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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?

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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 $0< |x-a| < \delta \Rightarrow |f(x)-L| < \epsilon$).

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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 $f(x_1)>0$.

Here’s a picture that might help you see things more clearly: http://en.wikipedia.org/wiki/Thomae%27s_function

Your $a_n$’s are irrational, so $f(a_n)=0$. The function is continuous on the irrationals. You can show this using the $\epsilon$-$\delta$ definition of continuity. Once you’re given an irrational $x$ and an $\epsilon>0$, there is an integer $n>1/\epsilon$, and there are only finitely many rational numbers in, say, $(x-1,x+1)$ having denominator smaller than $n$ is lowest terms. Thus there is a closest one to $x$, and you can use this to find your $\delta$.

Using similar reasoning, you can show that the limit at each rational is 0, but, as AD. points out, without knowing this there is a quick way to see that the function is discontinuous at each rational number.

I was isterested in Riemann function and saw it,I would like to write another proof:

Let $x_{ 0 }=\frac { m }{ n } $ is rational and satisfying $f\left( { x }_{ 0 } \right)=\frac { 1 }{ n } $.It is obvious that rational number $\left( \frac { { q }_{ m+1 } }{ { q }_{ n } } \right) $ converges to $\frac { m }{ n } ={ x }_{ 0 }$ in $q\rightarrow \infty $.However ,$\lim _{ q\rightarrow \infty }{ f\left( \frac { { q }_{ m+1 } }{ { q }_{ n } } \right) =\lim _{ q\rightarrow \infty }{ \frac { 1 }{ q_{ n } } =0 } } $ which shows every rational point $\frac { m }{ n }$ is discontinuous.

Now assume that $\alpha $ is arbitrary irrational number and take artibrary rationla sequences such as $\left( { x }_{ q } \right) =\left( \frac { m_{ q } }{ n_{ q } } \right) $ which congerges to $\alpha $.Then $\lim _{ q\rightarrow \infty }{ n_{ q }=\infty } $ and

$$\lim _{ q\rightarrow \infty }{ f\left( x_{ q } \right)=\lim _{ q\rightarrow \infty }{ f\left( \frac { { m }_{ q } }{ { n }_{ q } } \right)=\lim _{ q\rightarrow \infty }{ \frac { 1 }{ { n }_{ q } } =0=f\left( \alpha \right) } } } $$

Hence $f\left( x \right) =0$ for every irrational $x$ so $\lim _{ q\rightarrow \infty }{ f\left( { x }_{ q } \right) =f\left( \alpha \right) =0 } $ which shows function is contious in every irrational point $x$

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