# Diameters of and distances between sets in metric spaces

I know that: If $\DeclareMathOperator{\diam}{diam}(X,d)$ is a metric space and $A\subset X$ is bounded, then there $\sup \{ d(a,a'):a,a'\in A \}$, called the diameter of the set $A$ and is denoted by $\diam A$.

But I didn't know how to prove the follows property:

Let $(X,d)$ is a metric space. Show that the subsets $A,B\subset X$ are valid following property:

1. $\diam \overline A= \diam A$;

2. $d(\overline A, \overline B)=d(\overline A, B)=d(A, \overline B)=d(A,B)$.

1. The diameter is a supremum. Taking supremum over a larger set results in a greater value. Hence, $\diam \overline{A}\ge \diam A$
2. To prove the reverse inequality $\diam \overline{A}\le \diam A$, take two points $a,b\in \overline{A}$ such that $d(a,b)\ge \diam\overline{A}-\epsilon$. There are points $a',b'\in A$ such that $d(a,a')\le\epsilon$ and $d(b,b')\le \epsilon$ (why? use the definition of closure). Show that $d(a',b')\ge \diam\overline{A}-3\epsilon$. The triangle inequality will help. Since $\diam A\ge d(a',b')$ and $\epsilon$ was arbitrary, the conclusion follows.
3. The distance is an infimum. Taking infimum over a larger set results in a smaller value. Hence, $d( \overline{A},\overline{B})\le d(A,B)$, and the other two distances are in between these two.
4. To prove the reverse inequality $d( \overline{A},\overline{B})\ge d(A,B)$, take two points $a,b\in \overline{A}$ such that $d(a,b)\le d( \overline{A},\overline{B})+\epsilon$. As in part 2, there are points $a',b'\in A$ such that $d(a,a')\le\epsilon$ and $d(b,b')\le\epsilon$. Show that $d(a',b')\le d( \overline{A},\overline{B}) +3\epsilon$.