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Let $A$, $B$ be non-intersecting compact subsets of a Hausdorff topological space $M$ . How to prove that there exist a pair of open subsets $V\supset A$, $W\supset B$ satisfying $V\cap W=\emptyset$.

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Find "General Topology" by R.Engelking and you'll find the proof there in the chapter on compactness... –  W_D Sep 27 '13 at 15:04
    
See Martin's comment here for an outline of the argument. –  Cameron Buie Sep 27 '13 at 15:05

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For each point $a\in A$ and each point $b\in B$ there are disjoint open sets $U_{a,b}\ni a,\ V_{a,b}\ni b$. Fix $b\in B.$ Then the collection $\mathcal U_b=\{U_{a,b}\mid a\in A\}$ covers $A$. By compactness there are points $a_1,...,a_n\in A$ such that $A\subseteq U_b:=\bigcup_{i=1}^n U_{a_i,b}$. If you define $V_b=\bigcap_{i=1}^n V_{a_i,b}$, then $V_b$ is an open neighborhoood of $b$ which is disjoint from the open neighborhood $U_b$ of $A$. This means a point and a compact set can be separated by neighborhoods.
Now, $B$ is covered by $\{V_b\mid b\in B\}$, and again by compactness finitely many of these open sets suffice to cover $B$. By the same argument you get a union of finitely many $V_b$ that covers $B$ and an intersection of finitely many $U_{b_i}$ which is a neighborhood of $A$, and both sets are disjoint.

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