Motivation
I was thinking about how computers should deal with partial functions. An example of such situation: I'm doing arbitrary real number computation and a division by zero must not fall into an infinite loop. This can be accomplished by letting lazy evaluation extend the codomain by adding erroneous values. Thinking the domain and the codomain as topological spaces (dubbed by admissibility theory), the most trivial example of such extension is to add only one point whose the only neighborhood is the entire codomain-and-the-point.
Yet this example is too trivial. One would like to see what actually caused the error. Like "Is it a division by zero? Is it a jump discontinuity? Was this error avoidable?" And so on.
To make this an actual math problem, I had to make classification of partial functions and its singularities. Consequently, I had to find a partial function with a pathological nature.
Terminology
First, let me make a clarification of terminologies regarding partial functions.
Given a partial function $f : X \rightharpoonup Y$, $X$ is the domain, and $Y$ is the codomain.
The set of elements of $X$ whose function value exists is the support. Restricting $f$ to its support, we get a function; the underlying function of $f$.
An element of $X$ outside of $f$'s support is a singularity.
If $X$ and $Y$ are topological spaces, $f$ is said to be continuous iff its underlying function is continuous.
A singularity is removable iff it can be given a function value where the new underlying function is continuous.
Objective
To find a pathological example, I wanted to find a partial function $f$, where the followings are satisfied:
$f$ is continuous,
$f$'s support is dense in $f$'s domain,
$f$'s support is neither closed nor open in $f$'s domain,
None of $f$'s singularities is removable.
Attempt
My sketch of such $f : (0,1) \rightharpoonup \mathbb{R}$ goes like this. First, I define staircase functions $F_n : (0,1) \rightharpoonup \mathbb{R}$:
$$ F_n(x) = 2^{-n} × \lfloor 2^n × x \rfloor $$
whose discontinuities are replaced by singularities, making $F_n$ continuous.
Now give
$$ f(x) = \sum_{n=1}^\infty \left((-2)^{-n} × F_n(x)\right) $$
where $f$'s support is the intersection of the supports of $F_n$, resulting in $(0,1) \setminus \mathbb{Z}[½]$ (That is, numbers within $(0,1)$ whose binary expansion cannot terminate).
Yet I doubt whether this $f$ satisfies the desired conditions because:
Though uniformly converging sequence of continuous functions converges to a continuous function, I've only seen this only for functions whose domain is a Euclidean space.
And somehow, my intuition tells every singularity of $f$ is removable.
Am I on a right track?
How about this: enumerate the rationals as $r_1,r_2,\dots$, and let $f(x)=\sum_{n:r_n<x}2^{-n}$ for irrational $x$. I think you'll find $f$ has its support on the irrationals, which are dense in the reals, it is continuous on the support, and its singularities on the reals are all jump discontinuities.