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Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be defined as: $f(x)=\begin{cases}\frac{1}{\min\{q\in\mathbb{N}:qx\in\mathbb{Z} \} } &\text{ if $x\in\mathbb{Q}$}\\ 0 &\text{if$x\in\mathbb{R/Q}$} \end{cases} $

The task is now to find every point, in which f is continous. (0 is not an element of $\mathbb{N}$ in this case). If $x\in\mathbb{Z}$, $f(x)=1$ obviously, also if $x\in\mathbb{Q}$, so $x=\frac{a}{b}$ with $a\in\mathbb{Z}$ and $b\in\mathbb{N}$, then $f(x)=\frac{1}{b}$. So this function almost looks like the Dirichlet function, which is discontinuous in every point. Intuitively, I would say that this function is continous in $x=0$, my explanation is kinda "hand-wavey" but when x becomes really small, $q$ has to get really big, so $\frac{1}{q}$ has to become really small, almost 0. So if we have divergent series $a_n$ of rational numbers,the limit of $f(a_n)$ should be 0, right? Did I make a mistake? Could you give me some advice how to prove this statement? Thanks in advance!

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    $\begingroup$ As far as proving $f$ is continuous at 0, I think your ideas are going in the right direction. And proving discontinuity elsewhere should not present you with much more difficulty. You should make your arguments concise (for example, "really small, almost $0$" can be made more specific). $\endgroup$ – ZxJx Dec 7 '19 at 18:08
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    $\begingroup$ Does this answer your question? Why is this function continuous, unlike the Dirichlet function? $\endgroup$ – LinearAlgebruh Dec 7 '19 at 18:09
  • $\begingroup$ If the images under $f$ of rationals around zero become smaller (get closer to zero) when the rationals get closer to zero, you would need $f(0)=0$ for $f$ to be continuous. $\endgroup$ – Aldoggen Dec 7 '19 at 19:15