-1
$\begingroup$

I am taking an introductory real analysis course and I have difficulty understanding and solving the problem below .Is it trying to say that we have an infimum for the distance between every two points in $A$ and another arbitrary set? $A$ is compact and closed then prove there exist an $a$ member of $A$ and $b$ a member of $B$ such that $d(a,b) = d(x,y)$ ($x$ is a member of $A$ and $y$ a member of $B$).

$\endgroup$
0
$\begingroup$

For $A,B$ two non-empty sets, $d(A,B):= \inf \{d(x,y): x \in A, y \in B\}$ always exists, as we have a set of real numbers that is non-empty and bounded below (by $0$ trivially). The same holds for $d(x,B) = \inf \{d(x,b); b \in B\}$ as a special case.

However, we can note that the function $f_B: x \to d(x,B)$ is continuous for $x \in X$ (as $|d(x,B) - d(x', B)| \le d(x,x')$ by the triangle inequality, so the function is contractive hence uniformly continuous) and so $f_B$ assumes a minimum and maximum value on the compact subset $A$.

If we do not assume anything on $B$ we need not always have such $a,b$ as asked for: $A = [0,1], B = (1,2)$ is an example where $d(A,B)=0$ and $A$ is compact. If $B$ too is compact we do have such $a,b$, e.g. In many cases $B$ being closed is also sufficient besides $A$ compact.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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