$A-B = \{a-b: a\in A, b\in B\}$. Prove that $\sup(A-B) = \sup(A) – \inf(B)$
OK, let $x=\sup(A), y=\sup(B)$:
$a\in A \implies a\leq x$
$b\in B \implies b\leq y$
$a+b\leq x+y$ is a upper bound
Take $\varepsilon > 0$ and find $a,b$ s.t.:
$a>x-\dfrac {\varepsilon}{2}, b>y-\dfrac {\varepsilon}{2}$
$a+b>x+y-\varepsilon$
It means that every potential smaller upper bound $x+y-\varepsilon$ is not really an upper bound.
Therefore $\sup(A+B) = \sup(A) + \sup(B)$
$A-B = \{a-b: a\in A, b\in B\}$.
Proving that $\sup(A-B) = \sup(A) – \inf(B)$
Let $x=\sup \left( A\right), y=\inf \left( B\right)$
$a\in A\implies a\leq x$
$b\in B \implies b\geq y$
$a+b\geq x+y$
Best Answer
Define similarly
Hint 1: $\sup(X+Y)=\sup X+\sup Y$
Hint 2: $\sup(-Y)=-\inf Y$
Proving that $\sup(X+Y)=\sup X+\sup Y$ is psychologically simpler than the statement you have, but they're actually the same, in view of hint 2.