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): $$\inf A = 0\Rightarrow\sup A' = \sup A $$
I don't know how to proceed. My professor told me, as a hint, to find an upper bound for $A'$ which is smaller than $\sup A$ and I found the following: $\sup A -\inf A$
Any hints on where to go from here?
Suppose $\sup A'<\sup A$. Then there exists $a\in A$ such that $a>\sup A'\ge b$ for all $b\in A'$. In particular, we have $a>|a-c|$ for all $c\in A$. Since $\inf A=0$, for each $n\in\mathbb N^*$ there is some $c_n\in A$ such that $c_n<\frac an$. Thus, for all $n$ $$\sup A'\ge (a-c_n)\ge a-\frac{a}{n}.$$ Thus, $\sup A'\ge a$, a contradiction.