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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 $\epsilon_0$?

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  • $\begingroup$ You want an $\epsilon_0$ such that the values on the rationals and on the irrationals are "far from each other". For the irrationals, you know the value is $0$, and for the rationals in $(x-\epsilon_0,x+\epsilon_0)$, you know the value is in that interval. So you want that interval to stay away from $0$. $\endgroup$ – Daniel Fischer Jun 22 '14 at 15:57
  • $\begingroup$ You need to properly negate your definition of continuity: you have flipped the quantifiers correctly but the negation of $p \Rightarrow q$ is $p \land \neg q$, not $p \Rightarrow \neg q$. $\endgroup$ – Ian Jun 22 '14 at 16:26
  • $\begingroup$ @Ian Ah yes, fixed. $\endgroup$ – user40333 Jun 22 '14 at 16:38

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