real analysis - Proving a function is continuous on all irrational numbers

Let $\langle r_n\rangle$ be an enumeration of the set $\mathbb Q$ of rational numbers such that $r_n \neq r_m\,$ if $\,n\neq m.$

$$\text{Define}\; f: \mathbb R \to \mathbb R\;\text{by}\;\displaystyle f(x) = \sum_{r_n \leq x} 1/2^n,\;x\in \mathbb R.$$
Prove that $f$ is continuous at each point of $\mathbb Q^c$ and discontinuous at each point of $\mathbb Q$.

I find this question very challenging and have no idea even how to start off with the proof. Please suggest a proof or any hint.

Answer

This is a part of my answer here, but it should completely answer your questions too.

I use the notation

$$\lim_{y\to x^{+}}f(y)=f(x^+)$$ $$\lim_{y\to x^{-}}f(y)=f(x^-)$$

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There is a very nice way of constructing, given a sequence $\{x_n\}$ of real numbers, a function which is continuous everywhere except the elements of $\{x_n\}$ [That is, discontinuous on a countable set $A\in\Bbb R$]. Let $\{c_n\}$ by any nonnegative summable sequence [that is $\sum\limits_{n\geq 0} c_n$ exists finitely], and let $$s(x)=\sum_{x_n

What we do is sum through the indices that satisfy the said inequality. Because of absolute convergence, order is irrelevant. The function is monotone increasing because the terms are nonnegative, and $s$ is discontinuous at each $x_n$ because

$$s(x_n^+)-s(x_n^-)=c_n$$

However, it is continuous at any other $x$: see xzyzyz's proof with the particular case $c_n=n^{-2}$. In fact, this function is lower continous, in the sense $\lim\limits_{y\to x^{-}}f(y)=f(x^-)=f(x)$ for any value of $x$. If we had used $x_n\leq x$, it would be upper continuous, but still discontinuous at the $x_n$.

To see the function has the said jumps, note that for $h>0$, we have $$\begin{align} s(x_n^+)-s(x_n^-)&=\\ \lim_{h\to 0^+} s(x_k+h)-s(x_k-h)&=\lim_{h\to 0^+}\sum_{x_n

and we can take $\delta$ so small that whenever $0