# How can I show that this function is continuous at all irrational numbers?

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?

• You can use the same approach as here: math.stackexchange.com/a/3008948/42969 Nov 12, 2021 at 12:31
• @MartinR Can you please tell me how I should relate the infinite sum with what I have?
– user974205
Nov 12, 2021 at 12:34

Just use the definition of the limit. Let $$x\notin\mathbb{Q}$$ and $$\epsilon>0$$. There is some $$m_0\in\mathbb{N}$$ such that $$\frac{1}{2^m}<\epsilon$$ for all $$m\geq m_0$$. Now, take $$\delta=\min\{|x-q_1|, |x-q_2|,...,|x-q_{m_0}|\}>0$$. We'll show that if $$|x-y|<\delta$$ then $$|f(x)-f(y)|<\epsilon$$. Indeed, if $$y\notin\mathbb{Q}$$ then the statement is trivial, as we have $$|f(x)-f(y)|=0$$ in this case. If $$y\in\mathbb{Q}$$ then we have $$y=q_m$$ for some $$m$$. But since $$|x-y|<\delta$$ we must have $$m>m_0$$, and so:
$$|f(x)-f(y)|=\frac{1}{2^m}<\epsilon$$
So $$f$$ is indeed continuous at the point $$x$$.