Distance between Sets and their Closures

[Kindayr]

Homework Statement

Suppose $(X,d)$ is a metric space, and suppose that $A,B\subseteq X$. Show that $dist(A,B)=dist(cl(A),cl(B))$.

Homework Equations

$cl(A)=\partial A\cup A$.
$dist(A,B)=\inf \{d(a,b):a\in A,b\in B\}$

The Attempt at a Solution

Its clear that $dist(cl(A),cl(B))\leq \min\{dist(A,B),dist(A,\partial B),dist(\partial A,B),dist(\partial A,\partial B)\}\leq dist(A,B)$. I just can't exactly find a way to go the other direction of the inequality. I'm not looking for a direct solution, just some intuition.

 

[Kindayr]

Here's my attempt at a solution:

$dist(A,B)\leq dist(A,x)+dist(x,B)\implies \inf _{x\in cl(A)}\{dist(A,B)\}\leq \inf _{x\in cl(A)}\{dist(A,x)+dist(x,B)\}=\inf _{x\in cl(A)}\{dist(x,B)\}$. And, $dist(cl(A),B)\leq dist(cl(A),y)+dist(y,B)\implies \inf _{y\in cl(B)}\{dist(cl(A),B)\}\leq\inf _{y\in cl(B)}\{dist(cl(A),y)+dist(y,B)\}=\inf _{y\in cl(B)}\{dist(cl(A),y)\}=dist(cl(A),cl(B))$. Hence, $$dist(A,B)=\inf _{x\in cl(A)}\{\inf _{y\in cl(B)}\{dist(A,B)\}\}\leq dist(cl(A),cl(B)).$$

Does this work?

 

[Dick]

dist(cl(A),cl(B)) is clearly less than dist(A,B) just because A is contained in cl(A) and B is contained in cl(B). Do you agree with that? If so why? To go farther you need to deal with the definition of what cl(A) or boundary(A) in a metric space is. What is it?