Proof Checking for Homework: Tips and Tricks | Attached Question & Answer

[Artusartos]

Homework Statement

I attached my question and answer...

Homework Equations

The Attempt at a Solution

 

[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.

 

[SammyS]

Homework Statement

I attached my question and answer...

https://www.physicsforums.com/attachment.php?attachmentid=54064&d=1355849917

Homework Equations

The Attempt at a Solution

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.

 

[Artusartos]

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.

Thanks