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.

  • 1
    $\begingroup$ You have another problem there, $U^c$ may not be countable. $\endgroup$ – Klaus Jan 25 at 9:44
  • $\begingroup$ That is fine. The theorem works anyways. It is for any set of the reals. $\endgroup$ – The Bosco Jan 25 at 9:47
  • $\begingroup$ But you asked for a countable intersection. $\bigcap_{x\in U^c} I^c_x$ is not a countable intersection unless it is empty. $\endgroup$ – Klaus Jan 25 at 9:51
  • $\begingroup$ Why is that? I fail to see it $\endgroup$ – The Bosco Jan 25 at 9:53
  • 1
    $\begingroup$ 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. $\endgroup$ – Klaus Jan 25 at 10:06

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.


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