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I am having trouble showing the following.

Let $F \subset O \subset \mathbb{R}$, where $F$ is closed and $O$ is open. Prove that there is an open set $U$ such that $F \subset U$ and $\bar{U} \subset O$.

It seems so trivial, but I can't get a start on this question. Can I start with intervals?

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Is it somehow possible to give a slightly more meaningful title to this question? – Asaf Karagila Mar 22 '12 at 23:51
@AsafKaragila: How about this? – Nate Eldredge Mar 22 '12 at 23:57
up vote 3 down vote accepted

Every open subset of $\mathbb{R}$ is a countable union of disjoint open intervals, so you can at least start by considering $F\subset O=(a,b)$ for $a<b$.

Now let $a'=\inf F$ and $b'=\sup F$. Then $a'> a$, otherwise $F$ would not contain one of its accumulation points, and similarly $b'<b$.

This should be enough to help you find your new open set $\bar{U}$.

(It's maybe not immediate to go from this case to the general case, but I don't think it's too hard either)

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Since $O$ is open, for every $x \in F \subset O$ there exists $\varepsilon(x) > 0$ such that the ball $B_{\varepsilon(x)}(x)$ is contained in $O$. Then $U := \bigcup_{x \in F} B_{\varepsilon(x)/2}(x)$ satisfies $F \subset U \subset \overline U \subset O$.

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Does this work in every metric space? – you Mar 23 '12 at 0:04
Good question. Now I am not even sure whether it works in $\mathbb R$ because the closure of the union of open balls is not obviously equal to the union of the closed balls of the same radius... – marlu Mar 23 '12 at 1:00
I think that should read "not always equal": take the sequence of balls $(-1+\frac{1}{n},1-\frac{1}{n})$. The union of their closures is $(-1,1)$, but the closure of their union is $[-1,1]$. I also tried coming up with a counter example to your claim that "$\bar{U}$ is a strict subset of $O$" without success, but I haven't been able to prove it either... – you Mar 23 '12 at 2:12
@marlu: If $F$ is compact you can extract a finite open cover from your $U$ and then argue that the closure of the union of a finite number of open balls is the union of the closure. – Louis Mar 23 '12 at 11:32

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