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)?
Countability of boundary points
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real-analysis
1 Answers
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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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0And more generally, any perfect set is uncountable and is the boundary of an open set (its complement). – 2012-03-23
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0@Robert: You mean nowhere dense. I am fairly certain that $[0,1]$ is perfect and is not the boundary of its complement. :-) – 2012-03-23
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0What is a strictly open interval? – 2012-03-23
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0@Matt: strictly open as opposed to one-sided open, I assume. But I don't know for sure. – 2012-03-23
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0Oh I see. It's vocabulary from the question. – 2012-03-23
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1Sorry, yes, I meant nowhere dense perfect set. – 2012-03-24
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0On this matter of points in the Cantor set that are not endpoints of any of the complements of deleted middle thirds, not that some of them are rational numbers: in particular $1/4$ and $3/10$ are members of the Cantor set. – 2012-09-28