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Let $A$ and $B$ be two nonempty disjoint subsets of $\mathbb{R}^{n}$. Put $d(A,B)=inf\left \{ ||a-b||:a\in A, b\in B \right \}$.

a) Show that if one of the sets $A$ and $B$ is compact then $d(A,B)>0$.

b) Let $n=2$, $A=\left \{ (x,1/x):x>0 \right \}$ and $B=\left \{ (x,0):x>0 \right \}$. Show that the sets $A$ and $B$ are closed in $\mathbb{R}^{2}$, $A\cap B=\emptyset$ but $d(A,B)=0$.

My thoughts: Let $A$ be compact then it is not only closed but also bounded. Assume $d(A,B)=0$, so $inf||a-b||=0$. So $a=b$ for some $a$ and $b$. (I think $A$ and $B$ being closed is the reason of that, but I'm not sure how to show this.) but that contradicts the fact that the sets are disjoint.

for part b:$ d(A,B)=\sqrt{((x-x)^{2}+(1/x)^{2})}$, as $x$ increases distance decrases, since $lim_{x\rightarrow \infty}(1/x)^{2}=0$, $d(A,B)=0$. For showing their intersection is an empty set $1/x\neq 0$ if $x\in \mathbb{R}$. I need to show every convergent sequence converges to a point in the sets, for showing $A$ and $B$ are closed, but I couldn't find how to.

user242520
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    For (a) you need more than only one set compact. If $A = [0,1]$ and $B = (-\infty, 0)$, then $A$ is compact, but $d(A,B) = 0$. – aexl May 25 '15 at 11:44

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For part (b): Try showing that the sets are (finite intersections of) pre-images of closed sets under suitable continuous functions $\mathbb R^2\to \mathbb R$. (And I think it must be "$x\ge 0$" in the definition of $B$.)

For part (a): As the example in part (b) shows, you need more than $A$ and $B$ being closed to move from $d(A,B)=0$ to the existence of $a,b$ such that $a=b$. Compactness is crucial here. Hint: Suppose $A$ is compact. Show that, for each $a\in A$, there exists $\varepsilon_a > 0$ such that $d(\{a\},B)>\varepsilon_a$. Now what you would like to be able to do is take $\varepsilon = \min_{a\in A} \varepsilon_a$, and conclude $d(A,B)>\varepsilon$. But you can't do this because there might be infinitely many values of $\varepsilon_a$, so there might be no minimum, and if you take the infimum, well, that might be zero. If only you could restrict attention to only finitely many $a$!