Attempt to prove that a locally compact Hausdorff space is regular.

Question

I am trying to prove that a locally compact Hausdorff space is regular. I am wondering if I might be missing something in my solution, and if someone could clear up why I might need to adjust my proof? Thank you in advance.

Proof. Let $X$ be a locally compact Hausdorff space. We want to show that for each $\overline{E} \subseteq X$, and $x \in X \setminus E$, we have disjoint open sets $U, V \in \mathcal{T}$ such that $E \subset U$ and $x \in V$.

Fix $x \in X$. By local compactness, $\exists W \in \mathcal{T}$ such that $x \in W$ and $\overline{W}$ is compact. Since $\overline{W}$ is compact, there exists a family of open subsets $\{U_\alpha\}_{\alpha \in A}$ which covers $\overline{W}$, i.e., $\overline{W} \subseteq \{U_\alpha\}_{\alpha \in A}$. Since the arbitrary union of open sets is open, take $U = \{U_\alpha\}_{\alpha \in A}$. Then $x \in W \subseteq \overline{W} \subseteq U$.

Now take $z \in X \setminus U$. Since $X$ is Hausdorff, $\exists V \in \mathcal{T}$ such that $U \cap V = \emptyset$ and $z \in V$. Therefore, a locally compact Hausdorff space is regular. $\quad\blacksquare$

Answer 1

Your interpretation of "regular" is inadequate.

We want to show that for each $\overline{E} \subseteq X$, and $x \in X \setminus E$, we have disjoint open sets $U, V \in \mathcal{T}$ such that $E \subset U$ and $x \in V$.

Observe that if $x \in \overline{E} \setminus E$, then each open neigborhhod of $x$ intersects $E$.

What you have to show is

For each closed $E \subset X$ and each $x \in X \setminus E$, there are disjoint open sets $U, V \in \mathcal{T}$ such that $E \subset U$ and $x \in V$.

Also you proof does not work.

Fix $x \in X$. By local compactness, $\exists W \in \mathcal{T}$ such that $x \in W$ and $\overline{W}$ is compact. Since $\overline{W}$ is compact, there exists a family of open subsets $\{U_\alpha\}_{\alpha \in A}$ which covers $\overline{W}$, i.e., $\overline{W} \subseteq \{U_\alpha\}_{\alpha \in A}$. Since the arbitrary union of open sets is open, take $U = \{U_\alpha\}_{\alpha \in A}$. Then $x \in W \subseteq \overline{W} \subseteq U$.

This is essentially correct (except that you should write $\overline{W} \subseteq \bigcup_{\alpha \in A}U_\alpha$ and $U = \bigcup_{\alpha \in A}U_\alpha$), but it does not help you because you do not say anything about the choice of the $U_\alpha$. As you do it, one could take $\{U_\alpha\}_{\alpha \in A} = \{X\}$.

Also you final argument does not work.

Now take $z \in X \setminus U$. Since $X$ is Hausdorff, $\exists V \in \mathcal{T}$ such that $U \cap V = \emptyset$ and $z \in V$. Therefore, a locally compact Hausdorff space is regular.

This is wrong. If $z \in \overline U \setminus U$, no such $V$ exists.

Moreover, your proof does not refer to $E$, therefore it can never give a conclusion involving $E$.

Here is a correct proof.

Since $E$ is closed, $X \setminus E$ is an open neigbborhood of $x$. There exists an open neigborhood $W$ of $x$ such that $\overline W$ is compact. Then $\overline W \cap (X \setminus E)$ is an open neigborhood of $x$ in $\overline W$. Since $\overline W$ is regular (recall that compact Hausdorff spaces are regular), there exists an open neighborhood $G$ of $x$ in $\overline W$ such that $\overline G^{\overline W} \subset \overline W \cap (X \setminus E)$. Note that $\overline G^{\overline W} = \overline G$ because $ \overline W$ is closed in $X$.

Take an open $H \subset X$ such that $\overline W \cap H = G$ and define $V = W \cap H$. This is an open neigbborhood of $x$ in $X$. We have $V \subset G$, thus $\overline V \subset \overline G \subset \overline W \cap (X \setminus E) \subset X \setminus E$. Then $U = X \setminus \overline V$ is an open neigbborhood of $E$ in $X$ which is disjoint from $V$.