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Math Help - Upper bound proof

  1. #1
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    Upper bound proof

    Let x be an upper bound for A subset Real #'s.
    Prove that if x element A, then x = sup(A).

    Proof:
    Let x be an upper bound for A subset Real #'s.
    Assume x element A.
    a <= x for all a element A by definition of upperbound.
    Since x element A, x <= x.
    Therefore, since x is an upper bound for A and x <= x for all upper bounds x of A, x = sup(A).

    I was wondering if this was a valid proof.
    If not could you help correct it please.
    Thanks
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by tbyou87 View Post
    Let x be an upper bound for A subset Real #'s.
    Prove that if x element A, then x = sup(A).

    Proof:
    Let x be an upper bound for A subset Real #'s.
    Assume x element A.
    a <= x for all a element A by definition of upperbound.
    Since x element A, x <= x.
    Therefore, since x is an upper bound for A and x <= x for all upper bounds x of A, x = sup(A).

    I was wondering if this was a valid proof.
    If not could you help correct it please.
    Thanks
    I would do it this way:

    Suppose x in A is an upper bound for A, but that x is not the supremum of A.

    Then there exists another upper bound y for A, such that y<x. But this conradicts y being an upper bound for A, as there is an element x of A such that x>y.

    Hence if x in A, is an upper bound for A a subset of R, then x = sup(A).

    RonL
    Last edited by CaptainBlack; March 11th 2007 at 12:19 AM. Reason: Tidying up the last line
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tbyou87 View Post
    Let x be an upper bound for A subset Real #'s.
    Prove that if x element A, then x = sup(A).

    Proof:
    Let x be an upper bound for A subset Real #'s.
    Assume x element A.
    a <= x for all a element A by definition of upperbound.
    Since x element A, x <= x.
    Therefore, since x is an upper bound for A and x <= x for all upper bounds x of A, x = sup(A).

    I was wondering if this was a valid proof.
    If not could you help correct it please.
    Thanks
    well it sounded great up to the " Therefore, since x is an upper bound for A and x <= x for all upper bounds x of A, x = sup(A)." line.

    remember the definition of supremeum--it is the least upper bound. that is you have to show that two conditions hold:
    1) if x is the supremum for a set A, then a <= x for all a in A (you proved this nicely)

    2) if y is any other upperbound for A, then y>=x (you have yet to prove this.)

    i'd end with something like, "let y be another upperbound for A distinct from x. since x is the upperbound for A and is also an element of A, then x is also max(A), and therefore is y >= x. (since y has to be an upperbound outside the set). thus, since x >= a for all a in A, and x <= y for all y upperbounds of A, x is sup(A)."

    well, the language needs to be cleared up a bit, but i'd say something to that effect. no matter how you phrase it though, you have to show those two things.
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