Showing the modified Dirichlet function is discontinuous

[Usagi]

Show, using the $\epsilon-\delta$ definition of continuity, that the modified Dirichlet function, i.e., $f(x) = x$ if $x$ is rational and $f(x) = 0$ if $x$ is irrational, is discontinuous at all points $c \neq 0$

My attempt:

Is the following argument right (using the sequential definition of continuity?)

That is, consider any real $c \neq 0$, then, for $c\neq 0$, we can find some sequence $(x_n) \subset Q$ and $(y_n) \subset I$ such that $(x_n) \rightarrow c$ and $(y_n) \rightarrow c$ but $f(x_n) \rightarrow c$ and $f(y_n) \rightarrow 0$, thus $\lim_{x \rightarrow c} f(x)$ does not exist and hence is discontinuous at all points besides $0$.
Now how do I prove it using the definition? I.e, I need to show that $\exists$ $\epsilon_0 >0 \ \forall \ \delta >0 \ \exists x \in \mathbb{R}$, $|x-c| < \delta$ and $|f(x) - f(c)| \ge \epsilon_0$. How do I find the $\epsilon_0$?

 

[Evgeny.Makarov]

Is the following argument right (using the sequential definition of continuity?)

That is, consider any real $c \neq 0$, then, for $c\neq 0$, we can find some sequence $(x_n) \subset Q$ and $(y_n) \subset I$ such that $(x_n) \rightarrow c$ and $(y_n) \rightarrow c$ but $f(x_n) \rightarrow c$ and $f(y_n) \rightarrow 0$, thus $\lim_{x \rightarrow c} f(x)$ does not exist and hence is discontinuous at all points besides $0$.

I agree.

I.e, I need to show that $\exists$ $\epsilon_0 >0 \ \forall \ \delta >0 \ \exists x \in \mathbb{R}$, $|x-c| < \delta$ and $|f(x) - f(c)| \ge \epsilon_0$. How do I find the $\epsilon_0$?

Hint: In the proof of discontinuity of the regular Dirichlet function $\epsilon$ is chosen as $f(c)/2=1/2$ if $c\in\Bbb Q$ and $f(c')/2=1/2$ if $c\notin\Bbb Q$ and $c'\in\Bbb Q$ is some point close to $c$.