# Proof for Dirichlet Function and discontinuous

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 $|f(x) – f(x_0)| = 1$ ?

#### Solutions Collecting From Web of "Proof for Dirichlet Function and discontinuous"

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.

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: $$1=|1-c+c|\leq ???<1/2+1/2=1$$

$\forall x\in \mathbb{R}, \space \forall \delta \gt 0, \exists x_{1}\in\mathbb{Q},x_{2}\in\mathbb{R}-\mathbb{Q}$, 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.