For any subset $A\subseteq\mathbb{R^+}$, we define:
$$A':=\{|x-y| \mid x,y\in A,x\ne y\}.$$
By definition, $A'\subseteq\mathbb{R^+}$.
I showed the following:$$\sup A \text{ finite} \Longrightarrow \sup A'\le \sup A.$$
I now want to show (still assuming $\sup A$ finite): $$\sup A' = \sup A \Longleftrightarrow \inf A = 0.$$
I don't know how to proceed. For the first statement ($\Rightarrow$) I showed that $A$, being a subset of $\mathbb{R^+}$, is bounded from below by $0$ and therefore $\inf A$ exists. To show that $\inf A =0$, I considered proving:
$\forall\space\gamma>0$ such that $\gamma\in{R^{+*}}$, $\exists\space a\in A$ such that $a<\gamma$.
If $\gamma> \sup A$, then any element $a\in A$ is such that $a<\gamma$.
If $\gamma\le \sup A$, by the density of $\mathbb{Q}$ in $\mathbb{R}$, I can find a rational $r\in\mathbb{Q}$ such that $0<r<\gamma$. But I need to show that this $r\in A$.
Any hints?
Prove that $(\inf A>0 \land \sup A<\infty)\implies \sup A'<\sup A.$
Let $0<r=\inf A$ and $s=\sup A.$ For any $x,y\in A$ we have $x\leq s$ and $y\geq r$ so $$x-y\leq s-y\leq s-r$$ and we have $y\leq s$ and $x\geq r$ so $$y-x\leq s-x\leq s-r.$$ So $|x-y|=\max (x-y,y-x)\leq s-r.$
So $A'\subset (0,s-r]$ so $\sup A'\leq s-r<s=\sup A.$