 18:22
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    $\begingroup$ @catfish It should probably read "C is the collection of points ..." $\endgroup$ – zhw. Nov 11 '15 at 18:32
  • $\begingroup$ I think that once the $S$ is fixed, your $C$ is also fixed. In this case, $C$ has no choice but to become $R$. $\endgroup$ – user166305 Nov 11 '15 at 18:33
  • $\begingroup$ Not the way you wrote it. You should edit it. $\endgroup$ – zhw. Nov 11 '15 at 18:34
  • $\begingroup$ @zhw yes I'll make it 'the'. Thanks for the correction $\endgroup$ – user166305 Nov 11 '15 at 18:35
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Try this interpretation of the argument: Let $\cal I$ be the collection of all intervals $(r,s)$ with rational endpoints $r$ and $s$, that have the property that $(r,s)\cap S$ is countable. For any point $x$ that is in $S$ but fails to be in $C$, we see that $x$ must be in one of these intervals. Etc.

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