Prove that there exists an open set $V$ with compact closure such that $K⊆V⊆\overline V⊆ U$

Question

Suppose that $U$ is open in a locally compact Hausdorff space $X$ and $K\subseteq U$ is a compact set. Then there exists an open set $V$ with compact closure such that $K\subseteq V\subseteq\overline V\subseteq U$.

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Theorem 2.5

Suppose $X$ is a Hausdorff space, $K\subset X$ compact and $p\in K^c.$ Then there exists $U,V\in\tau$ such that $p\in U,K\subset V$ and $V\cap U=\varnothing$

Theorem 2.6

Let $\{K_\alpha\}$ be a collection of compact sets of a Hausdorff space. If $\displaystyle\bigcap_\alpha K_\alpha=\emptyset,$ then there exists $\alpha_1,\dots,\alpha_n$ such that $\displaystyle\bigcap_{k=1}^n K_{\alpha_k}=\emptyset.$

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I am having confusions with the proof, in the first paragraph we have $K\subseteq\ U_x\subseteq\overline U_x,\forall x\in K$, I think.

But $K$ is compact, so $K\subseteq\bigcup_{i=1}^n U_{x_i}\subseteq\bigcup_{i=1}^n\overline U_{x_i}.$This last set is compact. If $G=\bigcup_{i=1}^n U_{x_i},$ then $K$ lies in an open set with compact closure?

How will I know that $\overline G=\overline{\bigcup_{i=1}^n U_{x_i}}$ is compact?

I only know that $\bigcup_{i=1}^n\overline U_{x_i}$ is compact.

2nd qstn. where it says $K\subset W_p$ and $p\not\in\overline W_p$ is that because $K\subset W_p\subset\overline W_p$ and thus $p$ can't be in $\overline W_p$ ?

3rd qstn. This is a collection of compacts $\{C\cap\overline G\cap\overline W_p\}$ and is empty. All are compacts because $C$ and $\overline W_p$ are closed thus intersected with compact $\overline G$ will be compact, correct?

And is also empty because, suppose it's not. Then we would get a contradiction with $p\in C$ and $p\not\in\overline W_p$, right?

4th qstn. This contention $\overline G\cap\overline W_{p_1}\cap...\cap\overline W_{p_n}\subset U$ (not explicitly mentioned in the proof) is because if it weren't truth i.e. $p\in\overline G\cap\overline W_{p_1}\cap...\cap\overline W_{p_n}$ and $p\not\in U$, then $p\in U^c=C!$ with $C\cap \overline G\cap\overline W_{p_1}\cap...\cap\overline W_{p_n}=\emptyset$

Can anyone help me please?

Thank you.

Answer 1

First question: note that $\bigcup_{i=1}^n \overline{U}_{x_i}$ is a compact, and hence closed set (note: Hausdorff) containing $G$. The closure of $G$ is therefore contained in this set. It is a closed subset of a compact set, and hence compact.

Second question: According to the statement of Theorem 2.5, there exist disjoint open $U_p$ and $W_p$ such that $p \in U_p$ and $K \subseteq W_p$. Since $p \in U_p \subseteq X \setminus W_p$, we have that $p$ is in the interior of $X \setminus W_p$, which is to say, $p \notin \overline{W}_p$.

Third question: I don't really follow your reasoning here. The reason why I saw that the intersection was empty was because it was contained in $C = X \setminus U$, and every $p \in C$ fails to be in $\overline{W}_p$. That is, every point in $C$ is excluded from the intersection, hence no points can be in the intersection.

Fourth question: It's not really a question, but I agree!

Answer 2

The closure of that union is in that finite union of the closures which is known compact and a closed subset of a compact set is compact .