Let $\{q_k\}_1^\infty$ be an enumeration of the rationals, and $$f(x)=\sum_{k=1}^\infty\frac{1}{2^k}\chi_{(q_k,\infty]}(x).$$ I want to show this function is discontinuous at all rational points. I've seen discussion about this on mathexchange but then I really didnt fully understand any of the arguments that were presented.
I tried the following, let $r$ be rational number then there exists $n\in \mathbb{N}$ such that $r=q_n$. To show that $f$ is not continuous at $r$, I want to argue the left and right limit is not the same. First consider $\delta>0$ and $x-r<\delta$ then there exists $q_l\in \mathbb{Q}$ such that $r<q_l<x$. Then $f(x)-f(r)>f(x)-f(q_l)=\displaystyle \sum_{k:q_k<q_n}\frac{1}{2^k}-\sum_{k:q_k<q_l}\frac{1}{2^k}=\sum_{k:q_n< q_k<q_l}\frac{1}{2^k}=t>0$ where $t<\infty$ since the sum has to be countable.
On the other hand if we let $x<r$ then every $(q_k,\infty]$ contains $x$ also contains $r$ hence their difference approaches $0$ as $x\to r$.
Does this make sense at all? Thanks!