$$ f(x)=\begin{cases} \frac{1}{q},&\text{if $x=\frac{p}{q}$, $x\in\mathbb{Q}$, with $p\in\mathbb{Z}$ and $q\in\mathbb{N}$ and $p$ with $q$ is coprime;}\\ 0,&\text{if $x\in\mathbb{R}\setminus \mathbb{Q}$;} \end{cases} $$
I want to proof continuous at all irrational numbers.
My steps:
- I choose one irrational number $c\in\mathbb{R}\setminus \mathbb{Q}$ and look on his image $f(c) = 0$
- I take an arbitrary $\epsilon > 0$ and $V_{\epsilon}(0) = V_{\epsilon}(f(c))$ -- open neighborhood of $0$
- So for now, I want to create the open neighborhood $U_{\delta}(c)\subset \mathbb{R}$, for which true that $f(U_{\delta}(c))\subset V_{\epsilon}(f(c))$
Consider all such points $x$ around $c$, the image $f(x)$ of that is bigger that $\epsilon < f(x)$. So that points fall outside of $V_{\epsilon}(0)$. That is any $\frac{p}{q}$, $\frac{1}{q} \geq \epsilon$ $\Rightarrow$ $1\geq \epsilon q$ $\Rightarrow$ $\epsilon^{-1}\geq q$ that is $\frac{1}{\epsilon}\geq q$
$q$ bounded above by $\frac{1}{\epsilon}$ and $0$ below, because $q\in\mathbb{N}$
The set of $\{f(\frac{p}{q})\mid f(\frac{p}{q})\geq \epsilon\}$ is a finite set. And that if $0< q \leq \frac{1}{\epsilon}$ in $U_{\delta}(c)$ will be exist only finite number of rational numbers $\{\frac{p}{q} \mid p\in \mathbb{Z}, q\in\mathbb{N}, gcd(p,q) = 1\} = T$
So $T \cap U_{\delta}(c)$ -- finite. So we can find $\delta_T$ for which will be true, that $U_{\delta_T}(c) \cap T = \emptyset \Rightarrow f(U_{\delta_T}(c))\subset V_{\epsilon}(f(c))$.
All of these steps correct?
