# Can anybody check my proof?

1. Dec 18, 2012

### Artusartos

1. The problem statement, all variables and given/known data

I attached my question and answer...

2. Relevant equations

3. The attempt at a solution

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2. Dec 18, 2012

### jbunniii

It would be easier to give a good response if you would typeset your work here instead of pasting a photo.

Your proof is incorrect. The following statement is not necessarily true:

"Then for some $s_0 \in S$ such that $s_0 < M$, we have $cs_0 = \sup(cS)$."

There are two problems with this statement. First, there may not be any $s_0 \in S$ such that $s_0 < M$, for example, if $S$ contains exactly one point. Second, there may not be any $s_0$ which will give this exact equality: $cs_0 = \sup(cS)$. For example, consider $S = \{q \in \mathbb{Q}: q^2 < 2\}$ and $c = 1$. Then $\sup(cS) = \sqrt{2}$ and clearly this does not equal $cs_0$ for any $s_0 \in S$, because $S$ contains only rational numbers.

3. Dec 18, 2012

### SammyS

Staff Emeritus
Assuming that $\displaystyle \sup(\text{S})=M\ :$

If $\displaystyle \sup(c\text{S})\ne cM\,,\ \text{ then either }\ \sup(c\text{S})< cM\ \text{ or } \sup(c\text{S})> cM\ .$

If $\displaystyle \ \sup(c\text{S})> cM\,,\ \text{ then there exists}\ cs_0\in c\text{S}\ \text{ such that }\ cs_0>cM\ .\ \ \ ...$

That should quickly lead to a contradiction.

Then do the other case.

4. Dec 19, 2012

Thanks