Let $S\subset\mathbb{R}$ bounded Prove that $\mbox{diam}\left(S\right)=\mbox{diam}\left(\overline{S}\right)$ Let $S\subset\mathbb{R}$ bounded and $\mbox{diam}\left(S\right)=\sup\left\{ |x-y|:x,y\in S\right\} $ Prove that $\mbox{diam}\left(S\right)=\mbox{diam}\left(\overline{S}\right)$
I have trouble with the second inequality, any help is welcome! ;)
If $S$ is closed then $S=\overline{S}$ and that's it, otherwise:
$\leq)$ As $S$ is bounded then $\exists i,s:i=\inf S\wedge s=\sup S$ so diam exists and as $S\subseteq\overline{S}$ then $\mbox{diam}\left(S\right)\leq\mbox{diam}\left(\overline{S}\right)$ due to supremum property
$\geq)$ $\overline{S}=\left[i,s\right]$, $\exists\left\{ x_{n}\right\} _{n\in\mathbb{N}}\subset S,x_{n}\to t\in S:t=s-i$ I don't know if this is useful...
 A: Take any points $x,y\in\overline S$.
Then for any $\varepsilon>0$ there exist points $a,b\in S$ such that $|x-a|,|b-y|\le\frac\varepsilon2$.
Now just use the triangle inequality $$|x-y|=|(x-a)\!+\!(a-b)\!+\!(b-y)|\le|x-a|\!+\!|a-b|\!+\!|b-y|\le|a-b|+\varepsilon\le\text{diam }S+\varepsilon$$
A: Hints: 
To prove that $\mbox{diam} (\overline{S})\leq \mbox{diam}(S)$, it suffices to show that $|x-y|\leq \mbox{diam}(S)$ all $x,y \in \overline{S}$. Why? 
Let $x,y \in \overline{S}$. Since $x,y$ are in the closure of $S$ we can find points in $S$ arbitrary close to them, how? (If both $x,y$ are points of $S$ this is trivial, otherwise use the fact that they are limit points of $S$). Try to utilise this to show that $|x-y|\leq \mbox{diam}(S) + \epsilon$, where $\epsilon$ is arbitrary. 
A: Well before I start I should say this is a rather ugly argument and I'm sure there is a better one. 
$x, y \in S \implies x, y \in \overline S$ from which it follows that the distance between any two elements in $S$ is not greater than that in $\overline S$. Whence, $\mbox{diam}\left(S\right) \le \mbox{diam}\left(\overline{S}\right)$
Now suppose $\mbox{diam}\left(S\right) \lt \mbox{diam}\left(\overline{S}\right)$. Then there is $z_1, z_2 \in \overline S$ such that $ |z_1 - z_2| \gt |x - y|$ for every $x, y \in S$. Without loss of generality we may assume that $z_1 \not \in S$ and $z_1$ is greater than all elements in $S$ and $z_2$. Let $l = \sup |z_1 - x|$ for $x \in S$. $l \gt 0$ or else we would have a contradiction. Now consider the open set $I = (z_1 - l, z_1)$. The intersection of $\Bbb R \setminus I$ and  $\overline S$ returns a smaller closed set which contains $S$ saying that $\overline S$ is not its closure leading to a contradiction.
This suggests that $\mbox{diam}\left(S\right) = \mbox{diam}\left(\overline{S}\right)$
