Let $ (r_n)_{n \in \mathbb{N}} $ be an enumeration of the rationals in $[0,1]$.
Consider the map $$ f(x) := \sum_{k \text{ such that } r_k < x} \frac{1}{2^k} $$ with $f(0) = 0$.
Does this function jump at every rational in $(0,1]$ independently of the choice of enumeration of the rationals?
I suspect it does, however I think there is a counter example: if one chooses an enumeration that starts $(1/2, 1/3, 1/4, \dots)$, then for $x \in (1/2, 1]$ we have already summed over the first 'infinitely many' values of $k$ and as such will be arbitrarily close to $1$. Hence we cannot jump by any $\varepsilon >0$ as we are already less than $\varepsilon$ away from $1$. So for this enumeration there are no jumps after $x = 1/2$. Is this reasoning correct?
EDIT: as is pointed out in the comments, the key error is that the suggested enumeration fails.
Best Answer
If we are closer than any $\epsilon$ to $1,$ that means that we have enumerated all the rationals. Which means that the interval $[x, 1]$ has NO rationals in it. Do you think that is plausible?