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Thread: 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 $\displaystyle x$. To prove it's discontinuous at $\displaystyle \frac{p}{q}$, let $\displaystyle \epsilon=\frac{1}{2q}$. Since for all $\displaystyle \delta>0$, $\displaystyle (p/q-\delta,p/q+\delta)$ contains an irrational point, $\displaystyle \exists x$ such that $\displaystyle |f(p/q)-f(x)|=|1/q-0|=\frac{1}{q}>\frac{1}{2q}=\epsilon$.

    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 $\displaystyle \lim_{x\to x_0}f(x)=0$ at every point in $\displaystyle (0,1)$ (Hint: if a sequence of rational points $\displaystyle \left\{\frac{a_n}{b_n}\right\}\,,\,a_n\,,\,b_n\in \mathbb{Z}\,,\,\,b_n> 0\,\,\forall\,n\in\mathbb{N}$ , converges to an irrational number, then $\displaystyle \lim_{n\to\infty}b_n=\infty$

    Tonio
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