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Math Help - showing dirichlet's function is discontinuous on (0,1)

  1. #1
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    showing dirichlet's function is discontinuous on (0,1)

    I need help showing that Dirichlet's function is discontinuous on the interval (0,1).
    f(x)=
    {1/q if x=p/q in lowest terms with p,q in N
    0 if x is irrational}
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by dannyboycurtis View Post
    I need help showing that Dirichlet's function is discontinuous on the interval (0,1).
    f(x)=
    {1/q if x=p/q in lowest terms with p,q in N
    0 if x is irrational}
    This function is continuous at irrational x. To prove it's discontinuous at \frac{p}{q}, let \epsilon=\frac{1}{2q}. Since for all \delta>0, (p/q-\delta,p/q+\delta) contains an irrational point, \exists x such that |f(p/q)-f(x)|=|1/q-0|=\frac{1}{q}>\frac{1}{2q}=\epsilon.

    There's a good picture of this function on MathWorld.
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  3. #3
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    Quote Originally Posted by dannyboycurtis View Post
    I need help showing that Dirichlet's function is discontinuous on the interval (0,1).
    f(x)=
    {1/q if x=p/q in lowest terms with p,q in N
    0 if x is irrational}

    In fact f(x) is continuous at every irrational point, and discontinuous at rational ones. To prove this you'll prove that \lim_{x\to x_0}f(x)=0 at every point in (0,1) (Hint: if a sequence of rational points \left\{\frac{a_n}{b_n}\right\}\,,\,a_n\,,\,b_n\in \mathbb{Z}\,,\,\,b_n> 0\,\,\forall\,n\in\mathbb{N} , converges to an irrational number, then \lim_{n\to\infty}b_n=\infty

    Tonio
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