Let $A,B$ be two compact subsets of $X$ where $(X,d)$ is a metric space.

1.Show that $\exists a\in A; b\in B$ such that $d(a,b)=d(A,B)$

where $d(A,B)=\sup\{d(a,b):a \in A;b\in B\}$

2.Show that $\exists a\in A$ such that $d(a,B)=d(A,B)$

3.Show that $\exists a,b\in A$ such that $d(a,b)=diam (A)$ where $diam(A)=\sup\{d(a,b):a ,b\in A\}$

I can compute for 2 by considering the function $f:X\to \mathbb R$ by $f(x)=d(x,B)$ and then considering its restriction to the set $A$ which is compact and hence the function $f$ will attain its bounds.So there exists $a\in A$ such that $f(a)=d(A,B)\implies d(a,B)=d(A,B)$

Is this correct?

Similarly I can proceed for $3$

But I can't do it for 1 .Any help on how to consider the continuous function

  • $\begingroup$ To be completely pedantic, you're missing something from your statement. $A,B$ must be nonempty. :) $\endgroup$ – user223391 May 15 '15 at 6:04

Consider the metric space $(A \times B, D)$, where $D$ is the metric on $A\times B$ defined by $$D((x,y),(a,b)) := d(x,a) + d(y,b)$$

The function $f : (A \times B,D) \to \Bbb R$ given by $f(a,b) = d(a,b)$, for all $(a,b) \in A \times B$, is continuous. For

$$|f(x,y) - f(a,b)| \le D((x,y),(a,b))$$

for all $x,a\in A$ and $y,b\in B$. Since $f$ is a continuous map from a compact metric space, it attains it's supremum, i.e., there exists $(a_0,b_0) \in A \times B$ such that $$f(a_0,b_0) = \sup\{f(a,b): (a,b)\in A \times B\}.$$ That is, $d(a_0,b_0) = d(A,B)$.

  • $\begingroup$ will the problem hold only if $A\times B$ is given this metric ;if I take product metric? $\endgroup$ – Learnmore May 15 '15 at 7:09
  • $\begingroup$ @learnmore: You could choose a different metric, e.g. $D'((x,y),(a,b)) = \max\{d(x,a),d(y,b)\}$ will do. The point is to use an appropriate metric on $A \times B$ so that $f$ will be continuous (as a map between metric spaces). Then you can invoke the extreme value theorem to finish the proof. $\endgroup$ – kobe May 15 '15 at 13:17
  • $\begingroup$ the problem is if I use your metric then it becomes easy to show continuity;i cant show continuity for product metric;any help $\endgroup$ – Learnmore May 15 '15 at 14:37
  • $\begingroup$ is the problem independent of the metric used? $\endgroup$ – Learnmore May 15 '15 at 14:37
  • $\begingroup$ @learnmore, you're misunderstanding. You do not have to show that $f$ is continuous with respect to any metric. The problem was to show that there exists $a \in A, b\in B$ such that $d(a,b) = d(A,B)$. So certainly this problem depends on the metric $d$. You say that if you use my metric, then it becomes easy to show continuity -- this is why I used this method. $\endgroup$ – kobe May 15 '15 at 14:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.