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How can I prove $\sup(A+B)=\sup A+\sup B$ if $A+B= \lbrace a+b\mid a\in A, b\in B\rbrace $

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?



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

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  • 1
    $\begingroup$ Let $c$ some upper bound of $A + B$. Then for $a \in A$ $c-a$ is an upper bound of $B$, therefore ... $\endgroup$ – martini Mar 26 '12 at 11:42
  • $\begingroup$ I think that I saw at least two of these questions posted on the site before. $\endgroup$ – Asaf Karagila Mar 26 '12 at 11:49
  • $\begingroup$ 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 $(a_n+b_n)_n$ ? $\endgroup$ – 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

$\sup(A+B)\geq \sup A +\sup B$ $\blacksquare$

  • $\begingroup$ Isnt't there a dot missing prior to "for the other..."? $\endgroup$ – Fabian Schn. Nov 15 '17 at 11:33

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