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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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    @AsafKaragila: How about this?2012-03-22

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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.

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'.

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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    @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.2012-03-23