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Statement : In $\mathbb{R}^{n}$, if $U$ is open and $C \subset U$ is compact, show that there is a compact set $D$ such that $C \subset$ int$(D)$ and $D \subset U$.

I want to see if the proof is correct, and more over if it can be shortened. Also, what happens outside of $\mathbb{R^n}$? What hypotheses are essential for the result to hold?

My attempt : Let Bd$(C)$ denote boundary of C. Fix $y \in U$. By Hausdorff, for each $x \in C$, there are open balls $T_{x,y} \ni x$ and $V_{x,y} \ni y$ that don't intersect. Since Bd$(C)$ is a closed subset of $C$, finitely many of these cover Bd$(C)$. Close the finitely many balls, $\overline{T_{x,y}}$. Then, $D = C \cup {\overline{T_{x,y}}}$ is closed, bounded (because there are finitely many balls), hence compact, and contains $C$ in its interior.

Avijit Dikey
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2 Answers2

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It is not clear why you need $V_{x,y}$. Just choose any open set $T_x$ containing $x$ whose closure is a compact setcontained in $U$ and then choose a subcover for $\partial C$ as you have done.

One small error in your proof is you did not say that $\overline {T_{x,y}}$ is a subset of $U$.

geetha290krm
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  • Yes, I see what you mean, this is also at the heart of the other answer posted by @psl2Z. But how do I assert the existence of a compact set containing $x$? Maybe by considering smaller open balls contained inside $T_{x,y}$ (follows from the openness of U)? – Avijit Dikey Sep 01 '22 at 10:51
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    @AvijitDikey The closure of any open ball around $x$ is compact since it is closed and bounded. – geetha290krm Sep 01 '22 at 11:05
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If the statement holds true in some space $X$ then $X$ has to be locally compact, i.e. every point has a compact neighborhood.

If $X$ is locally compact and Hausdorff then the statement holds true, since every point has a local base of compact neighborhoods.

$\mathbb{R}^n$ is locally compact and Hausdorff so the statement is true. Any infinite dimensional normed space for example is not locally compact so the statement does not hold there.

psl2Z
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