Bounded set absolute value of the difference

Question

Let $\emptyset\neq A\subseteq\mathbb R$. Prove that $$ \sup(\left\{ \,\left|x-y\right|\;:\;x,y\in A\,\right\})=\sup A-\inf A$$ and prove the following as the first step: $$\sup(\left\{ \,\left|x-y\right|\;:\;x,y\in A\,\right\})=\sup(\left\{ \,x-y\;:\;x,y\in A\,\right\}). $$ I've tried proving the assumption using the strict definition of the supremum, but all I could do is prove the assumption without the needed "first step". any idea on how to do that?

Answer 1

Hint

$$\sup\limits_{\{(x,y)\in A^2\}}\vert x-y\vert= \sup\limits_{\{(x,y)\in A^2 \mid x \ge y\}}(x-y)= \sup\limits_{\{(x,y)\in A^2\}}(x-y)$$

for the first step.

And

$$x-y \le \sup A -\inf A$$ for any $(x,y) \in A^2$ for the conclusion.