The function defined by $$f(x)=\begin{cases}
\frac{1}{q} & x=\frac{p}{q}\in\mathbb{Q}, (p,q)=1 \\
0 & x\in\mathbb{I}
\end{cases}$$ is continuous at every irrational in $]0,+\infty[$, but discontinuous at every rational in $]0,+\infty[$.
Proof that it's continuous at every non-negative irrational:
Let $b>0$ be irrational and $\varepsilon>0$. We have to show that there exists a $\delta>0$ such that $$|x-b|<\delta\implies |f(x)-f(b)|<\varepsilon.$$ By the archimedian property, there exists a natural number $n_0$ such that $\frac{1}{n_0}<\varepsilon$.
The interval $(b-1,b+1)$ contains finitely many rationals with denominator smaller than $n_0$.
Therefore, we can choose a $\delta>0$ sufficiently small such that the interval $(b-\delta,b+\delta)$ does not contain any rational with denominator smaller than $n_0$.
It follows that, for $|x-b|<\delta$ with $x>0$, we have $$|f(x)-f(b)|\leq\frac{1}{n_0}<\varepsilon.$$
I was able to understand the proof, but how can I show that the interval $(b-1,b+1)$ contains finitely many rationals with denominator smaller than $n_0$?
Best Answer
Well.. this is an interval of length $2$. Fractions with denominator $2$ are within a $1/2$ distance from each other, so no more than $4$ of those within the given interval. Fractions with denominator $3$ are $1/3$ distance from each other, so no more than $6$ etc. Fractions with denominator less than a given $n_0$ are a finite number by the same reasoning.