TITLE: How can I prove $\sup(A+B)=\sup A+\sup B$ if $A+B=\{a+b\mid a\in A, b\in B\}$
QUESTION [46 upvotes]: If $A,B$ non empty, upper bounded sets and $A+B=\{a+b\mid a\in A, b\in B\}$, how can I prove that $\sup(A+B)=\sup A+\sup B$?

REPLY [3 votes]: $$\forall ~~a\in A,~~b\in B, a\le \sup A~~ \text{and }~~b\le \sup B \implies a+b \le a\le \sup A +\sup B$$
hence $$\sup A +B\le \sup A +\sup B$$
Now we know there exist $a_n\in A$ and $b_n\in B$ such that $$\sup A=\lim_{n\to\infty} a_n$$
and 
$$\sup B =\lim_{n\to\infty}b_n$$
then 
$$\sup A+\sup B =\lim_{n\to\infty} a_n+\lim_{n\to\infty} b_n =\lim_{n\to\infty} a_n+b_n \le \sup A+B$$
whence 
$$\sup A+\sup B = \sup A+B$$