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Math Help - continuity

  1. #1
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    continuity

    continuity-dirichlet.jpg
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  2. #2
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    Re: continuity

    That argument would work fine to *suggest* that there is a discontinuity. However, it would not *prove* that there is a discontinuity. There are many continuous functions such that f(1.41)=1.41, f(1.414...)=0 and f(1.42)=1.42.
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  3. #3
    Super Member redsoxfan325's Avatar
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    Re: continuity

    The standard argument uses the observation that for any \epsilon>0, there are only finitely many points a_i such that f(a_i)>\epsilon.

    Spoiler:
    So given an irrational point x, let \delta=\min\{|x-a_1|,|x-a_2|,...,|x-a_n|\}. Then if |x-y|<\delta, |f(y)-f(x)|=f(y)<\epsilon, so f is continuous at the irrational point x.


    It's straightforward to show f is discontinuous at any rational point just using the fact that any interval contains an irrational number.

    Spoiler:
    If x\in\mathbb{Q}, then f(x)=c>0, so letting \epsilon=c/2 shows that since any small interval around x contains an irrational point y, |f(x)-f(y)|=c > c/2.
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