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Suppose we have a bounded open set $O \subset R^n$. Let $F$ be a closed subset in the subspace $O$ which is bounded away from the boundary of $O$. That is, there exists $\delta > 0$, for every $y \in F$, we have \begin{align*} \|x - y\| > \delta, \end{align*} for all $x \in \partial O$ (boundary of $O$).

I am thinking whether this will imply $F$ is compact. Boundedness is not an issue and we only need to make sure $F$ is closed in $\mathbb R^n$. Intuitively, I think the set that is closed in subspace $O$ but not in $\mathbb R^n$ must have the property: $\partial F \cap \partial O \neq \emptyset$, where we consider $F$ and $O$ as both subsets of $\mathbb R^n$. I made up this condition and wonder whether we can prove this? If this condition is not correct, what kind of reasonable condition would guarantee $F$ is compact?

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Let $(x_n)_{n\in\mathbb N}$ be a convergent sequence of elements of $F$ and let $x$ be its limit. I will prove that $x\in F$. If $x\in O$, then $x$ in $F$, since $F$ is a closed subset of $O$. And if $x\not\in O$, then $x\in\partial O$, since $(x_n)_{n\in\mathbb N}$ is a sequence of elements of $O$. But this is impossible, since $F$ is bounded away from the boundary of $O$.

Since each convergent sequence of elements of $F$ converges to an element of $F$, $F$ is closed (in $\mathbb{R}^n$). And obviously (as you wrote) $F$ is bounded. Therefore, $F$ is compact.

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  • $\begingroup$ Why is O a closed subset of F? $\endgroup$ – William Elliot May 16 '18 at 6:11
  • $\begingroup$ @WilliamElliot I made I mistake. What I meant was that $F$ is a closed subset of $O$. $\endgroup$ – José Carlos Santos May 16 '18 at 6:12
  • $\begingroup$ Thanks for your answer. Is the intuition I mentioned in the problem right: a subset $E$ is closed in subspace $O$ but not closed in $\mathbb R^n$ if and only if $\partial E \cap \partial O \neq \emptyset$? $\endgroup$ – user1101010 May 16 '18 at 6:14
  • $\begingroup$ @iris2017 If $O$ is bounded, that is correct. $\endgroup$ – José Carlos Santos May 16 '18 at 6:16
  • $\begingroup$ @JoséCarlosSantos: Thanks. $\endgroup$ – user1101010 May 16 '18 at 6:18

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