I want to show that any two disjoint compact sets in a Hausdorff space $X$ can be separated by disjoint open sets. Can you please let me know if the following is correct? Not for homework, just studying for a midterm. I'm trying to improve my writing too.
My work:
Let $C$,$D$ be disjoint compact sets in a Hausdorff space $X$. Now fix $y \in D$ and for each $x \in C$ we can find (using Hausdorffness) disjoint open sets $U_{x}(y)$ and $V_{x}(y)$ such that $x \in U_{x}(y)$ and $y \in V_{x}(y)$. Now the collection $\{U_{x}: x \in C\}$ covers $C$ so by compactness we can find some natural k such that
$C \subseteq \bigcup_{i=1}^{k} U_{x_{i}}(y)$
Now for simplicity let $U = \bigcup_{i=1}^{k} U_{x_{i}}(y)$, then $C \subseteq U$ and let $W(y) = \bigcap_{i=1}^{k} V_{x_{i}}(y)$. Then $W(y)$ is a neighborhood of $y$ and disjoint from $U$.
Now consider the collection $\{W(y): y \in D\}$, this covers D so by compactness we can find some natural q such that $D \subseteq \bigcup_{j=1}^{q} W_{y_{j}}$.
Finally set $V = \bigcup_{j=1}^{q} W_{y_{j}}$, then $U$ and $V$ are disjoint open sets containing $C$ and $D$ respectively.
What do you think?
Best Answer
This is a very good start, but there is a slight problem with your argument: as you change $y$, your $U$ changes as well (since $U$ is constructed in terms of $y$); you should really call it $U(y)$.
Your construction gives you an open neighborhood $W(y)$ of $y$ for each $y$; $W(y)$ is disjoint from $U(y)$. But for all you know, $W(y)$ may fail to be disjoint from $U(y')$ with $y'\neq y$.
So you really still have a bit more to go before you are done.