Show $\sup{A}-\inf{B}=\sup\{a-b:a\in A, b\in B\}$

Question

Let $A, B \subset\mathbb{R}$ be bounded sets. Show $$\sup{A}-\inf{B}=\sup\{a-b:a\in A, b\in B\}$$

Answer 1

We have $$\sup A\geq a\quad,\quad\forall a\in A\qquad;\qquad \inf B\leq b\quad,\quad\forall b\in B$$ so $$\sup A-\inf B\geq a-b\quad,\quad\forall a\in A,b\in B$$ hence we deduce $$\sup A-\inf B\geq S:=\sup\{a-b:a\in A, b\in B\}$$ Now let $\epsilon>0$ there's $a\in A$ s.t. $a\geq \sup A-\epsilon/2$ and there's $b\in B$ s.t. $b\leq \inf B+\epsilon/2$ then $$\sup A-\inf B-\epsilon=(\sup A-\epsilon/2)-(\inf B+\epsilon/2)\leq a-b$$ hence $$\sup A-\inf B= S$$

[edit: conclusion is that sup A - inf B ≤ a-b, not a+b]