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Thread: Proving a function is discontinuous

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

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

    I know that I should find a sequence $\displaystyle (x_{n})$ where $\displaystyle \lim_{n\to \infty}x_{n} = x$, where $\displaystyle x_{n}$ is rational for even $\displaystyle n$ and irrational for odd $\displaystyle n$ (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 $\displaystyle \mathbb Q$ with limit in $\displaystyle \mathbb R - \mathbb Q$

    For example


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

    etc

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

    so


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


    But any such sequence works! For example a sequence of rationals with $\displaystyle \pi $ as limit.
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    Senior Member Pinkk's Avatar
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    But that is for specific values of $\displaystyle x$, is there any way to have a generalized case? Because the general idea and hint that the book suggests is that when you have $\displaystyle x_{n}$ be rational for even $\displaystyle n$ and irrational for odd $\displaystyle n$, so that the sequence $\displaystyle f(x_{n})$ does not converge at all. Because to me, it seems that all you've shown is that the function is discontinuous for irrational $\displaystyle x$ (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 $\displaystyle x\in \mathbb{Q}$, then $\displaystyle \lim_{n\to \infty}(x + \frac{1}{\sqrt{2} + n}) = x$, but all the terms of the sequence $\displaystyle x_{n} = x + \frac{1}{\sqrt{2} + n}$ are irrational.
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    Given any $\displaystyle a\in \mathbb{R}$ there is a sequence of rationals $\displaystyle \left(q_n\right)\to a$.
    There is a sequence of irrationals $\displaystyle \left(\gamma_n\right)\to a$.
    Now define a new sequence $\displaystyle x_n = \left\{ {\begin{array}{rl} {q_n ,} & {\text{n is even}} \\ {\gamma _n ,} & {\text{n is odd}} \\ \end{array} } \right.$
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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 $\displaystyle a~\&~a+1 $ there is a rational number $\displaystyle q_1$.
    Between $\displaystyle a~\&~a+0.5 $ there is a rational number $\displaystyle q_2$.
    In general, between $\displaystyle a~\&~a+2^{-n+1} $ there is a rational number $\displaystyle q_n$.
    Thus $\displaystyle \left(q_n\right)\to a$.

    The same for the irrational case.
    Last edited by Plato; Mar 4th 2010 at 03:35 PM. Reason: signs
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  7. #7
    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 $\displaystyle f(x)=1$ if $\displaystyle x\in \mathbb{Q}$ and $\displaystyle f(x)=0$ if $\displaystyle x\notin \mathbb{Q}$. Show that $\displaystyle f$ is discontinuous for all $\displaystyle x\in \mathbb{R}$

    I know that I should find a sequence $\displaystyle (x_{n})$ where $\displaystyle \lim_{n\to \infty}x_{n} = x$, where $\displaystyle x_{n}$ is rational for even $\displaystyle n$ and irrational for odd $\displaystyle n$ (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 $\displaystyle x\in\mathbb{R}$ since both $\displaystyle \mathbb{Q},\mathbb{R}-\mathbb{Q}$ are dense in $\displaystyle \mathbb{R}$ there exists sequences of points $\displaystyle \{q_n\},\{i_n\}$ which lie in both. Now, if we assume that $\displaystyle f$ is continuous at $\displaystyle x$ then $\displaystyle 1=\lim\text{ }f(q_n)=f(x)=\lim\text{ }f(i_n)=0$. Contradiction.



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