Is there a name for the open set there is some compact set that contains the set?

Also, if $X$ is a locally compact Hausdorff space, and $C$ be some compact set. If there exist a open set $U$ that contains $C$, I want to show that there exist a open set $D$ such that $C\subset D$ and $D\subset A$ for some compact set $A$.

I am thinking of covering the set $C$ with some finite open set $U_n$ with compact set $C_n$ that contains it since the space is locally compact. Then let $A=\cup U_n$ and $D=\cup C_n$ is a compact set such that $C \subset D$.

But this set $D$ can be not contained in $U$

I'm wondering if someone can help with the proof and the nomenclature.


2 Answers 2


It's not about the nomenclature (though in your setting such an open set is said to be "relatively compact"), but the idea of the proof of what you want.

You can choose for each $x \in C$ an open set $O_x$ such that $$x \in O_x \subseteq \overline{O_x} \text{ (which is compact) } \subseteq U$$

because in a locally compact Hausdorff space the compact neighbourhoods (or open sets with compact closure, aka relatively compact open sets) of $x$ form a local base.

$C$ being compact allows us to find a finite subcover of $O_x$'s and the union of these can be your $D$, and its closure (which you could call $A$) is compact (as the finite union of the corresponding $\overline{O_x}$) and of course sits inside $U$ still.

Confusing symbol choices: I'd say:

for a compact $C$ with an open set $C \subseteq U$ we can find an open set $V$ such that $\overline{V}$ is compact and such that $C \subseteq V \subseteq \overline{V} \subseteq U$.

There is no need for an extra name for the compact set $\overline{V}$. If $D \subseteq A$ and $A$ is compact we can just use that $\overline{D}$ is compact (as a closed subset of $A$, we do use Hausdorffness in that $A$ is closed, so $\overline{D} \subseteq \overline{A} = A$, etc). So in a Hausdorff space "an open set that is a subset of a compact set" is the same as "an open set with compact closure", and that last is (as I already mentioned) called a relatively compact set in this context, or maybe shorter " a compact neighbourhood (of $C$)".


I think the first thing you ask about is getting at the definition of local compactness. To me the phrase "locally compact" would initially mean that every point in a space has an open neighborhood containing a compact set. But this leads nowhere, as the point itself is compact. Instead you want to be able to say that the point is contained in some compact set with some extra space, i.e. the point should be in the interior of the compact set. An equivalent condition (which I think requires Hausdorff) is that every point has an open neighborhood with compact closure.

As an example, an infinite graph with finite valence is locally compact but not compact. An infinite graph with infinite valence at some vertex is not locally compact, because every neighborhood of the vertex contains portions of infinitely many edges.


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