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Math Help - Function prove

  1. #1
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    Function prove

    Show that the function f defined by f(x)=0 if x is irrational and f(x)=1 if x is rational does not have a limit at any point.

    Any suggestions?
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  2. #2
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by paperclip View Post
    Show that the function f defined by f(x)=0 if x is irrational and f(x)=1 if x is rational does not have a limit at any point.

    Any suggestions?
    Let x_{n}=\bigg(1+\frac{1}{n}\bigg)^{n} for all n\in\mathbb{N}, then we know that x_{n}\in\mathbb{Q} for all n\in\mathbb{N} and x_{n}\to\mathrm{e}\in\mathbb{Q}^{c} as n\to\infty.
    Therefore, we have f(x_{n})=1 for all n\in\mathbb{N} but \lim\nolimits_{n\to\infty}f(x_{n})=f(\mathrm{e})=0  \neq1.
    Thus f is not continuous at \mathrm{e}.

    However, you can give a good proof for the general case since for any \xi\in\mathbb{R}, you may find \{\sigma_{n}\}_{n\in\mathbb{N}}\subset\mathbb{Q} and \{\zeta_{n}\}_{n\in\mathbb{N}}\subset\mathbb{Q}^{c  } such that \sigma_{n},\zeta_{n}\to\xi as n\to\infty.

    I hope this can help you...
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  3. #3
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    Quote Originally Posted by paperclip View Post
    Show that the function f defined by f(x)=0 if x is irrational and f(x)=1 if x is rational does not have a limit at any point.

    Any suggestions?
    You can solve this using the facts that:

    (a) for any s,t \in \mathbb{R}, \exists r \in \mathbb{Q} \ : \ s < r < t

    (b) for any r \in \mathbb{Q}^c, there exists a sequence \left\{a_n\right\}_{n=1}^{\infty} such that \forall n \in \mathbb{N}, a_n \in \mathbb{Q} and  a_n \rightarrow r
    That is to say, for any irrational number there is a sequence of rationals that converges to it.
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