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Suprema and Infima

  1. Jan 25, 2005 #1
    I need to show that if sup A < sup B, then there is an element in B that is an upper bound of A.

    Well I know that if sup B is in B, then sub B is the element in B that is an upper bound of A. But I don't know how to show this is true if sup B is not in B.
     
  2. jcsd
  3. Jan 25, 2005 #2

    Hurkyl

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    You try a proof by contradiction?
     
  4. Jan 25, 2005 #3

    AKG

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    You need to prove simply that [itex]\exists b \in B[/itex] such that [itex]\sup A \leq b \leq \sup B[/itex]. Assume not: then [itex]\sup A[/itex] is an upper bound of B, which contradicts the fact that [itex]\sup A < \sup B[/itex], and you're done.
     
  5. Jan 25, 2005 #4
    AKG--

    I'm not sure I understand your reasoning. What you say makes sense but that doesn't prove that such a b exists. Or does it and I just don't get it which is equally as possible.
     
  6. Jan 25, 2005 #5

    Hurkyl

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    Suppose b doesn't exist. Can you prove a contradiction? (such as sup A = sup B)
     
  7. Jan 25, 2005 #6
    But what about the case where sup B is not in B? Then I don't think I can get that contradiction.
     
  8. Jan 25, 2005 #7

    Hurkyl

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    Then, try to find two sets A and B where b doesn't exist, but sup A < sup B!

    We're lucky, we already know this will be impossible, but it still might give you insight.

    (As always, it often helps to write out what the definitions say too)
     
  9. Jan 25, 2005 #8
    Okay solid. Thanks.
     
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