For $B=\{|x-y|: x,y\in A\}$, show that $\sup B = \sup A - \inf A$ and find $\inf B$ I'm trying to solve this exercise:

Let $A$ be a non-empty set $\mathbb R$-bounded. 
  Let $B=\{|x-y|: x,y\in A\}$.
  
  
*
  
*Prove that $B$ has a least upper-bound and a greatest lower-bound.
  
*Prove that $\sup B = \sup A - \inf A$.
  
*Find $\inf B$.
  

I've completed part 1, but how should I tackle parts 2 and 3?
 A: Given $x, y \in A$, $x \le \sup A$ and $y \ge \inf A$, thus $x - y \le \sup A - \inf A$ Similarly $y - x \le \sup A - \inf A$. Therefore $$|x - y| \le \sup A - \inf A.$$ Since this is true for all $x,y \in A$, the set $\{|x - y| : x, y\in A\}$ is bounded above by $\sup A - \inf A$. Consequently, $$\sup B \le \sup A - \inf A.$$ On the other hand, given $\varepsilon > 0$, $\inf A + \frac{\varepsilon}{2}$ is not a lower bound for $A$ and $\sup A - \frac{\varepsilon}{2}$ is not an upper bound for $A$. Hence, there exists $x, y \in A$ for which $\inf A + \frac{\varepsilon}{2} > y$ and $\sup A - \frac{\varepsilon}{2} < x$. Thus $$\sup A - \inf A - \varepsilon = \left(\sup A - \frac{\varepsilon}{2}\right) - \left(\inf A + \frac{\varepsilon}{2}\right) < x - y \le |x - y| \le \sup B.$$ Since $\sup A - \inf A \le \sup B + \varepsilon$ for all $\varepsilon > 0$, we have $$\sup A - \inf A \le \sup B.$$ Hence $$\sup B = \sup A - \inf A.$$
You can see that $\inf B = 0$ by noting that $(1)$ $|x - y| \ge 0$ for all $x,y \in A$ and $(2)$ $0 \in B$ by taking an $x\in A$ and writing $0 = |x - x|$.
