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Let $V=\cup_n I_n$ be a countable union of intervals in $\mathbb{R}$. Is the set of boundary points of $V$ countable? What if the intervals are strictly open (does this even make a difference)?

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up vote 11 down vote accepted

No. Every open subset of $\mathbb{R}$ is the countable union of (strictly) open intervals (you can make them disjoint if you want). The complement of the Cantor set has the Cantor set as boundary, which is uncountable.

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And more generally, any perfect set is uncountable and is the boundary of an open set (its complement). – Robert Israel Mar 23 '12 at 21:29
@Robert: You mean nowhere dense. I am fairly certain that $[0,1]$ is perfect and is not the boundary of its complement. :-) – Asaf Karagila Mar 23 '12 at 21:48
What is a strictly open interval? – Rudy the Reindeer Mar 23 '12 at 21:54
@Matt: strictly open as opposed to one-sided open, I assume. But I don't know for sure. – t.b. Mar 23 '12 at 21:56
Sorry, yes, I meant nowhere dense perfect set. – Robert Israel Mar 24 '12 at 1:30

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