# Can closed sets in real line be written as a union of disjoint closed intervals?

It is known that open sets in real line can be written as a countable union of disjoint open intervals. (link) I'm curious that if there is similar statements for closed sets in real line.

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Not as a union, no, as nowhere dense closed sets (such as the Cantor set) demonstrate. However, one could use complements to arrive at an analogous result for closed sets. –  Jonathan Y. Nov 8 '13 at 14:12
If a closed interval is defined to be a set $[a,b]$ with $a< b$, then the most obvious counterexample would be a singleton in $\mathbb{R}$. If you allow $a\leq b$ then we still get a counterexample from more exotic spaces such as the cantor space as @JonathanY. points out. –  Dan Rust Nov 8 '13 at 14:25
The title of your question makes no sense. Any closed set is a union of one closed set (itself). Please fix the title so it describes your question better. Evidently someone figured out what you meant and gave a good answer but you should improve the title for the sake of posterity. –  Stefan Smith Nov 9 '13 at 0:38

Moreover Sierpinski proved that if $\Bbb R$ (or any Baire space) is the countable union of disjoint closed sets then exactly one is non-empty. So you cannot get a nontrivial result using union of disjoint closed sets.
Nice. Also of some interest, perhaps, is that every closed set is a $G_\delta$ set, i.e., the intersection of countably many open sets. –  Jonathan Y. Nov 8 '13 at 14:30