# Proving that any closed subset $U \subset \mathbb{R}$ is a countable intersection of open sets

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.

• You have another problem there, $U^c$ may not be countable. – Klaus Jan 25 at 9:44
• That is fine. The theorem works anyways. It is for any set of the reals. – The Bosco Jan 25 at 9:47
• 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 at 9:51
• Why is that? I fail to see it – The Bosco Jan 25 at 9:53
• 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 at 10:06

## 1 Answer

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.