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Math Help - Theorem 4.2 in Rudin POMA

  1. #1
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    Theorem 4.2 in Rudin POMA

    Can someone with the book ready to hand explain how he proves the converse? He just says taking {\delta_n}=1/n, we thus find a sequence......

    Sometimes he's a little too presuming of the reader's perception.
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  2. #2
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    Re: Theorem 4.2 in Rudin POMA

    Thanks to this choice, we can find a sequence \{x_n\} such that d(x_n,p)\leq\frac 1n and d(f(x_n),f(p))\geq \varepsilon, hence we have shown that we don't have the part after the "if and only if".
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    Re: Theorem 4.2 in Rudin POMA

    Quote Originally Posted by boromir View Post
    Can someone with the book ready to hand explain how he proves the converse? He just says taking {\delta_n}=1/n, we thus find a sequence......
    You may be mixing apples and oranges.
    He means for each \delta_n=\tfrac{1}{n} we can find a p_n such that d(p_n,p)<\delta_n but d(f(p_n),q)\ge\epsilon .
    So (p_n)\to p but f(p_n)\not\to q.
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    Re: Theorem 4.2 in Rudin POMA

    I think you mean q instead of F(p). So essentialy that is chosen since 1/n-> 0 (allowing x_n to be arbitarily close to 0) and x_n consists of all 'rogue' values of x is so far as they make F(x) more than epsilon away from q. Is this correct?
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    Re: Theorem 4.2 in Rudin POMA

    Quote Originally Posted by boromir View Post
    I think you mean q instead of F(p). So essentialy that is chosen since 1/n-> 0 (allowing x_n to be arbitarily close to 0) and x_n consists of all 'rogue' values of x is so far as they make F(x) more than epsilon away from q. Is this correct?
    Look at reply #3.
    The sequence (p_n) is close to p not 0.
    But the sequence f(p_n) is not close to q.
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