Is every countable set a jump discontinuity set? Every set that is the jump discontinuity set of a real function is at most countable. Is the converse true? That is, for any at most countable set S of real numbers, is there a function f that has jump discontinuities at precisely S?
 A: Yes. Let $a_0,a_1,\dots$ enumerate the set (in an injective fashion), and define 
 $$ f(x)=\sum\{2^{-n}\mid a_n<x\}, $$
that is, we add all numbers $2^{-n}$ where $n$ varies over all natural numbers such that $a_n<x$. (This is a standard example.)
We have that $f(x)$ converges for all $x$, and it is easy to see that it is continuous everywhere except at the $a_i$, where it has jump discontinuities. to see this, consider for any $x$, the two side limits $\lim_{t\to x^{-}}f(t)$ and $\lim_{t\to x^{+}}f(t)$, which exist since $f$ is increasing by definition. If $x=a_n$ for some $n$, these two expressions disagree (by $2^{-n}$), while they coincide otherwise.   
In slightly more detail, note that for any $N$, if $\epsilon>0$ is chosen sufficiently small, then any $a_i$ in $(x-\epsilon,x+\epsilon)\setminus\{x\}$ must have index $i>N$, so if $x$ is not one of the $a_i$, the sums defining $f(t)$ for $t<x$ or $t>x$ disagree only on a number which is at most $\sum_{n>N}2^{-n}$, which can be made as small as desired, while if $x=a_i$ for some $i$, then the sums disagree by $2^{-i}+$ a negligible amount which is at most $\sum_{n>N}2^{-n}$.
