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Supremum of sets

  1. Aug 13, 2011 #1
    1. The problem statement, all variables and given/known data
    Let A be a set of real numbers that is bounded above and let B be a subset of real numbers such that A (intersect) B is non-empty.
    Show that sup (A(intersect)B) <= sup A


    3. The attempt at a solution
    I don't know how to start but tried this...
    Let C = A (intersect) B
    So sup C = sup (A (intersect) B)

    Then I thought of trying to prove it by contradiction,
    show sup C > sup A leads to a contradiction.

    Since for all a in A, a <= sup A.
    a < sup C,
    can I say that this leads to a contradiction as there exist an a that is larger than c because not all elements in A are in C...

    but it seems a bit weak...
     
  2. jcsd
  3. Aug 13, 2011 #2

    HallsofIvy

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    The crucial point here is that [itex]C= A\cap B[/itex] is a subset of A. But it does NOT follow that C is a proper subset of A. It might happen that B= A, in which case [itex]A\cap B= A\cap A= A[/itex].

    It is not just that a< sup a< sup C but that there must exist a member, x, of C such that [itex]sup A< c\le sup C[/itex]
     
  4. Aug 13, 2011 #3
    Hmm... but is my approach correct?? Because I am totally clueless about this now...
     
  5. Aug 13, 2011 #4

    vela

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    You can certainly come up with a proof by contradiction, but HallsofIvy's point is that your proposed contradiction isn't really a contradiction. As he noted, if C=A, then all elements of A are in C, so you can't assume there's an element x∈A that's not in C.

    Let a = sup(A) and c = sup(C), and assume c>a. Show that a is an upper bound of C and explain why this leads to a contradiction.
     
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