Let $A$ and $B$ be upper bounded subsets, and let $A+B$ be the subset of numbers $x+y$ with $x \in A$ and $y \in B$. Prove that $\sup(A+B)=\sup(A)+ \sup(B)$.
Let $a =\sup A$ and $b =\sup B$, it is easy to see that $x+y \le a+b$.
Because $a =\sup A$ the for all $\epsilon>0$ exist $a_{0} \in A$ that satisfy $$a-\frac{\epsilon}{2} <a_{0} \le a$$ And $b =\sup B$ the for all $\epsilon>0$ exist $b_{0} \in B$ that satisfy $$b-\frac{\epsilon}{2} <b_{0} \le b$$ Then $$a+b - \epsilon < a_{0} + b_{0} \le a+b $$
We conclude that $a+b $ is the supremum of $A+B$, and $\sup(A+B)= \sup A + \sup B$. Am I right?