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Math Help - supremum on closed section proof

  1. #1
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    supremum on closed section proof

    f is continues on [a,b]
    A={x in [a,b]|f(x)=f(a)}
    prove that f(x) has a maximum
    f(supA)=f(a)
    ??


    first of all there are two kinds of maximums here
    1st a supremum which is inside [a,b] is called maximum

    and maximum of the function values
    correct??
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: supremum on closed section proof

    Quote Originally Posted by transgalactic View Post
    f is continues on [a,b]
    A={x in [a,b]|f(x)=f(a)}
    prove that f(x) has a maximum
    f(supA)=f(a)
    ??


    first of all there are two kinds of maximums here
    1st a supremum which is inside [a,b] is called maximum

    and maximum of the function values
    correct??


    Extreme value theorem - Wikipedia, the free encyclopedia
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  3. #3
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    Re: supremum on closed section proof

    Quote Originally Posted by transgalactic View Post
    f is continues on [a,b]
    A={x in [a,b]|f(x)=f(a)}
    prove that f(x) has a maximum
    f(supA)=f(a)?
    Because \left( {\forall x \in A} \right)\left[ {x \leqslant b} \right], you know that \beta=\sup\{A\} exists and \beta\in[a,b].
    If it is true \beta\in A you are done.
    Else If you assume \beta\notin A that leads to a contradiction to the continuity of f.

    BTW: This not the Extreme value theorem.
    Last edited by Plato; August 4th 2011 at 12:58 PM.
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