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I think I don't understand how it works.. I found some proofs.. okay, let's see:

Well I'd like to show that the function,

$$f(x) = \begin{cases} 0 & x \not\in \mathbb{Q}\\ 1 & x \in \mathbb{Q} \end{cases}$$

is discontinuous.

Now with epsilon-delta-definition:

Let's choose an $\varepsilon < 1$, for example $\varepsilon := 1/2$. And: $\delta > 0$ .

I have to show that $|f(x) - f(x_0)| > \varepsilon$

So if $x_0 \not\in \mathbb{Q}$ let $x$ be rational in $(x_0- \delta, x_0+ \delta)$

if $x_0 \in \mathbb{Q}$ let $x$ be irrational in $(x_0- \delta, x_0 + \delta)$

In the end it is $|f(x) - f(x_0)| = 1 > 1/2$.

Why is $|f(x) - f(x_0)| = 1 $ ?

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    $\begingroup$ By definition of the function. In both cases the difference is between the function at a rational and at a irrational and the distance between these values is $1$. $\endgroup$ – user70962 Jun 11 '13 at 8:27
  • $\begingroup$ @BabakS. Okay, done. $\endgroup$ – user70962 Jun 11 '13 at 8:32
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There always exist two distinct points: one is rational and the other is irrational in the interval $(x_0-\delta, x_0+\delta)$. By the definition of Dirichlet function, we can get the conclusion.

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Let's have a look from the first point. Assume that the function $D(x)$ has limit $c$ at some point say $x_0$. Then if we choose $\epsilon=1/2$ then we have a number $\delta$ such that $$0<|x-x_0|<\delta\to|D(x)-c|<1/2$$ You do know as you point that every deleted neighborhood $0<|x-x_0|<\delta$ contains a rational point say $x_1$ and another point which is irrational say $x_2$. So regarding to what we have achieved, $$|D(x_1)-c|=|1-c|<1/2,~~~|D(x_2)-c|=|0-c|<1/2$$ and so we have reached to point of meeting a nice contradiction: $$1=|1-c+c|\leq ???<1/2+1/2=1$$

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  • $\begingroup$ My, you were busy at work here at MSE today! +1! $\endgroup$ – Namaste Jun 12 '13 at 0:08
  • $\begingroup$ @amWhy: For two or three consecutive days, I was in a member of a funeral ceremony. $\endgroup$ – mrs Nov 15 '13 at 8:56
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$\forall x\in \mathbb{R}, \space \forall \delta \gt 0, \exists x_{1}\in\mathbb{Q},x_{2}\in\mathbb{R}-\mathbb{Q}$, s.t. $x_{1},x_{2} \in (x-\delta, x+\delta)$.

This is because the density of rational number and irrational number in real line. I guess you should learn more about the construction of real number field.

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For the question asked why is $|f(x)-f(x_0)|=1$ Just imagine the coordinate system. Here $x_0$ is irrational and $x$ is rational (or vice versa) . So $f(x_0)=0$ and $f(x)=1$ . So its obvious.

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