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SupA<SupB, What can you say?

  1. Jan 21, 2009 #1
    1. The problem statement, all variables and given/known data
    If Sup A<Sup B Then show that there exists a b[tex]\in[/tex]B which serves as an upper bound for A.
    First off, I am not looking for a complete solution but rather a hint.

    2. Relevant equations
    SupA-[tex]\epsilon[/tex]<a for some a[tex]\in[/tex]A
    SupB-[tex]\epsilon[/tex]<b for some b[tex]\in[/tex]B

    3. The attempt at a solution
    The only thing I have succeeded in so far is "locating one element of each set"
    SupA-[tex]\epsilon[/tex]<a[tex]\leq[/tex]SupA for some a[tex]\in[/tex]A
    SupB-[tex]\epsilon[/tex]<b[tex]\leq[/tex]SupB for some b[tex]\in[/tex]B

    I know that if I can demonstrate that SupA[tex]\leq[/tex]SupB-[tex]\epsilon[/tex], then I can be done. Equivalently, if I can show that b[tex]\geq[/tex]SupA for some b[tex]\in[/tex]B I will also be done. However, right now this has me caught.
  2. jcsd
  3. Jan 21, 2009 #2
    I think I might have a solution. Redefine Epsilon so that it is less than or equal to SupB-SupA. Then SupA[tex]\leq[/tex]SupB-[tex]\epsilon[/tex]. We can still guarantee an element of b SupB-[tex]\epsilon[/tex] and Sup B (same for a). Thus this element of b[tex]\geq[/tex]SupA and so b is an upper bound. If this incorrect, please tell me. Sorry to those of you who thought long and hard.
  4. Jan 22, 2009 #3
    Firstly I assume we're talking about real numbers.

    The key property we want to use: for every [tex]\varepsilon>0[/tex] there is a [tex]b\in B[/tex] such that [tex]\sup{B} - \varepsilon < b \leq \sup{B}[/tex].

    Now what positive number to choose for epsilon? It's given (albeit subtly) in the question: we know that sup(B)-sup(A) > 0.
  5. Jan 22, 2009 #4
    Many thanks, but I think I realized this just before you answered (in my post). Yeah, it was about the reals; what a great feeling when I recognized this. Again, Many thanks for being the first to help. Thread closed, because my homework has been turned in.
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