Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\alpha + \beta = \sup E$

share|improve this question
    
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 Answer 1

$1$. 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 $$x+y > \alpha + \beta - \epsilon$$

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.