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Math Help - strictly increasing function continous only at irrationals

  1. #1
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    strictly increasing function continous only at irrationals

    hi!

    I'm looking for a strictly increasing function on I = [0, 1] that is continuous only at the irrational numbers in I.

    After some thought, I came up with the following arrangement.

    Enumerate the rational numbers in I as such: q_1, q_2, q_3, ... , q_n (with n a natural number) such that q_k < q_{k+1}

    Let f(x) = 2^k     if x = q_k (for 1 <= k <= n)
    and f(x) = (2^k + x) if x is irrational and q_k < x < q_{k+1}

    Is the function continuous at x irrational? I think so.
    For any Epsilon > 0, choose Delta = min (x - q_k, q_{k+1} - x) * 0.5 so that for any x' satisfying |x' - x| < Delta we have |f(x')-f(x)| < Epsilon.

    Is the function discontinuous at x rational? I think that can be shown using similar reasoning.

    Am I close? If not, can I modify my function to meet the requirement?
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  2. #2
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    Re: strictly increasing function continous only at irrationals

    I'm very suspicious of your ability to enumerate the rationals in increasing order but I'm no expert.
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  3. #3
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    Re: strictly increasing function continous only at irrationals

    Thanks for your reply romsek. I'm not an expert either but I think you're right. I have to rethink this.
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  4. #4
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    Re: strictly increasing function continous only at irrationals

    Here is an such a function due to Walter Rudin.
    Suppose that (c_n) is a sequence of points in [0,1] such that \sum\limits_{n = 1}^\infty  {{c_n}} converges. Note that convergence is absolute so the terms can be rearranged.

    Now define \forall (x \in [0,1])\left[ {{E_x} = \{c_n}|q_n < x\} \right] i.e the rational terms so that q_n<x.

    Then define f(x) = \left\{ {\begin{array}{rl}{\sum\limits_{t \in {E_x}} t ,}&{{E_x} \ne \emptyset }\\{0,}&{{E_x} = \emptyset }\end{array}} \right.

    Now that function is not quite what you asked for, but is should give you some ideas.
    Last edited by Plato; January 20th 2014 at 06:08 AM.
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  5. #5
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    Re: strictly increasing function continous only at irrationals

    Hi again,

    In my class, we were shown the following example of a strictly increasing function on (0, 1) that is continuous at the irrationals and discontinuous at the rationals:

    Let g_{1}, g_{2}, g_{3}, ..., g_{m} be a listing of all rationals in (0, 1) and
    f(x) = \Sigma_{ \big\{n | g_{n} < x\big\} }  \frac{1}{2^{n}}

    I was wondering whether the above could be easily extended to the case of I = [0, 1].
    I would need to include the rational end points g_{1} = 0 and 1 in the list.

    I think the function would need to be modified because otherwise I would get f(0) = 0 and \lim_{x \rightarrow 0^{+}}  f(x) = 0
    and similarly f(1) = 1 and \lim_{x \rightarrow 1^{-}}  f(x) = 1

    Will the following function work for the [0, 1] case?

    f(x) =\begin{cases}-2 & x = 0\\ \Sigma_{ \big\{n | g_{n} < x\big\} }  \frac{1}{2^{n}} & x  \in  (0, 1) \\ 2 & x = 1\end{cases}

    Thanks again.
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