$A$ is compact and closed then prove ...

Question

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 $B$).

Answer 1

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.