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This question already has an answer here:

Let $A,B$ be subsets of $\mathbb{R}$. Prove that $\sup{(A+B)} = \sup{A}+\sup{B}$.

I think in order to solve this we are going to have to use the mathematical definition of supremum. Maybe we can break this up into $4$ cases: $A$ is finite, $B$ infinite; etc. This would allow us to relate supremum to the maximal value of a set and make it easier to work with.

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marked as duplicate by user147263, Shailesh, YoTengoUnLCD, Silvia Ghinassi, Joel Reyes Noche Mar 2 '16 at 4:04

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  • $\begingroup$ If B is infinite, does it have a supremum? Otherwise, using the definition of supremum is a good idea. $\endgroup$ – Sean I Mar 2 '16 at 3:00
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$A \leq \sup A$, $B \leq \sup B$, so $\sup A + \sup B \geq \sup(A+B)$

Also, $A + B \leq \sup (A+B)$, so $A \leq \sup(A+B) - B$, so $\sup A \leq \sup (A+B) -B$, then $\sup B \leq \sup (A+B) - \sup A$. Done

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If $x\in A+B$, then $\exists a \in A$, and $\exists b\in B$ such that $x=a+b$. We know that $a \leq sup(A)$ and $b \leq sup(B)$, therefore $x = a + b \leq Sup(A) +Sup(B)$. This means $Sup(A+B) \leq Sup(A)+Sup(B)$.

Now take an arbitrary $k$ satisfying $k < Sup(A) + Sup(B) $, then we can find $k_A$ and $k_B$ such that $k = k_A+k_B$, together with $k_A < Sup(A)$ and $k_B< Sup(B)$. Then by definition we know that $\exists a \in A$ and $\exists b \in B$ such that $a > k_A$ and $b>k_B$. Therefore $ a+b > k$ and hence $Sup(A+B) \geq Sup(A)+Sup(B)$.

As a result we have $Sup(A+B) = Sup(A)+Sup(B)$

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  • $\begingroup$ Using words instead of symbols yields a much more readable writing. $\endgroup$ – YoTengoUnLCD Mar 2 '16 at 3:32

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