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Math Help - Showing a function divereges at C

  1. #1
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    Showing a function divereges at C

    f: R->R is defined by f(x)=x if x is rational, and f(x)=0 is irrational.


    I need to show f does not have a limit at c if c is not equal to 0. I've shown this true if c is rational, but I'm having a problem with if it's irrational, any help?
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  2. #2
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    Suppose that \varepsilon  > 0\,\& \,c \ne 0 then due to density
    \left( {\exists q \in \mathbb{Q}} \right)\left( {\exists i \in \Re \backslash \mathbb{Q}} \right)\left[ {\left\{ {q,i} \right\} \subseteq \left( {c - \varepsilon ,c + \varepsilon } \right)} \right].
    Thus in any epsilon neighborhood of c there is a rational and an irrational number.
    Thus on that neighborhood of a nonzero c the function is both c & 0.
    Last edited by Plato; October 2nd 2008 at 03:59 PM.
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