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Can anybody check my proof?

  1. Dec 18, 2012 #1
    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
     

    Attached Files:

  2. jcsd
  3. Dec 18, 2012 #2

    jbunniii

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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.
     
  4. Dec 18, 2012 #3

    SammyS

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    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.
     
  5. Dec 19, 2012 #4
    Thanks
     
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