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Math Help - [SOLVED] Fixed point

  1. #16
    Senior Member Dinkydoe's Avatar
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    Bruno, I fixed my proof. Hopefully convincing now

    So we assumed f had no fixpoint and S=\left\{x:f(x)>x\right\}, and  T= \left\{x:f(x)< x\right\}.

    As shown before we know that S,T are closed under f

    Now let u= \sup(S). Then we have u\in T. Suppose that u\in S then f(u)>u but since f(u)\in S this contradicts the fact that u= \sup(S).

    Now since u\in T we have f(u)< u and we can find x_n\in (f(u), u)\cap S ( this is clearly true since u = \sup(S)

    But then we have f(u)<x_n<f(x_n)<u and this is a contradiction.
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  2. #17
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Dinkydoe View Post
    Bruno, I fixed my proof. Hopefully convincing now

    So we assumed f had no fixpoint and S=\left\{x:f(x)>x\right\}, and  T= \left\{x:f(x)< x\right\}.

    As shown before we know that S,T are closed under f

    Now let u= \sup(S). Then we have u\in T. Suppose that u\in S then f(u)>u but since f(u)\in S this contradicts the fact that u= \sup(S).

    Now since u\in T we have f(u)< u and we can find x_n\in (f(u), u)\cap S ( this is clearly true since u = \sup(S)

    But then we have f(u)<x_n<f(x_n)<u and this is a contradiction.
    That works!
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