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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.

Thank you for your help!

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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.

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  • $\begingroup$ (to add an example) Consider $\{0\}\cup\{1\}$ which is closed but not an interval. $\endgroup$ – Alberto Takase Feb 20 '19 at 9:15
  • $\begingroup$ @AlbertoTakase Yes, but that is not a countable intersection of open intervals. $\endgroup$ – Brahadeesh Feb 20 '19 at 9:16
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    $\begingroup$ I only wanted to give an example of the last remark in your answer. $\endgroup$ – Alberto Takase Feb 20 '19 at 9:18
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    $\begingroup$ @RobertShore Not any set of isolated points can be a countable intersection of open intervals, only singletons. And singletons are also (trivially) closed intervals. $\endgroup$ – Brahadeesh Feb 20 '19 at 9:37
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    $\begingroup$ @LeAnhDung Glad to be of help :) $\endgroup$ – Brahadeesh Feb 20 '19 at 9:38

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