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 $ ?
$\endgroup$ 24 Answers
$\begingroup$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: $$1=|1-c+c|\leq ???<1/2+1/2=1$$
$\endgroup$ 0 $\begingroup$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.
$\endgroup$ $\begingroup$$\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.
$\endgroup$ $\begingroup$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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