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Consider the function $$f(x)= \begin{cases}0&\text{if }x\in\Bbb{R}\setminus\Bbb{Q}~\text{or}~x=0\\\frac{1}{q}&\text{if }x=\frac{p}{q}, p\in\Bbb{Z},q\in\Bbb{N},(p,q)=1\end{cases}$$

Of course this function is discontinuous at rational numbers. To show it is continuous at irrational numbers I need: If $x_n\to x$, with $x_n=\frac{p_n}{q_n}$, and $x_n\ne x,n\in\Bbb{N}$, then $\lim_{n\to\infty} q_n=\infty$.

How to do that?

theREALyumdub
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QED
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  • Since $q_n \geq 1$, this limes cannot be $0$. You mean $\lim_{n \rightarrow \infty} q_n = \infty$? – p4sch May 03 '18 at 18:56
  • Sorry.. That was a typo. Edited it likewise. Obviously for continuity i need the limit to be $\infty$ – QED May 03 '18 at 18:59
  • You should also use the enviroment cases, as I have edited ... – p4sch May 03 '18 at 19:00

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Let us assume that $(q_n)_{n \in \mathbb{N}}$ has a bounded subsequence - say w.l.o.g. that $(q_n)_{n \in \mathbb{N}}$ is already bounded. Then we have $$|q_n x - p_n | \longrightarrow 0.$$ So $(p_n)_{n \in \mathbb{N}}$ is also bounded. Since both are integers, there exists a subsequence $(n_k)_{k \in \mathbb{N}}$ such that $p_{n_k} = p$ and $q_{n_k} = q$. (We have only finite number of possible values, but infinity many elements.) Therefore, we find that $$x= \frac{p}{q},$$ i.e. $x$ is rational.

p4sch
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