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$$f(x) = \begin{cases}0\,;\,x\in\mathbb Q\\ 1\,;\,x\in\mathbb Q^c\end{cases}$$ Show that $\lim\limits_{x\to0}f(x)$ doesn't exist using epsilon delta definition. How would I separate rational and irrational numbers if the definition deals with open intervals?

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  • $\begingroup$ Please learn how to format your questions properly. $\endgroup$
    – Aweygan
    Aug 21, 2016 at 13:06
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    $\begingroup$ Let $\epsilon=1/2$. $\endgroup$
    – Vim
    Aug 21, 2016 at 13:10
  • $\begingroup$ Vim. Why would it be 1/2? $\endgroup$
    – Math
    Aug 21, 2016 at 13:26
  • $\begingroup$ Because $1/2$ works for what you want to show. You can also use $1/3$ or $2/7$, etc. The point is you just need to find one specific epsilon that doesn't satisfy the epsilon delta definition. $\endgroup$
    – user307169
    Aug 21, 2016 at 13:34
  • $\begingroup$ You ask how to "separate" rational and irrational numbers. What do you mean by that, and why do you want it? (The fact that there are both rational and irrational numbers in every open interval makes the requested proof easy, not difficult.) $\endgroup$
    – David K
    Aug 21, 2016 at 13:52

4 Answers 4

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We are to prove that, for every $l \in \mathbb{R}$ there is some $\varepsilon > 0$ such that for every $\delta > 0$ there is some $0 < |x| < \delta$ such that $|f(x) - l| \geq \varepsilon$. (You sufficiently familiar with such a statement?)

Let $l \in \mathbb{R}$. If $l \geq 1$, then for all $x \in \mathbb{Q}$ we have $|f(x) - l| = l$; taking $\varepsilon := l/2$ suffices. If $0 < l < 1$, let $m := \min \{ l, 1-l \}$. Then for all $x \in \mathbb{R}$ we have $|f(x) - l| \geq m$; taking $\varepsilon := m/2$ suffices. If $l \leq 0$, then for all $x \in \mathbb{Q}^{c}$ we have $|f(x) - l| = 1 + |l|$; taking $\varepsilon := (1+|l|)/2$ suffices.

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Hint.

Open intervals around $0$ are of the form $(-\delta, \delta)$. What can you say about rationals and irrationals in those intervals? What might the limit be if it were to exist?

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$$|f(\text{rational})-f(\text{irrational})|>\frac12$$

EDIT (Answer to the comment)

Take $\epsilon=\frac12$. For every $\delta>0$ there exists an irrational number $t$ in $(-\delta,\delta)$. Then $$|f(t)-f(0)|>\epsilon$$

We have shown the negation of the statement "$f$ is continuous at $0$".

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  • $\begingroup$ Sorry. But what is the point? $\endgroup$
    – Math
    Aug 21, 2016 at 13:24
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the limit does not exist since this holds. $\forall \delta \in \mathbf{Q}: f(\delta)=0 \land f(\delta\frac{(1+\pi)}{5})=1$ which means that for any $\delta$ in the rationals there is a smaller irrational and so the function near 0 just oscillates between 0 and 1

EDIT. choose an arbitrary rational $\delta$ arbitrarily close to 0

let's say $\delta=0.01$

now if we multiply $\delta$ by $\frac{(1+\pi)}{5}$ it becomes $0.01\frac{(1+\pi)}{5}=0.00828...$ which an irrational smaller (in absolute sense) than the $\delta$ we first chose

Evaluating your function in both these numbers gives for $\delta$ the value 0 and for $\delta\frac{(1+\pi)}{5}$ gives the value 1.

this argument works for any rational $\delta$ we pick, so near zero we have demonstrated that the function oscillates.

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  • $\begingroup$ it would be a good exercise for you to translate this in $\epsilon-\delta$ form. $\endgroup$
    – Frank
    Aug 21, 2016 at 13:32
  • $\begingroup$ Sorry, could you give a step by step reason why that mathematical statement would tell you that it oscillates. $\endgroup$
    – Math
    Aug 21, 2016 at 13:34
  • $\begingroup$ ok. i'm gonna edit the answer. $\endgroup$
    – Frank
    Aug 21, 2016 at 13:35
  • $\begingroup$ edited, hope it's more transparent now :) $\endgroup$
    – Frank
    Aug 21, 2016 at 13:43
  • $\begingroup$ Now try to formulate this new understanding in $\epsilon-\delta$ form, you have all the tools now. $\endgroup$
    – Frank
    Aug 21, 2016 at 13:50

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