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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

(and nobody accept answers. Why??)

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?

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  • $\begingroup$ 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). $\endgroup$
    – Aweygan
    Aug 9, 2016 at 21:39
  • $\begingroup$ @Aweygan I had the wrong definition. I have since changed it. $\endgroup$
    – Olórin
    Aug 9, 2016 at 21:41
  • $\begingroup$ 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 $C$). $\endgroup$
    – Aweygan
    Aug 9, 2016 at 21:50
  • $\begingroup$ 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? $\endgroup$
    – Olórin
    Aug 9, 2016 at 21:53
  • $\begingroup$ the third point of your first proof is wrong. If $C$ is closed and $U$ open then $U\cap C$ is not necessarily open. $\endgroup$
    – Masacroso
    May 18, 2018 at 15:18

2 Answers 2

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Suppose $X$ is a locally compact Hausdorff space and $Y \in \Gamma(X)$. I'll prove that $Y$ is a locally compact Hausdorff space with the subspace topology.

I

Suppose $x,y \in Y$ and $x \not= y$.

  • Then $\exists V_i \in \Gamma(X),i \in V_i$ where $i=x,y$ and $V_x \cap V_y = \phi$. This follows from X being Hausdorff.
  • For $i=x,y$, put $G_i = Y \cap V_i$. Then $G_i \in \Gamma(Y)$, $i \in G_i$ and $G_x \cap G_y=\phi$.

Hence $Y$ is Hausdorff.

II

Let $x \in Y$. We have:

  1. $\{x\} \subset X$. $\{x\}$ is $X-$compact.
  2. By theorem 2.7 of RCA Rudin:Th 2.7

$\;\;\;\;\exists V \in \Gamma(X), \{x\} \subset V \ \subset cl_X(V) \subset Y \ \text{ such that } cl_X(V)\text{ is $X-$compact.}$

  1. $cl_X(V)$ is $Y-$compact as well. Also $V \in \Gamma(Y)$

  2. As $cl_X(V) = cl_X(V) \cap Y$, $cl_X(V)$ is $Y-$closed.

  3. By 2 and 4, $cl_Y(V) \subset cl_X(V)$.

  4. As closed subsets of compact sets are compact,$cl_Y(V)$ is $Y-$compact.

Now we have all the ingredients ready. We have:

  • $V \in \Gamma(Y)$ such that $x \in V$.
  • $cl_Y(V)$ is $Y-$compact. Therefore Y is locally compact.
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  • $\begingroup$ That 's unlucky. I need to prove this to show theorem 2.7. Not in the reverse order. $\endgroup$
    – Mariana
    Nov 6, 2021 at 2:45
  • $\begingroup$ Then take a look at the theorem's proof, does it use the fact. $\endgroup$
    – somitra
    Nov 6, 2021 at 5:43
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    $\begingroup$ Your step 3, closure of V is Y-compact. Here it uses the fact that closed subset of Hausdorff space, is compact. And closure of V is closed. And you already show Y is Hausdorff. So closure of V is Y-compact. I write down this because previously I was thinking bigger space-compact can imply smaller space-compact, but that was WRONG. $\endgroup$
    – Mariana
    Nov 6, 2021 at 12:53
  • $\begingroup$ >> bigger space-compact can imply smaller space-compact. That's correct as far as closure w.r.t $X$ is concerned. I think you are confusing it with closure w.r.t Y. $\endgroup$
    – somitra
    Nov 9, 2021 at 7:34
  • $\begingroup$ Yes, you are right. I didn't learn with closure w.r.t. Y $\endgroup$
    – Mariana
    Nov 9, 2021 at 15:31
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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.

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    $\begingroup$ 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. $\endgroup$
    – Masacroso
    May 18, 2018 at 20:13

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