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Let $X$ be Hausdorff and $A,B \subseteq X$ be disjoint compact subspaces of $X$. Prove that there are $U$ and $V$ open disjoint sets in $X$, $A\subseteq U$ and $B\subseteq V$.

I know that: $A$ and $B$ are closed in $X$.

But I have no idea how to prove statement using that $X$ is Hausdorff since with that information I know what to do only with points :/

Maybe just an idea?

Meow
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    The statement that a compact set is closed, can be made stronger: A compact subset $A$ and a point $x\notin A$ have disjoint neighborhoods. Try to prove that first. – Stefan Hamcke Jun 16 '13 at 22:12
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    Done that. =) My try is this: We first do that for every point in $A$ and set $B$. Union of that neighbourhoods (we take open neighbourhoods) is cover of $A$ and then there is finite subcover- we take $U$ is union of elements of that subcover. And for $V$ we take intersection of matching neighbourhoods of $B$, (and matching means that neighbourhoods we got when we used this statement for point from that finite subset and set $B$). – Meow Jun 16 '13 at 22:35
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    Surely you meant to assume that $A$ and $B$ are disjoint. – Andreas Blass Jun 16 '13 at 22:47
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    Well done! A simple formula is: Compact subsets behave like points. Here they resemble the property of two points being separable by disjoint neighborhoods. – Stefan Hamcke Jun 16 '13 at 22:53
  • @AndreasBlass: It must be =) – Meow Jun 17 '13 at 11:50
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    Possible duplicate of Disjoint compact sets in a Hausdorff space can be separated – Arnaud D. May 31 '17 at 10:18

2 Answers2

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Fix $a \in A$. For every $x \in B$, we can find disjoint neighborhoods $U_x(a)$ of $a$ and $V_x(a)$ of $x$. Since $B$ is compact, the open cover $\{V_x(a) : x \in B\}$ has a finite subcover $\{V_{x_1}(a) , \dots, V_{x_n}(a)\}$. Let $V(a) = \bigcup_{i = 1}^n V_{x_i}(a)$ and $ U(a) = \bigcap_{i=1}^n U_{x_i}(a)$

Then $U(a)$ and $V(a)$ are disjoint neighborhoods of $a$ and $B$. The open cover $\{U(a) : a \in A\}$ of $A$ has a finite subcover $\{U(a_1), \dots, U(a_k\}$. Let

$$U = \bigcup_{i = 1}^k U(a_i)$$ $$V = \bigcap_{i = 1}^k V(a_i)$$

It follows that $U$ and $V$ are disjoint neighborhoods of $A$ and $B$.

Ink
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Hint:

  1. Fix $b\in B$ use the Hausdorff property on the elements of $A$ and compactness of $A$ to construct construct $U_b,V_b$ such that $$A\subset U_b,\qquad b\in V_b, \qquad U_b\cap V_b=\emptyset$$

  2. Now $$\bigcap_{b\in B} U_b\supset A$$ but $\bigcap_{b\in B} U_b$ need not be open, because it might not be a finite intersection. Finish off using compactness of $B$.

AD - Stop Putin -
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