# A property about locally compact Hausdorff spaces

A space $$X$$ is said to be locally compact if for every point $$x$$ of $$X$$ there exists an open set $$U$$ and a compact set $$K$$, such that $$x\in U\subseteq K$$.

What I want to show is that (which I think is true)

If $$X$$ is a locally compact Hausdorff space, then for every point $$x$$ of $$X$$ and every compact set $$K$$ containing $$x$$ there exists an open set $$U$$, such that $$x \in U \subseteq K$$.

Things which might help:

1. A locally compact Hausdorff space is regular.

2. If $$X$$ is a locally compact Hausdorff space, then for every point $$x$$ of $$X$$ and every open set $$U$$ containing $$x$$ there exists a compact set $$K$$, such that $$x \in K \subseteq U$$.

Here is the proof of 2)

Let $$x$$ be a point of $$X$$ and let $$U$$ be an open neighbourhood of $$x$$. Then $$X\setminus U$$ is closed and $$x \notin X\setminus U$$. Since $$X$$ is regular, there exist open sets $$U_1$$ and $$U_2$$ such that $$x\in U_1$$ and $$X\setminus U \subseteq U_2$$ and $$U_1\cap U_2=\emptyset$$. Let $$K_1=X\setminus U_2$$. Then $$K_1$$ is closed and $$x\in U_1\subseteq K_1 \subseteq U$$. Since $$X$$ is locally compact, there exists an open set $$V$$ and a compact set $$K_2$$ such that $$x \in V \subseteq K_2$$. Since $$X$$ is Hausdorff, we have $$K=K_1 \cap K_2$$ is a compact set, such that $$x\in K \subseteq U$$.

Edit: If I can replace compact set with compact neighborhood in 2), then my purpose will suffice.

Edit 2: In proof of 2) $$x \in U_1 \cap V \subseteq K \subseteq U$$. Hence $$K$$ is a compact neighborhood of $$x$$.

• What is your definition of neighborhood?
– GuPe
Commented Jun 24, 2021 at 1:26
• @GuachoPerez I should have used "compact set" instead of "compact neighborhood" in my question. Otherwise, I could take the interior of K to be my open set U. I will edit my question. Commented Jun 24, 2021 at 1:32
• If $K$ is only compact then the claim is false. Just take $K = \{x\}$.
– GuPe
Commented Jun 24, 2021 at 1:33
• @GuachoPerez Thanks for your input. I think I can edit my proof of 2) to get the desired result without using the question that I had asked. Thanks for clearing my basic definitions of my neighborhood and open sets. Commented Jun 24, 2021 at 1:39

Your claim is false. Let's take $$\mathbb{R}$$ as our topological space for this example. $$1 \in [0,1] \subset \mathbb{R}$$, but no open (in $$\mathbb{R}$$) set $$U \subset [0,1]$$ containing 1.
If $$x \in K$$, $$K$$ compact, either $$x \in \operatorname{int}(K)$$ and then trivially $$U = \operatorname{int}(K)$$ works as your open set, or $$x \notin \operatorname{int}(K)$$ and then no such $$U$$ can exist (becuase it would make $$x$$ an interior point of $$K$$). So your proprosed property is not really relevant: it either trivially holds or cannot hold anyway (e.g. as in $$K=\{x\}$$ when $$x$$ is not an isolated point of $$X$$).
It has nothing to do with local compactness at all. Local compactness (as you define it) just says that every point is in the interior of at least one compact $$K$$, nothing more. But in general it won't be an interior point of any compact set it's in, as you wanted!