One of the supplementary problems from Schaum's General Topology states that if you have a compact subset of a Hausdorff space and a point not in the compact subset, there exist disjoint, open sets containing the point and the compact subset, respectively. The proof from the book is as follows:
Now, it's nice that the author is rigorous and careful about his proof but I postulate that there is an easier proof to this.
Proof
Let $A$ be a compact subset of a Hausdorff space $X$ and let $p \in X$ with $p \not\in A$.
Consider $a \in A$. Since $X$ is Hausdorff, for each $a \in A$ and $p \not\in A$, there exist disjoint open sets $G_a$ and $G_p$ such that $a \in G_a$ and $p \in G_p$.
Since the union of open sets is open, $\bigcup \{G_a\}$ is open.
Since $G_a \cap G_p = \varnothing$ for every $G_a$, $G_p \cap \bigcup\{G_a\}= \varnothing.$
We have shown now that $p \in G_p$ and $a \in A \subset \bigcup\{G_a\}$ and $G_p \cap \bigcup\{G_a\} =\varnothing$, which is what we were trying to prove.
I like my proof because it is simpler but I noticed that I completely failed to use compactness in my proof. I'm not seeing why compactness is even an essential part of the proof here. Because, if we know that each $x_1$ and $x_2$ in a Hausdorff space $X$ have disjoint open sets containing each, can't you just take each $x_i$ in the compact set, take the union of each open sets $U_{x_i}$ containing each $x_i$ and claim that $x_i \in \bigcup U_{x_i}$ and that $U_p \bigcap U_{x_i} = \varnothing$?
Best Answer
You need compactness because in your argument the neighborhood $G_p$ that you construct depends on a!
You say(paraphrasing): Given $a\in A$ and $p\notin A$, there exist disjoint neighborhoods $G_a$ and $G_p$ of $a$ and $p$...but this $G_p$ depends on $a$, so really the notation should be $G_{p,a}$ and then the neighborhood of $p$ that you want in the end is $\cap_{a\in A} G_{p,a}$, but this is a possibly infinite intersection of open sets, unless...