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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?

Arturo Magidin
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Marinela
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  • There’s a slight mistake: you cannot guarantee that $a_0\lt a$ or that $b_0\lt b$. You can only guarantee that $a_0\leq a$ and $b_0\leq b$. (What if $A$ and $B$ are singletons?) Also, I think the final displayed equation is missing a + before the $\epsilon$. – Arturo Magidin Mar 05 '21 at 01:22
  • I tagged this [solution-verification] because the question itself is a duplicate: https://math.stackexchange.com/questions/4551/how-can-i-prove-supab-sup-a-sup-b-if-ab-ab-mid-a-in-a-b-in-b – Arturo Magidin Mar 05 '21 at 01:23
  • @ArturoMagidin yeah, that was my mistake! I dind't read right , but if they are singletons we have the equality right? – Marinela Mar 05 '21 at 01:59
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    But the point is that you cannot find an $a_0$ satisfying $a-\frac{\epsilon}{2}\lt a\lt a_0$, because the set would not contain any $a$ less than $a_0$. Or you could have a that has “gaps” and your $\epsilon$ happens to fall between the maximum of the set (which would also be the supremum) and the “next” largest element. So you need to replace those $\lt$ with $\leq$. – Arturo Magidin Mar 05 '21 at 02:01
  • But $a_{0} \in A$ and we can have that $a-\frac{\epsilon}{2}<a_{0} \le a$, because $a$ is the supreme – Marinela Mar 05 '21 at 02:37
  • Yes; that’s why with the $\leq$ the statement is right, while with the $\lt$ the statement need not be true. – Arturo Magidin Mar 05 '21 at 02:38

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