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Let $K_{1}, K_{2}\subset \mathbb{R}^{n}$ compact sets, such that $K_{1}\cap K_{2}=\emptyset$. Prove that, there are open sets $U_{1}, U_{2}\subset \mathbb{R}^{n}$ disjoint such that $K_{1}\subset U_{1}$ and $K_{2}\subset U_{2}$.

My ideia is: how $K_{1}$ and $K_{2}$ are disjoint sets, then $d(K_{1},K_{2})>0$ and $K_{1}\subset \cup_{i=1}^{m}A_{i}$, $K_{2}\subset\cup_{j=1}^{n}B_{j}$, where $A_{i}$, $B_{j}$ are open sets covering $K_{1}$ and $K_{2}$, respectively (Borel-Lebesgue Theorem). Now, i take $U_{1}=:\cup_{i=1}^{m}A_{i}$ and $U_{2}:=\cup_{j=1}^{n}B_{j}$, but how can i conclude that $U_{1}$ and $U_{2}$ are a disjoint sets?

Does anyone have any ideas?

FakeAnalyst56
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MárioBezerra
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1 Answers1

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If $r=d(K_1,K_2)$, try covering $K_1$ and $K_2$ with open balls of radius $r/2$. Take the covering composed of putting an open ball of radius $r/2$ at each point, and use compactness to get a finite subcover. Do so for both $K_1$ and $K_2$. It must be that these finite coverings (by open sets) are disjoint, for if $x$ lies in both coverings, then $d(x,K_1)<r/2$ and $d(x,K_2)<r/2$, so that $d(K_1,K_2)<r$, contradicting the choice of $r$.

FakeAnalyst56
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