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Math Help - Proving a function is discontinuous

  1. #1
    Senior Member Pinkk's Avatar
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    Proving a function is discontinuous

    Let f(x)=1 if x\in \mathbb{Q} and f(x)=0 if x\notin \mathbb{Q}. Show that f is discontinuous for all x\in \mathbb{R}

    I know that I should find a sequence (x_{n}) where \lim_{n\to \infty}x_{n} = x, where x_{n} is rational for even n and irrational for odd n (or vice versa), but I do not know how to find such a sequence.
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  2. #2
    Member mabruka's Avatar
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    You just need to find one sequence in \mathbb Q with limit in \mathbb R - \mathbb Q

    For example


    x_1=1
    x_2=1+\frac{1}{2}
    x_3=1+\frac{1}{2+\frac{1}{2}}

    etc

    This sequence is obviously made of rationals but its limit is \sqrt{2}

    so


    \lim f(x_n)=\lim 1 = 1  \not = 0 = f( \lim x_n) = f(\sqrt{2})


    But any such sequence works! For example a sequence of rationals with \pi as limit.
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  3. #3
    Senior Member Pinkk's Avatar
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    But that is for specific values of x, is there any way to have a generalized case? Because the general idea and hint that the book suggests is that when you have x_{n} be rational for even n and irrational for odd n, so that the sequence f(x_{n}) does not converge at all. Because to me, it seems that all you've shown is that the function is discontinuous for irrational x (because I believe there is a lemma that states that any real number is the limit of a sequence of rational numbers).

    Edit: Nevermind, if we have x\in \mathbb{Q}, then \lim_{n\to \infty}(x + \frac{1}{\sqrt{2} + n}) = x, but all the terms of the sequence x_{n} = x + \frac{1}{\sqrt{2} + n} are irrational.
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    Given any a\in \mathbb{R} there is a sequence of rationals \left(q_n\right)\to a.
    There is a sequence of irrationals \left(\gamma_n\right)\to a.
    Now define a new sequence x_n  = \left\{ {\begin{array}{rl}   {q_n ,} & {\text{n is even}}  \\   {\gamma _n ,} & {\text{n is odd}}  \\ \end{array} } \right.
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  5. #5
    Senior Member Pinkk's Avatar
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    Thanks. I know it's not the topic, but could you perhaps prove that? We never proved in my class that any real number is the limit of a sequence of rationals or irrationals.

    Edit: D'oh, I guess the sequence I suggested in my previous post shows the irrational part of the proof, but how is the rational part proven?
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  6. #6
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    Quote Originally Posted by Pinkk View Post
    Thanks. I know it's not the topic, but could you perhaps prove that? We never proved in my class that any real number is the limit of a sequence of rationals or irrationals.
    Theorem: Between any two numbers there is a rational number and an irrational number.
    So between a~\&~a+1 there is a rational number q_1.
    Between a~\&~a+0.5 there is a rational number q_2.
    In general, between a~\&~a+2^{-n+1} there is a rational number q_n.
    Thus \left(q_n\right)\to a.

    The same for the irrational case.
    Last edited by Plato; March 4th 2010 at 03:35 PM. Reason: signs
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    Senior Member Pinkk's Avatar
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    Ah okay, gotcha, that makes sense. Thanks again.
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Pinkk View Post
    Let f(x)=1 if x\in \mathbb{Q} and f(x)=0 if x\notin \mathbb{Q}. Show that f is discontinuous for all x\in \mathbb{R}

    I know that I should find a sequence (x_{n}) where \lim_{n\to \infty}x_{n} = x, where x_{n} is rational for even n and irrational for odd n (or vice versa), but I do not know how to find such a sequence.
    I'm not sure this is what Plato did...it looks different...

    Let x\in\mathbb{R} since both \mathbb{Q},\mathbb{R}-\mathbb{Q} are dense in \mathbb{R} there exists sequences of points \{q_n\},\{i_n\} which lie in both. Now, if we assume that f is continuous at x then 1=\lim\text{ }f(q_n)=f(x)=\lim\text{ }f(i_n)=0. Contradiction.



    Alternatively, assume that f is continuous at x and WLOG assume that f(x)=1. Clearly then 0\notin B_{\frac{1}{2}}(f(x)) but by continuity we should be able to find some B_{\delta}(x) such that f\left(B_{\delta}(x)\right)\subseteq B_{\varepsilon}(f(x)). See a problem?
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