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I'm reading Rudins "Complex Analysis". There's a theorem that goes like this.

Theorem: Let $U$ be open, in a locally compact Hausdorff space. Let $K\subset U$ and $K$ be compact. Then there exist an open set $V$ such that $$K\subset V\subset \overline{V}\subset U $$ Proof:

Because every point of $K$ has a neighbourhood that's closure is compact, and we can cover $K$ with finite amount of such neighbourhoods, then there exist a set $G$, that has a compact closure, and is a subset of $K$. If $U=X$ then we can take $V=G$.

If not, define $C$ as the complement of $U$. From theorem $2.5$, for every point $p\in C$, there exist an open set $W_p$, such that $K\subset W_p$, and $p\notin\overline{W_p}$.

And the proof goes on.

Theorem $2.5$: Suppose that $X$ is a Hausdorff space, $K\subset X$ is a compact set, and $p\in K^c$. Then there exist open sets $U$ and $W$, such that $p\in U$, $K\subset W$, and $U \cap W = \emptyset$.

My question is, how we can use this theorem in the proof, so that we can state the existence of such sets $W_p$. Mainly, where does $p\notin\overline{W_p}$ come from. Thank you.

Another question that might help me. If $U$ and $V$ are open, $U\cap V=\emptyset$ holds, does $\overline{U}\cap V=\emptyset$?

I've found a similar question here. In the comments, Andreas says that if $p$ was in closure of $W_p$, then every neighbourhood of $p$ would have to meet $W_p$. Why is that?

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  • $\begingroup$ Re: "Another question" : If $V$ is open then its complement is $V^c$ is closed and disjoint from $V $. So if $ V $ is open and is disjoint from $U$ then $V\cap \overline U\subset V \cap \overline V^c=$ $V \cap V^c=\emptyset.$... Another way to see this is that if $V$ is open and $U\cap V= \emptyset,$ then every $p\in V $ has a nbhd (namely, $V $ ) that's disjoint from $U,$ so $p\not \in \overline U.$ $\endgroup$ Commented Jul 22, 2018 at 19:10
  • $\begingroup$ The theorem, with $\overline V$ compact, directly: (1). In a Hausdorff space if $C$ is compact and $p\not\in C$ than $p, C$ are completely separated. (2). If $D$ is open and $ \overline D$ is compact then $\partial D= \overline D$ \ $ D$ is compact. (3). We use (1) and (2) to prove that a compact $T_2$ space is regular..... Then for each $p \in K$ let $V_p, U_p$ be open sets containing $p$ such that $\overline V_p$ is compact and $\overline U_p\subset U. $ Now $A=\{V_p\cap U_p: p\in K\}$ is an open cover of $K.$ Let $B$ be a finite sub-cover of $A$ and let $V=\cup B.$ $\endgroup$ Commented Jul 22, 2018 at 19:28
  • $\begingroup$ ERRATUM: In my previous comment , (3) should say that a LOCALLY compact $T_2$ space is regular. I ran out of editing time . $\endgroup$ Commented Jul 22, 2018 at 19:35
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    $\begingroup$ @DanielWainfleet Thank you! I get it now. $\endgroup$
    – Habrum
    Commented Jul 22, 2018 at 20:03

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Use the Hausdorff condition. For each $a\in K$ there are disjoint open neighbourhoods $U_a$ and $V_a$ of $p$ and $a$ respectively. Finitely many of the $V_a$ will cover $K$; let $W$ be their union, and $U$ the intersection of the corresponding $U_a$.

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  • $\begingroup$ Yes, that's how the proof of Theorem $2.5$ is stated. My question is how to use this theorem $\endgroup$
    – Habrum
    Commented Jul 22, 2018 at 16:50
  • $\begingroup$ The point is that $U$ and $W$ are disjoint. $\endgroup$ Commented Jul 22, 2018 at 17:02
  • $\begingroup$ $U$ and $W$ are disjoint, does that imply that, for example, $\overline{U}$ and $W$ are disjoint? $\endgroup$
    – Habrum
    Commented Jul 22, 2018 at 17:03
  • $\begingroup$ $W$ is open $\endgroup$ Commented Jul 22, 2018 at 17:04

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