For any $E\subseteq\mathbb{R}^n$ we define $\text{diam}(E)=\sup\{|x-y|:x,y\in E\}$ with the Euclidean distance.
Let $\Omega\subseteq \mathbb{R}^n$ be a bounded open set, I want to prove that $\text{diam}(\partial\Omega)=\text{diam}(\Omega)$.
I am able to prove the $\leq$ inequality, what about the other?
I would like to avoid (if possible) any kind of compactness/continuity argument.
Thanks
Since $\Omega$ is bounded, its closure $\overline\Omega$ is compact. Take any pair of sequences $x_i,y_i \in \Omega$ such that $d(x_i,y_i)$ converges to $\text{diam}(\Omega)$. Passing to subsequences which converge, and using that $\overline\Omega$ is closed and bounded and therefore compact, it follows that there exist $(x,y) \in \overline\Omega$ such that $d(x,y) = \text{diam}(\Omega)$.
Now we want to prove that $x,y \in \partial\Omega = \overline\Omega - \Omega$. Arguing by contradiction, suppose, say, that $x \not\in \partial\Omega$. It follows that $x \in \Omega$. Since $\Omega$ is open, pick $r>0$ such that the open ball $B(x,r)$ is a subset of $\Omega$.
Consider the line segment $\overline{xy}$. From $y$, travel along this line segment to $x$, and then continue along the same straight line to the point which is $r/2$ beyond $x$, arriving at a point $x' \in B(x,r) \subset \Omega$. We thus have $d(x',y) = d(x,y) + d(x,x') = d(x,y) + r/2$.
Using that $y_i$ converges to $y$, we may find a sufficiently large $i$ so that $d(y_i,y') < r/4$. By the triangle inequality we have $$d(x',y_i) \ge d(x',y) - d(y,y_i) \ge d(x,y) + r/2 - r/4 = \text{diam}(\Omega) + r/4 > \text{diam}(\Omega) $$ However, $x',y_i \in \Omega$, and we obtain a contradiction.
For any $d<\operatorname{diam}(\partial\Omega)$ there exist points $a,b\in\partial\Omega$ with $|a-b|>d$. There are points $a',b'\in\Omega$ with $|a-a'|$ and $|b-b'|$ smaller than $\frac12(|a-b|-d)$. Then $|a'-b'|>d$ and hence $d<\operatorname{diam}(\Omega)$. This shows $\operatorname{diam}(\Omega)\ge \operatorname{diam}(\partial \Omega)$.
For any $d<\operatorname{diam}(\Omega)$ there exist points $a,b\in\Omega$ with $|a-b|>d$. Consider the line through $a$ and $b$ and (just to simplify notation) identify it with $\Bbb R$, where $a<b$. There must be a point belonging to $\partial\Omega$ in the interval $(-\infty,a]$ and one in the interval $[b,\infty)$, thus showing $\operatorname{diam}(\partial\Omega)\ge |b-a|>d$. This shows $\operatorname{diam}(\Omega)\le \operatorname{diam}(\partial \Omega)$.
