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Suppose $X$ is a compact Hausdorff space and $C,D$ are disjoint closed subsets of $X$. I want to show that there exist open disjoint $U,V$ with $C\subseteq U, D\subseteq V$.

Since $C,D$ are disjoint, we can take a point $c\in C$ and for any point $d\in D$ find a pair of disjoint open sets $U_{c,d}, V_{c,d}$ such that $c\in U_{c,d}, d\in V_{c,d}$. Taking the union of $U_{c,d}$'s we can form an open cover of $C$: $$ C\subset \bigcup_{c\in C, d\in D} U_{c,d} $$ It is also possible to get an open cover of $D$ in a similar way, but I don't know how to use compactness or the fact that the sets $C,D$ are closed.

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  • $\begingroup$ Start by proving that you can separate a point from a closed set and, then, use this result to prove what you want. $\endgroup$
    – user40276
    Commented Nov 21, 2013 at 6:12

2 Answers 2

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Do it in two steps. What you’re trying to show is that every compact Hausdorff space is normal; start by showing that every compact Hausdorff space is regular.

Let $X$ be a compact Hausdorff space, $C$ a closed subset of $X$, and $p$ a point not in $C$. $X$ is Hausdorff, so for each $x\in C$ there are open sets $U_x$ and $V_x$ such that $p\in U_x$, $x\in V_x$, and $U_x\cap V_x=\varnothing$. Let $\mathscr{V}=\{V_x:x\in C\}$; clearly $\mathscr{V}$ is an open cover of $C$. $C$, being a closed subset of a compact space, is compact, so there is a finite $F\subseteq C$ such that $=\{V_x:x\in F\}$ covers $C$. Let $U=\bigcap_{x\in F}U_x$ and $V=\bigcup_{x\in F}V_x$; then $U$ and $V$ are open, $p\in U$, $C\subseteq V$, and $U\cap V=\varnothing$.

Now let $C$ and $D$ be disjoint closed subsets of $X$. For each $x\in D$ there are open sets $U_x$ and $V_x$ such that $x\in U_x$, $C\subseteq V_x$, and $U_x\cap V_x=\varnothing$. Use a slight modification of the argument in the preceding paragraph to find disjoint open nbhds of $C$ and $D$.

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You can use the fact that closed subsets of a compact Hausdorff space are themselves compact. Then you can construct a finite collection of subsets $U_c$, so that each $U_c$ is disjoint from every other $V_d$. The finiteness of each collection allows you to separate each of the $U_c$'s from each of the $V_d$'s, so that the respective unions are disjoint from each other.

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