What is the proof for sup{|x−y|, (x, y) ∈ A 2} = sup A−inf A?

Question

So my question is about what is the proof that:

$$\sup\{|x−y| : (x, y) \in A^2 \} = \sup A − \inf A$$

Please provide a detailed demonstration.

Thanks in advance.

Answer 1

Suppose $x,y \in A$, then $\inf A \le y$, $x \le \sup A$ hence $x-y \le \sup A - \inf A$ for all $x,y \in A$ and so $|x-y| \le \sup A - \inf A$ for all $x,y \in A$ and so $\sup_{x,y \in A} |x-y| \le \sup A - \inf A$.

Let $\epsilon>0$. Choose $x',y' \in A$ such that $x' > \sup A -\epsilon$, $y' < \inf A + \epsilon$. Then $\sup_{x,y \in A} |x-y| \ge x'-y' \ge \sup A -\inf A - 2 \epsilon$. Hence we have equality.