Is there a monotonic function discontinuous over some dense set? 
Can we construct a monotonic function $f : \mathbb{R} \to \mathbb{R}$ such that there is a dense set in some interval $(a,b)$ for which $f$ is discontinuous at all points in the dense set?  What about a strictly monotonic function?

My intuition tells me that such a function is impossible.
Here is a rough sketch of an attempt at proving that such a function does not exist: we could suppose a function satisfies these conditions.  Take an $\epsilon > 0$ and two points $x,y$ in this dense set such that $x<y$.  Then, $f(x)<f(y)$ because if they are equal, then the function is constant at all points in between, and there is another element of $X$ between $x$ and $y$, which would be a contradiction.  Take $f(y)-f(x)$.  By the Archimedean property of the reals, $f(y)-f(x)<n\epsilon$ for some $n$.
However, after this point, I am stuck.  Could we somehow partition $(x,y)$ into $n$ subintervals and conclude that there must be some point on the dense set that is continuous?

Such a function is possible.
Let $\Bbb Q=\{q_n:n\in\Bbb N\}$ be an enumeration of the rational numbers, and define
$$f:\Bbb R\to\Bbb R:x\mapsto\sum_{q_n\le x}\frac1{2^n}\;.\tag{1}$$
The series $\sum_{n\ge 0}\frac1{2^n}$ is absolutely convergent, so $(1)$ makes sense. If $x<y$, there is some rational $q_n\in(x,y)$, and clearly $f(y)\ge f(x)+\frac1{2^n}$, so $f$ is monotone increasing. However, $f$ is discontinuous at every rational:
$$\lim_{x\to {q_n}^-}f(x)=\sum_{q_k<q_n}\frac1{2^k}<\sum_{q_k\le q_n}\frac1{2^k}=f(q_n)\;.$$
Thus, $f$ is discontinuous on a set that is dense in $\Bbb R$ (and in every open interval of $\Bbb R$).
