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?
Best Answer
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 $U_a$.