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I got stuck in this proof.

First, we know that if $U$ is closed, then the complement $U^c$ is open.

Therefore, what I tried to do is use one of my previous results in another question. This is, that every open set in $\mathbb{R}$ is a countable union of disjoint open intervals.

Using the previous statement, simple set theory and De Morgan's Laws, we get

$$U = (U^c)^c = \bigg( \bigcup_{x\in U^c} I_x\bigg)^c = \bigg( \bigcap_{x\in U^c} I^c_x\bigg)$$

where $I_x$ are the intervals of the open set $U^c$. My problem is that we know that the intervals $I_x$ of $U^c$ are open, and, therefore the intervals $I^c_x$ are closed, which is the opposite of what I was trying to obtain.

I know there's another answer. Nevertheless, it is completely different to what I am trying to do. I want to know why this method does not work.

The Bosco
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    You have another problem there, $U^c$ may not be countable. – Klaus Jan 25 '19 at 09:44
  • That is fine. The theorem works anyways. It is for any set of the reals. – The Bosco Jan 25 '19 at 09:47
  • But you asked for a countable intersection. $\bigcap_{x\in U^c} I^c_x$ is not a countable intersection unless it is empty. – Klaus Jan 25 '19 at 09:51
  • Why is that? I fail to see it – The Bosco Jan 25 '19 at 09:53
  • You assumed $U^c$ to be open. Open sets are never countable unless they are empty. – Klaus Jan 25 '19 at 09:54
  • I am not saying the set is countable, I am invoking this theorem: https://proofwiki.org/wiki/Open_Sets_in_Real_Number_Line – The Bosco Jan 25 '19 at 09:55
  • You intersect over all $I_x^c$, where $x \in U^c$, i.e. over uncountably many intervals. – Klaus Jan 25 '19 at 09:57
  • For example, this source uses the same argument https://www.math.wustl.edu/~victor/classes/ma4111/s3.pdf – The Bosco Jan 25 '19 at 10:00
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    No, check their indices. They always have an intersection over countably many sets ($k = 1$ to $\infty$). You intersect over uncountably many sets, but claim you want a countable intersection. – Klaus Jan 25 '19 at 10:06
  • The cause of the confusion is that you index your intervals over $U^c$, which in general is uncountable. It seems implicit in your argument that $I_x$ is any interval around $x$ that is contained in $U^c$. The resulting collection is not countable. The solution set that you cite uses another result, stating that any open set is a countable union of intervals (in $\mathbb R$). It is simply misleading to index this union over $x$. – o.h. Jan 25 '19 at 10:09
  • You say "the intervals $I_x^c$". But they are not intervals. – TonyK Jan 25 '19 at 10:23

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Every open interval $(a,b)$ is a countable union of closed (but not disjoint) intervals, for instance $\bigcup_{n\ge 3}[a+(b-a)/n,b-(b-a)/n]$. And a countable union of countable unions is a countable union. So every open set in $\Bbb R$ is a countable union of closed intervals. Take it from there.

TonyK
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