Consider the following function:
$f(x)=\left\{\begin{array}{ll} 0 & \text { if } x \text { is irrational } \\ \frac{1}{d} & \text { if } x\left(=\frac{n}{d}\right) \text { is rational. } \end{array}\right.$
Where $n,d\in \mathbb{Z}$, $d>0$ and $x\in \mathbb{R}$.
I am trying to prove that this function is continuous at irrational numbers. Is the following correct?
Consider an arbitrary irrational number, $a$. Consider an arbitrary $\epsilon>0$. Consider an arbitrary $x\in \mathbb{R}$. Suppose first that $x$ is rational. There exists some interval, $(N,N+1), N\in\mathbb{N}$, such that $a\in (N, N+1)$. Given the result that there are only finitely many rational numbers in $(N, N+1)$ such that $f(r)\geq\epsilon$, there must exist a $\delta>0$ such that $|x-a|<\delta \implies f(x)<\epsilon$. But, by the definition of $f$, $f(x)<\epsilon \implies |f(x)|<\epsilon$ (since $f$ is strictly positive for all rational numbers) $\implies |f(x)-f(a)|<\epsilon$ (since $f$ is zero for all irrational numbers).
