Suppose $f(x) = \begin{cases} 0 \ \ \text{if} \ x \in \mathbb{R}- \mathbb{Q} \newline \frac{1}{q} \ \ \text{if} \ x \in \mathbb{Q} \ \text{and} \ x = \frac{p}{q} \ \text{in lowest terms} \end{cases}$
(i) Is $f$ continuous on the irrationals?
(ii) Is $f$ continuous on the rationals?
For (i) you could use the sequence definition of continuity? Maybe try $a_n = \frac{\sqrt{2}}{n}$ and show that $a_n \to 0$ but $f(a_n) \not \to 0$? So its discontinuous on the irrationals?
For (ii) I don't see why there are $1+2+ \cdots + (q-1)$ rational numbers? I know that we need to use this fact to choose an appropriate $\delta$ (e.g $0< |x-a| < \delta \Rightarrow |f(x)-L| < \epsilon$).
Best Answer
It's an well known function, continuous at all irrationals and discontinuous at all rationals. I will give sketch of the proof below.
First let's see why it's continuous at irrationals. To be continuous at an irrational point $x_0$, we need to show that $\forall\delta>0\exists\epsilon>0:|x-x_0|<\epsilon\Rightarrow |f(x)-f(x_0)|<\delta$.
Notice that within any range of length 1, in particular within $[x_0-1/2,x_0+1/2)$, $f(x)$ takes the value 1/2 exactly once, 1/3 twice, 1/4 thrice, and so on, taking 1/k exactly k+1 times. So if we remove these points upto say $\lceil1/\delta\rceil$, we will be removing only finitely many points, and we will be left with points on which $f(x)<1/\lceil1/\delta\rceil\le\delta$. Notice that $x_0$ being irrational was not one of the points that was removed. So find the closest point $a$ to $x_0$ that was removed, and call the distance $|x_0-a|$ as $\epsilon$.
In the same way, one can argue that the function is discontinuous for rationals. If $x_1$ is a rational point, notice that if we remove all the points of the form $p/q$ upto $0\lt q\le N$, any interval around $x_1$ will contain all points with $f(x)\lt 1/N$. So we can form intervals around $x_1$ with arbitrarily small $f$. Hence $f$ is discontinuous at $x_1$ since $f(x_1)>0$.