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Math Help - Proof of supremum

  1. #1
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    Proof of supremum

    Prove that if f is a bounded function on a nonempty set S, then |sup f| is less than or equal to sup|f|.
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  2. #2
    Member mabruka's Avatar
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    Obviously


    f(x)\leq|f(x)| for all x, so it holds for supremum:


    \sup_{x} f(x)\leq \sup_x |f(x)|


    Then we have


    - \sup_x |f(x)| \leq \sup_{x} f(x) \leq \sup_x |f(x)|



    hence

     |\sup_{x} f(x) | \leq \sup_x |f(x)|
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