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Let $A$,$B$ be subsets of $\mathbb{R}$ that are nonempty and bounded above. By axiom of completeness, we know $\alpha = \sup A$ and $\beta = \sup B$ exists.

Let $$E = \{x+y \text{ such that }x \in A \text{ and } y \in B\}$$

Prove that $\alpha + \beta = \sup E$

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I reordered the sentences and typeset the question. For some basic information about writing math on this site see e.g. here, here, here and here. – user17762 Feb 7 '13 at 0:12
Thanks I appreciate it. I've been wondering where to get some info on how to format questions. – Math Student Feb 7 '13 at 0:38

$1$. First show that $\alpha + \beta$ is an upper bound. This should be straightforward.

$2$. Next argue out why $\alpha + \beta$ should be the least upper bound i.e. consider $\alpha + \beta - \epsilon$ for some $\epsilon > 0$ and argue out why it cannot be an upper bound by producing elements $x \in A$ and $y \in B$ such that $$x+y > \alpha + \beta - \epsilon$$

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Could you consider some arbitrary upper bound of E, and call it me. This, for any e in E, e<=m. So for any x in A and y in B, x+y <= m. But since x and y are arbitrary and you already proved alpha + beta is an upper bound, alpha + beta <= m. Thus, alpha + beta = supE. Is this correct? – Math Student Feb 7 '13 at 1:07

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