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Let $A$ and $B$ be two non-empty bounded subsets of $\mathbb{R}$. Write $A + B = \{x + y : x \in A, y \in B\}$. Show that $\sup(A + B) = \sup A + \sup B$.

This is a problem from an assignment from my analysis course. the two definitions I know of supremum is that it is the least upper bound and also for all $\epsilon >0 \sup(S)-\epsilon < a$ for some $a$ belonging to the set $S$.

I am unable to prove this using the above definitions (though the second one is actually a theorem that can be proven from the previous definition). I have already tried writing the definition of supremum and saying that for all $\epsilon >0$, there exists $a+b$ belonging to $A+B$ such that $\sup(A+B)<a+b$ But I am unable to proceed further.

Can anyone please provide a right approach to solve this problem and what lemmas must be proven.

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    $\begingroup$ What have you attempted? $\endgroup$ – ncmathsadist Sep 23 '18 at 19:19
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    $\begingroup$ Can you at least see why $\sup A+\sup B$ is an upper bound of $A+B$? Given that, can you see how to proceed? $\endgroup$ – Crosby Sep 23 '18 at 19:22
  • $\begingroup$ I have edited the question to show how much I have actually tried. To be honest I am beginner to analysis and have not done a lot of progress on this question. $\endgroup$ – NKB Sep 23 '18 at 19:23
  • $\begingroup$ @Crosby Yep I see it's an UB. $\endgroup$ – NKB Sep 23 '18 at 19:23
  • $\begingroup$ @NKB Okay, now can you show it is the least upper bound? What do you know about $\sup A$ and $\sup B$, by definition. $\endgroup$ – Crosby Sep 23 '18 at 19:24
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You wrote in the comments that you see why $supA+supB$ is an upper bound of the set $A+B$. Now you just have to prove it is the least upper bound. Alright, so fix $\epsilon>0$. Then you know there exists $a\in A$ such that $a>supA-\frac{\epsilon}{2}$ and there exists $b\in B$ such that $b>supB-\frac{\epsilon}{2}$. So then $a+b\in A+B$ and $a+b>supA-\frac{\epsilon}{2}+supB-\frac{\epsilon}{2}=supA+supB-\epsilon$. Hence anything smaller than $supA+supB$ is not an upper bound of $A+B$.

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