# Can every closed subset of real numbers be written as the countable (or finite) union of disjoint closed intervals?

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.

• The corresponding thing that is true is that any closed set is the intersection of countably many closed intervals. – Joe Johnson 126 Oct 21 '11 at 16:40
• @Joe: What? Any intersection of closed intervals is itself a closed interval. – Chris Eagle Oct 21 '11 at 16:42
• @Joe: That intersection is empty. – Chris Eagle Oct 21 '11 at 16:52
• @Joe: As Chris points out, that is not true. The real "dual" is that any closed set can bet written as the countable intersection of sets of the form $(-\infty,a]\cup[b,+\infty)$, which are just the complements of the open intervals. – Thomas Andrews Oct 21 '11 at 16:56
• 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. – Thomas Andrews Oct 21 '11 at 17:15

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