# Proving $\sup(A+B)=\sup(A)+\sup(B)$ [duplicate]

here's a homework question I'm currently working on:

Let $A,B \subset \mathbb{R}$ non-empty sets bounded from above and from below. Show that $A+B$ is upper bounded and that $\sup(A+B)=\sup(A)+\sup(B)$

$A+B=\{a+b:a\in A, b \in B\}$

It was pretty easy to show to $A+B$ is upper-bounded by $\sup(A)+\sup(B)$, but I'm not quite sure how to prove that this is also the supremum. Any hints?

Thanks!

## marked as duplicate by lhf, Asaf Karagila♦, Martin Sleziak, user21436, Rudy the ReindeerMar 26 '12 at 11:58

• Let $c$ some upper bound of $A + B$. Then for $a \in A$ $c-a$ is an upper bound of $B$, therefore ... – martini Mar 26 '12 at 11:42
• I think that I saw at least two of these questions posted on the site before. – Asaf Karagila Mar 26 '12 at 11:49
• Take $(a_n)_n \subset A$ to be the sequence such that $\lim_n a_n = \sup(A)$, similarly for $b_n$. What can you say about $(a_n+b_n)_n$ ? – dtldarek Mar 26 '12 at 11:52

Well $a+b\leq \sup(A)+sup(B)$ then $\sup(A+B) \leq \sup (A)+\sup(B)$ for the other inequality consider $a_{\epsilon}\in A$ such that $a_{\epsilon}>\sup A- \epsilon /2$ and $b_{\epsilon}\in B$ such that $b_{\epsilon}>\sup B- \epsilon /2$ then
$\sup(A+B)\geq a_{\epsilon}+b_\epsilon>\sup A +\sup B-\epsilon$ for any $\epsilon>0$ then
$\sup(A+B)\geq \sup A +\sup B$ $\blacksquare$