Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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. –  Daniel 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

1 Answer 1

up vote 8 down vote accepted

Closed sets are the inverse of open sets. So the correct way to write it would be the intersection of closed sets; however intersecting disjoint sets is boring. So the correct way to think about it would be that every closed set can be written as an intersection of closed sets whose complements are pairwise disjoint.

If you are indeed interested in unions then note that every closed set is the union of itself, trivially. If you want to ask about the union of closed intervals, then the answer is no. Since some closed sets (e.g. the Cantor set) does not contain any interval. Unless you consider singletons as closed intervals, in which case you would need uncountably many and the question becomes somewhat moot (every set is the union of singletons).

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.