0
$\begingroup$

$ \mathbb{Q} $ is countable, so we can list its elements. Let

$ \mathbb{Q} = \{r_1 , r_2, ...\}$

Define $f: \mathbb{R} \to \mathbb{R} $ by the following rule:

$ f(x) = \begin{cases} 0 & x \notin \mathbb{Q}\\ \frac{1}{n} & x=r_n \in \mathbb{Q} \end{cases}$

Find the set of discontinuities $D_f$ of $f$.

So far, I have looked at Dirichlet's function proof about how his function was discontinuous, but I don't know where to start on this one. Could someone give me a hint?

$\endgroup$
1
  • $\begingroup$ Is there any more info about the ordered list of rationals? I've seen it with f(r)=1/n for n the denominator of r in reduced terms. $\endgroup$
    – jdods
    May 27, 2016 at 3:06

1 Answer 1

2
$\begingroup$

If $x$ is rational, then $f(x) >0$ and hence there are points $y$ arbitrarily close such that $f(y) = 0$.

Suppose $x$ is irrational and let $\epsilon>0$. Note that $I=\{ n | {1 \over n} \ge \epsilon \}$ is finite. Choose $\delta>0$ such that $B(x,\delta) \cap \{ r_n \}_{n \in I}$ is empty. Then $f(y) < \epsilon$ for all $y \in B(x,\delta)$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .