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Thread: 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 $\displaystyle \epsilon>0$, there are only finitely many points $\displaystyle a_i$ such that $\displaystyle f(a_i)>\epsilon$.

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


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

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