Artusartos
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Check my proof for this homework problem
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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 [itex]s_0 \in S[/itex] such that [itex]s_0 < M[/itex], we have [itex]cs_0 = \sup(cS)[/itex]."
There are two problems with this statement. First, there may not be any [itex]s_0 \in S[/itex] such that [itex]s_0 < M[/itex], for example, if [itex]S[/itex] contains exactly one point. Second, there may not be any [itex]s_0[/itex] which will give this exact equality: [itex]cs_0 = \sup(cS)[/itex]. For example, consider [itex]S = \{q \in \mathbb{Q}: q^2 < 2\}[/itex] and [itex]c = 1[/itex]. Then [itex]\sup(cS) = \sqrt{2}[/itex] and clearly this does not equal [itex]cs_0[/itex] for any [itex]s_0 \in S[/itex], because [itex]S[/itex] contains only rational numbers.
Your proof is incorrect. The following statement is not necessarily true:
"Then for some [itex]s_0 \in S[/itex] such that [itex]s_0 < M[/itex], we have [itex]cs_0 = \sup(cS)[/itex]."
There are two problems with this statement. First, there may not be any [itex]s_0 \in S[/itex] such that [itex]s_0 < M[/itex], for example, if [itex]S[/itex] contains exactly one point. Second, there may not be any [itex]s_0[/itex] which will give this exact equality: [itex]cs_0 = \sup(cS)[/itex]. For example, consider [itex]S = \{q \in \mathbb{Q}: q^2 < 2\}[/itex] and [itex]c = 1[/itex]. Then [itex]\sup(cS) = \sqrt{2}[/itex] and clearly this does not equal [itex]cs_0[/itex] for any [itex]s_0 \in S[/itex], because [itex]S[/itex] contains only rational numbers.
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Assuming that [itex]\displaystyle \sup(\text{S})=M\ :[/itex]Artusartos said: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
If [itex]\displaystyle \sup(c\text{S})\ne cM\,,\ \text{ then either }\ \sup(c\text{S})< cM\ \text{ or } \sup(c\text{S})> cM\ .[/itex]
If [itex]\displaystyle \ \sup(c\text{S})> cM\,,\ \text{ then there exists}\ cs_0\in c\text{S}\ \text{ such that }\ cs_0>cM\ .\ \ \ ...[/itex]
That should quickly lead to a contradiction.
Then do the other case.
Artusartos
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SammyS said:Assuming that [itex]\displaystyle \sup(\text{S})=M\ :[/itex]
If [itex]\displaystyle \sup(c\text{S})\ne cM\,,\ \text{ then either }\ \sup(c\text{S})< cM\ \text{ or } \sup(c\text{S})> cM\ .[/itex]
If [itex]\displaystyle \ \sup(c\text{S})> cM\,,\ \text{ then there exists}\ cs_0\in c\text{S}\ \text{ such that }\ cs_0>cM\ .\ \ \ ...[/itex]
That should quickly lead to a contradiction.
Then do the other case.
Thanks