Closed and open subsets of locally compact Hausdorff space are locally compact

These questions have been asked to death, but I found the proof for "open subsets of locally compact Hausdorff space are locally compact" too tedious in each of the answers I have sampled on MSE, for example:

In a locally compact Hausdorff space, why are open subsets locally compact?

Open subspaces of locally compact Hausdorff spaces are locally compact

I think the proof for "closed subsets of locally compact Hausdorff space are locally compact" is very brief and elegant. I want to obtain a similar proof for open subsets.

The definition of locally compact I am sing is:

A space $(X, \mathfrak{T})$ is locally compact if $\forall x \in X$, $\exists K, U \subseteq X, K$ is compact, $U$ is open, s.t. $x \in U \subseteq K$

(Proof 1: Closed subsets of locally compact Hausdorff space are locally compact)

Proof:

1. Let $(X,\mathfrak{T})$ be a locally compact Hausdorff space.

2. Then $\forall x \in X, \exists K$, $K$ compact, $U \in \mathfrak{T}$ s.t. $x \in U \subseteq K$.

3. Let $C$ be closed, then $\forall x \in C, x \in U \cap C \subseteq C \cap K \subseteq C$.

4. Since $U \cap C$ is open, $C \cap K$ is compact, therefore $C$ is locally Hausdorff. The end.

However, I can't figure out why I cannot complete the proof for "open subsets ..." in a few lines similar to above.

(Proof 2: Open subsets of locally compact Hausdorff space are locally compact)

Proof Attempt:

1. Let $(X,\mathfrak{T})$ be a locally compact Hausdorff space.

2. Then $\forall x \in X, \exists K$, $K$ compact, $U \in \mathfrak{T}$ s.t. $x \in U \subseteq K$.

3. Let $V$ be closed, then $\forall x \in V, x \in U \cap V$

Now we just need to produce a compact set containing $U\cap V$ that is contained in $V$

I think to proceed I need to use the following Lemma:

Lem:

Given $(X, \mathfrak{T})$ Hausdorff, then it is locally compact iff every point is contained in an open set with compact closure

Then $\forall x \in V, x \in U \cap V \subset \overline{ (U \cap V)} \subseteq V$

All I am left to do is to show that $\overline{ (U \cap V)} \subseteq V$. Can anyone provide suggestion as to how to show this?

• The second point of your proof 1 is incorrect. The Cantor middle-thirds set contradicts that statement (it is a compact set with empty interior). – Aweygan Aug 9 '16 at 21:39
• @Aweygan I had the wrong definition. I have since changed it. – Olórin Aug 9 '16 at 21:41
• Alright that's better, but still not phrased in a spectacular manner. Similarly for your third and fourth points. You should be specific about which topology you are referring to when mentioning openness, compactness, etc (either the topology on $X$, or the induced topology on $C$). – Aweygan Aug 9 '16 at 21:50
• Oh I see, so you mean the closure in the last line should be relative to $V$. So $x \in U \cap V \subseteq \overline U \cap V \subseteq V$, where $\overline U \cap V$ is compact....actually wait that is true/not true? – Olórin Aug 9 '16 at 21:53
• the third point of your first proof is wrong. If $C$ is closed and $U$ open then $U\cap C$ is not necessarily open. – Masacroso May 18 '18 at 15:18

Ok, so we have a locally compact Hausdorff space $X$, and we want to show that any $U \in \tau_{_X}$ is also locally compact in the subspace topology. By the lemma you cite, it suffices to show that for each $x \in U$, there is a compact neighborhood of $x$ that is contained entirely inside $U$.

Since $x$ is already guaranteed a compact neighborhood $K \subset X$ due to local compactness, this will serve as our starting point. The intersection of a closed set and a compact set is itself compact, so our strategy going forward will be to find a (finite) collection of closed neighborhoods $\{ C_i \}$ of $x$ together with $K$ such that their intersection is a (necessarily compact) proper subset of $U$.

A good step in the right direction is to take $K \cap \overline{U}$, which clips off a ton of "extra" points in $K \setminus U$.

To get a third closed set $S$ so that $S \cap K \cap \overline{U} \subset U$, we can do the following: note that $\partial U \cap K$ is compact because $\partial U$ is closed (where $\partial U$ denotes the boundary of $U$), and since $X$ is Hausdorff, we can cover $\partial U\cap K$ with open sets that are "far" from $x$. Let $T$ be the union of the open sets contained in the finite subcover this will necessarily admit, and let $S = T^c$ (which is a closed neighborhood of $x$).

$S \cap K \cap \overline{U}$ is a compact neighborhood of $x$ contained in $U$.

Footnote: Of course, we don't need to be laborious above w.r.t. compactness when transitioning between the normal topology and the subspace topology on $U$ since if $S$ compact in a topological space $X$, then $S \cap Y$ is compact in any given subset $Y$ under the subspace topology.

• in the last part you are saying that if $S$ is compact in $X$ then $S\cap Y$ is compact in any subspace $Y\subset X$, but this is not true. Check by example this. More over: you also need to take in account that is possible that $x\in\partial U$, so the above strategy doesnt work in general. – Masacroso May 18 '18 at 20:13