If $\{q_{n}\}$ be an ennumeration of rational numbers then how can I show that the function defined below
$f(x)=\begin{cases}0&\text{if }x\in \mathbb{R}\setminus\mathbb{Q}\\\frac{1}{2^{m}}& x\in\mathbb Q,x=q_{m}, \,m \geq1\\\end{cases}$
is continuous at all irrational numbers and discontinuous at all rationals.
I have already shown that it is discontinuous at rationals by sequential criteria for continuity. I also need to show that it is continuous at all irrationals.
I realize that this is similar to Thomae Function. But there we have a concept of order. But here the $q_{n}$'s are just an arbitrary ennumeration. Then how should one prove this?. Is what I am required to show even true?