# Is it true that every closed set is a countable intersection of open intervals?

Motivation:

We know that every open set is a countable union of open intervals with rational endpoints and that every open interval is a countable union of closed intervals. Hence every open set is a countable union of closed intervals. It follows by De Morgan's laws that every closed set is a countable intersection of open sets.

I would like to ask if we can prove a stronger result that

every closed set is a countable intersection of open intervals.

The answer is no, since an intersection of intervals is also an interval. Thus, if a closed set were to be a countable intersection of open intervals, it would have to be a closed interval, but there are closed sets that are not intervals.

For example (as @AlbertoTakase mentions in the comments below), consider the set $$\{0\} \cup \{1\}$$. This is a closed set since finite subsets of $$\Bbb{R}$$ are closed, but it is clearly not a closed interval. Hence, it cannot be written as a countable intersection of open intervals.

• (to add an example) Consider $\{0\}\cup\{1\}$ which is closed but not an interval. Feb 20, 2019 at 9:15
• @AlbertoTakase Yes, but that is not a countable intersection of open intervals.
– user279515
Feb 20, 2019 at 9:16
• I only wanted to give an example of the last remark in your answer. Feb 20, 2019 at 9:18
• @RobertShore Not any set of isolated points can be a countable intersection of open intervals, only singletons. And singletons are also (trivially) closed intervals.
– user279515
Feb 20, 2019 at 9:37
• @LeAnhDung Glad to be of help :)
– user279515
Feb 20, 2019 at 9:38