Prove that $\mathrm{diam} (\overline{E})=\mathrm{diam}(E)$ Prove that $\mathrm{diam} (\overline{E})= \mathrm{diam} (E)$, where $E\subset (X,d)$, a metric space, and $\mathrm{diam}$ is the diameter of a set which is defined to be $\mathrm{diam}(E)$= $\sup \{d(p,q):\;p,q\in E\}$.
 A: HINT: $\newcommand{\cl}{\operatorname{cl}}\newcommand{\diam}{\operatorname{diam}}$Obviously $\diam E\le\diam\cl E$, so it suffices to show that $\diam\cl E\le\diam E$. If not, there are points $p,q\in\cl E$ such that $d(p,q)>\diam E$. Use definition of the closure to show that this implies that there are points $x,y\in E$ such that $d(x,y)>\diam E$, an obvious contradiction.
A: Obviously $\mathrm{Diam}\overline E \geq \mathrm{Diam}E$
Let $\epsilon >0$,
now from definiton of the 
$\mathrm{Diam}\overline E= \sup\{\mathrm{d}(x,y) | x,y \in \overline E\} $
there are $x_0, y_0 \in \overline{E}$
such that 
$$\fbox{1}\, \,\mathrm{Diam}\overline E \leq \mathrm{d}(x_0,y_0)+ \frac{ \epsilon}{3}$$
Now we can find   $$x_1, y_1 \in E, \mathrm{d}(x_0,x_1) \leq \frac{\epsilon}{3},\, \,\mathrm{d}(y_0,y_1) \leq \frac{\epsilon}{3} $$
So from triangle inequality we get
$$\fbox{2}\, \,\mathrm{d}(x_0, y_0) \leq \mathrm{d}(x_0, x_1)+\mathrm{d}(x_1,y_1)+\mathrm{d}(y_1, y_0)\leq \frac{\epsilon}{3}+\mathrm{Diam}(E)+\frac{\epsilon}{3}$$
Now combining $\fbox{1}$ and $\fbox{2}$ we get
$$\mathrm{Diam}\overline E \leq \mathrm{Diam}(E) +\epsilon.$$
Since $\epsilon $ was arbitary conclude
$$\mathrm{Diam}\overline E \leq \mathrm{Diam}(E)$$
