I know several questions related to this already exist on this site, but I didn't find those particularly helpful. In some places, I wish to clarify my line of thought, and in others, I need some help.
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$U$ is open, so $C = U^c$ is closed. $\overline G, \overline{W_p}$ are closed by definition. So, $C\cap \overline G\cap \overline{W_p}$ is also closed. We know that $\overline G$ is compact, and that $C \cap \overline G\cap \overline{W_p} \subset \overline G$. Can we use this to conclude that $C \cap \overline G\cap \overline{W_p}$ is compact, since a closed subset of a compact set is also compact?
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I understand that $$V = G\cap W_{p_1}\cap\ldots\cap W_{p_n}$$ is open, since it is a finite intersection of open sets. I must show that $\overline V$ is compact. How does the author conclude $$\overline V \subset \overline G\cap \overline {W_{p_1}}\cap\ldots\cap \overline {W_{p_n}}$$
I think this is how: $\overline V$ is the smallest closed set containing $V$, in that every other closed set containing $V$ must be a superset of $\overline V$. $V \subset \overline G\cap \overline {W_{p_1}}\cap\ldots\cap \overline {W_{p_n}}$ is verified easily, since $A\subset \overline A$ for every set $A$, and inclusion is preserved under intersections. Is that right? -
How does $\overline V \subset \overline G\cap \overline {W_{p_1}}\cap\ldots\cap \overline {W_{p_n}}$ help us conclude that $\overline V$ is compact? Here's what I'm thinking: $\overline G$ is compact, and $\overline G\cap \overline {W_{p_1}}\cap\ldots\cap \overline {W_{p_n}} \subset \overline G$. Moreover, $\overline G\cap \overline {W_{p_1}}\cap\ldots\cap \overline {W_{p_n}}$ is closed, since it is an intersection of closed sets. A closed subset of a compact set is also compact, so $\overline G\cap \overline {W_{p_1}}\cap\ldots\cap \overline {W_{p_n}}$ is compact. We can apply a similar argument to $\overline V$ and $\overline G\cap \overline {W_{p_1}}\cap\ldots\cap \overline {W_{p_n}}$ to conclude that $\overline V$ is compact – correct?
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Not a technical doubt as such, but could you provide some intuition as to why the author chose to prove the theorem in the way he did? In particular, what may have motivated bringing $W_p$'s etc. in the picture?
Proof attached for reference:
Thanks a lot!
Best Answer
I'll go through the proof, adding explainers:
So $K \subseteq U$ is given, where $K$ is compact and $U$ is open. The space we're in is locally compact and Hausdorff.
The first and easiest case is when $U=X$ itself: every $x \in K$ has an open neighbourhood $U_x$ with compact closure, by definition of local compactness. $\{U_x: x \in K\}$ is an open cover of the compact $K$ so finitely many also cover $K$, say $\{U_{x_1}, \ldots, U_{x_n}\}$ so $K \subseteq \bigcup_{i=1}^n U_{x_i}$. We can use $V = \bigcup_{i=1}^n U_{x_i}$, which is open and whose closure is $\bigcup_{i=1}^n \overline{U_{x_i}}$, a finite union of compact sets, hence compact (this is not mentioned in the text before but follows easily from the definition of compactness). So this $V$ is then as required.
We don't need Hausdorffness in this case. This is the proof as sketched in the first paragraph. He keeps one thing from this case: we always know that $K$ has an open neighbourhood $G$ with compact closure (in fact the $V = \bigcup_{i=1}^n U_{x_i}$ will do). He's going to use that in the second case too.
So we can assume we're in the case $U \neq X$ so that $C= X\setminus U$ is non-empty.
For each $p \in U$ we have $p \notin K$ so that 2.5 applies which says
So this extends Hausdorffness from two points we can separate by open sets to a compact set and a point we can separate, which is a handy improvement.
So for $p \in U$ we have open (unnamed) $U_p$ containing $p$ and some $W_p$ containing $K$ such that $W_p \cap U_p = \emptyset$. Then $U_p$ is a witness to the fact that $p \notin \overline{W_p}$ and we just use that fact and forget $U_p$ itself.
So in all we indeed have $K \subseteq W_p$ and $p \notin \overline{W_p}$ and $W_p$ open.
Every set $C \cap \overline{W_p} \cap \overline{G} \subseteq \overline{G}$ is thus a closed subset of $\overline{G}$ which is compact by 2.4 as $\overline{G}$ is compact. That covers your 1.
The intersection $\bigcap \{ C \cap \overline{W_p} \cap \overline{G} \mid p \in C\}$ is empty: if $x$ were in it, it would be a point of $C$ so one of the $p \in C$ but then it cannot be in $C \cap \overline{W_p} \cap \overline{G}$ for that $p$ (as $p \notin \overline{W_p}$ for each $p$ by construction).
So Rudin then applies 2.6 to conclude we have $p_1, \ldots p_n$ in $C$ so that
$$C \cap \overline{G} \cap \overline{W_{p_1}} \cap \ldots \cap \overline{W_{p_n}} = \emptyset\tag{1}$$
Then $$V= G \cap W_{p_1} \cap \ldots \cap W_{p_n}$$
is open and contains $K$, as a finite intersection of open sets that all contain $K$.
It's also clear that
$$\overline{V} \subset \overline{G} \cap \overline{W_{p_1}} \cap \ldots \cap \overline{W_{p_n}} $$
as the right hand side is closed and contains $V$. But more is true: the right hand intersection is compact as it is still a closed subset of the compact $\overline{G}$ so 2.4 applies still. And anotehr application of 2.4 then concludes that $\overline{V}$ is also compact.
So you're correct about 2 and 3 too.
Finally note that $(1)$ can be rewritten as
$$\overline{G} \cap \overline{W_{p_1}} \cap \ldots \cap \overline{W_{p_n}} \subset X\setminus C = U$$
which allows us to indeed say $\overline{V} \subseteq U$, as we needed.
Rudin didn't need to set it up this way, using the complement of $U$ and the intersection of compact sets being empty, but it works with minimal theorems needed to prove as "lemma's": 2.4 to 2.6 suffice, and are needed later anyway. The "intuition" could be: we can do it for $K$ and one point outside $U$ with $W_p$ but we use the compactness to extend it from one point to the whole of $U^\complement$, using the one compact $\overline{G}$ to embed everything inside a compact context so we can do this (so all closures of $W_p$ become compact in the intersection with $\overline{G}$).
In a general topology book, maybe he would have shown this fact for a point and an open neighbourhood $U$ first, and then apply a compactness argument to extend to $K$, just as in the first case. All ways lead to the same results though.