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Homework Help: Sup A < sup B

  1. Sep 5, 2010 #1
    1. The problem statement, all variables and given/known data
    If sup A<sup B, then show that there exists an element b[tex]\in[/tex]B that is an upper bound for A.

    2. Relevant equations

    3. The attempt at a solution
    This one is really frustrating me. I'm having trouble even beginning.
    I know sup A implies s[tex]\leq[/tex]b where c is an upper bound for A.
    Similarily sup B implies s[tex]\leq[/tex] d where d is an upper bound for B.
  2. jcsd
  3. Sep 5, 2010 #2
    [tex]sup B= sup A+ \epsilon[/tex]

    Remember what sup b is and it's closeness to elements of B.
    Last edited: Sep 5, 2010
  4. Sep 5, 2010 #3
    I have no clue how you got to that.
  5. Sep 5, 2010 #4
    Well if x < y that means y= x+b where b= y-x.
  6. Sep 5, 2010 #5
    so x<x+b
    So that statement alone shows that b is an upper bound?
  7. Sep 5, 2010 #6
    I made an error while writing the latex. Please review my edit.
  8. Sep 5, 2010 #7
    [tex]sup B- \epsilon= sup A[/tex]
    Since sup B is the least upper bound what can we say about
    [tex]sup B- \epsilon_{0}[/tex] ?
    Is it in B?

    The point it that if we pick [tex] \epsilon_{0} [/tex] small enough we can find a number that is both in B and is an upper bound of A.
  9. Sep 6, 2010 #8
    [tex]sup B- \epsilon_{0}[/tex] must be an upper bound also?
  10. Sep 6, 2010 #9
    Do you know the definition of [tex] sup(B)[/tex] ?

    Given that sup(B) is the least upper bound this means that it is the smallest number that is an upper bound of the set B.

    That means given any [tex]\epsilon_{0}[/tex], no matter how small, there exist [tex]b_{0}\inB[/tex] such that
    [tex]b_{0} > sup(B) -\epsilon_{0}[/tex].

    Lets start again.

    [tex] sup(B) - sup(A) = \alpha >0[/tex]

    If we pick any [tex]\epsilon_{0}[/tex] such that [tex]\epsilon_{0} < \alpha[/tex] we can find an element in B such that
    [tex]b_{0} > sup(B) -\epsilon[/tex]. So [tex]b_{0}[/tex] is an upper bound of A.
  11. Sep 6, 2010 #10


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    You could prove this by contradiction as well. Assume there is no element in B that is an upper bound for A, and then show it leads to the conclusion sup(A)>sup(B), which contradicts the condition sup(A)<sup(B).
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