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Possible Duplicate:
How can I prove $\sup(A+B)=\sup A+\sup B$ if $A+B=\{a+b\mid a\in A, b\in B\}$

Prove:

If $A$ and $B$ are nonempty subsets of $\mathbb{R}$ and are both bounded above, then $$\operatorname{lub}(A+B)=\operatorname{lub}(A)+\operatorname{lub}(B).$$

I am thinking I need to utilize the fact that $A+B=\{a+b: a \in A, b \in B\}$, I'm just not sure how this gives a proof.

Thanks.

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marked as duplicate by mrf, Davide Giraudo, Asaf Karagila, Paul, Hagen von Eitzen Feb 4 '13 at 10:25

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First we show that $\sup(A + B) \leq \sup(A) + \sup(B)$:

Let $\gamma$ in $A + B$ be arbitrary. We see that $$\gamma = a + b \quad \text{where} \quad a \in A \text{ and } b \in B$$ Now, we know that $$\sup(A) \geq a \quad \text{and} \quad \sup(B) \geq b$$ so it follows that $$\gamma = a + b \leq \sup(A) + \sup(B)$$ Since $\gamma$ was arbitrary, this holds for all elements of $A+B$. We must have $$\sup(A + B) \leq \sup(A) + \sup(B)$$

Next, we show that $\sup(A + B) \geq \sup(A) + \sup(B)$:

Well, assume the contrary. Then we have $$ \sup(A + B) < \sup(A) + \sup(B)$$ This implies that for all $\gamma \in A + B$ there exists some $a \in A$ and some $b \in B$ such that $\gamma < a + b$. But $(A + B) \ni \sigma = a + b$, so this is a contradiction.

Putting these two together, we see that $\sup(A + B) = \sup(A) + \sup(B)$ as we wanted.

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