[Math] Closed and open subsets of locally compact Hausdorff space are locally compact

Question

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.

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

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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?

Answer 1

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:

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

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

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

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

6. 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.