Prove $\sup\{|x -y| \mid x,y \in S\} = \sup S -\inf S$ Is the following a proof or an explanation?
Let $ S \subseteq\mathbb{R} $ be a nonemtpy and bounded set. We will define $ \operatorname{diam}\left(S\right) = \sup\{ \left|x - y\right| \mid x, y \in S \} $. We shall prove $ \operatorname{diam}\left(S\right) = \sup S -\inf S $.
Proof:
By the definition of supremum and infimum, $ \sup S $ is the minimum upper bound of $ S $ and $ \inf S $ is the maximum lower bound. Therefore, the distance from $ \sup S $ to $ \inf S $ is the greatest among the distances between elements in $ S $, whether $ \sup S, \inf S \in S $ or not, and therefore the statement is proven.
 A: This is not a valid proof, it is too vague.
In order to show what you want, you need to use the fact that:
$$ y \geq \inf \mathbf{S}, \forall y \in \mathbf{S}, $$
$$-y \leq  - \inf \mathbf{S}, \forall y \in \mathbf{S}, $$
and,
$$ x \leq \sup \mathbf{S}, \forall x \in \mathbf{S}. $$
So now we have, that without loss of generality, assume $x\geq y$, then
$$ |x-y|\leq x-y \leq \sup \mathbf{S} - y \leq \sup \mathbf{S} - \inf \mathbf{S}$$
A: By translation, we can assume that $\inf S = 0$.
And by refletion, we can assume that $\sup S = 0$ or $\sup S > 0$.
If $\sup S = 0$, then $S = \{0 \}$. It is a trivial case.
Assume that $\sup S > 0$. We know that if $s \in S$ then $ 0 \le s \le \sup S$.
Thus $\sup\{|x -y| \mid x,y \in S\} \le \sup S$.
There is a non-negative sequence $(a_n)_{n \in \mathbb{N}}$ such that
$\lim a_n = 0$. Also, there is a non-negative sequence $(b_n)_{n \in \mathbb{N}}$
such that $\lim b_n = \sup S$. Because $\lim (b_n - a_n) = \sup S$,
we can say that $\sup\{|x -y| \mid x,y \in S\} \ge \sup S$.
From the above two fact, we obtain the desired result.
