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I want to show that if $C\subseteq \mathbb{R}$ is closed, then $C$ can be written as the union of countably many (or finitely many) disjoint closed intervals.

Note: If $C$ is itself a closed interval then this is trivially true, a bunch of people I have asked say $[0,1]$ is a counterexample, but it is not because $[0,1]=\cup\{[0,1]\}$ which is a finite union of disjoint closed intervals.

I know a similar theorem is true for open intervals.

  • 1
    To comment on the problem at hand, it is a common error when first dealing with the idea of closed and opened sets to think that "open sets are to open intervals as closed sets are to closed intervals." This is not the case. It is true that open intervals are open sets, and closed intervals are closed sets, but that is, for most practical purposes, the extent of the similarity.2011-10-21

2 Answers 2

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The Cantor set is a union of uncountably many points, but contains no closed intervals. So, it cannot be written was the union of countably many disjoint closed intervals.

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This result is false: any uncountable closed subset of $\mathbb{R}$ with empty interior (e.g. the standard Cantor set) gives a counterexample.