0
$\begingroup$

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 am having trouble with this problem because I do not see where the compactness comes in. To me it seems that since $|a-b|$ is a metric we have that it must be greater than or equal to zero. Therefore if we assume the negative, i.e $|a-b|=0$ then by defintion of a metric we must have $a=b$ which is a contradiction to the fact that A and B are disjoint. What is wrong and how should I go about this. Thank you!

  • 0
    To see what can happen without compactness consider the two parts of the graph of $x \mapsto 1/x^2$.2012-12-13
  • 0
    Or just let $n=1$ and $A=\{0\}$ and $B=(0,1)$.2012-12-13
  • 0
    I see from those examples we need compactness but where did I do something wrong in my explanation. It must be wrong since I never used compactness. Can you give me a hint on how to start the proof for this? Thank you!2012-12-13
  • 1
    The negative would be that however small you choose $\delta$ there will be points in $A$ and $B$ that are less than $\delta$ apart. That does not mean that their distance is zero.2012-12-13

3 Answers 3