Thread: Showing the modified Dirichlet function is discontinuous

1. Showing the modified Dirichlet function is discontinuous

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 $\epsilon_0$?

2. Re: Showing the modified Dirichlet function is discontinuous

You have your definition backwards. Given an $\epsilon$ you must find a $\delta$.

In this case suppose $\epsilon < |x| \in \mathbb{Q}$

As irrationals are dense in $\mathbb{Q}$ any $\delta$ neighborhood about x contains an irrational number c and so $\forall \delta > 0~~\exists c \in (x-\delta, x+\delta) \ni |f(x)-f(c)|=|x-0|=|x|$

and thus

$|f(x)-f(c)| > \epsilon$ and thus $f$ is discontinuous at $x$.

Likewise as rationals are dense in $\mathbb{R}$ any $\delta$ neighborhood about an irrational number x contains a rational number c and the result is the same.