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Let A and B be nonempty compact subsets of $\mathbb{R}^n$ with $A \cap B = \emptyset$, then there exists a $\delta > 0$ such that for all $a \in A$ and $b \in B$, $|a-b| > \delta$.

I can not seem to prove this. I have tried many things and been given many hints with no luck. Can someone show me how this one is done please? Thank you!

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    No not homework.$A$problem that I am trying to work through that I have had no luck with. I have tried many things that all came out wrong. I am trying to teach myself analysis and most of the hints have been topological stuff I don't understand. I feel like there should be a way to write a contradiction proof with sequences but I can't figure it out.2012-12-14

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$A\times B$ (with product topology) is compact, $f\colon A\times B\to \mathbb R$, $(a,b)\mapsto |a-b|$ is continuous, hence the minimum is assumed at some $(a_0,b_0)$. Then $f(a_0,b_0)>0$ as otherwise we would have $a_0=b_0$. Let $\delta=\frac12 f(a_0,b_0)$.