What would be a counterexample for a point which is a limit point of isolated singularities? Let $G$ be open in $\mathbb{C}$ and $f:G\rightarrow \mathbb{C}$ be a function.
Define $D$ as the set of points in $G$ at which $f$ is complex-differentiable. That is, $p\in D$ iff $\lim_{z\to p} (f(z)-f(p))/(z-p)$ exists. (This does not mean that $f$ is holomorphic on some neighborhood of $p$, but it literally means that $f$ is complex-differentiable at a single point $p$)
Assume that for each $\alpha\in G\setminus D$ is an isolated singularity of $f$.
Then, is $D$ open?
I have proven that $D$ is open when the isolated singularities are all non-removable. (It can be proven easily and it is direct)
However, Is $D$ open in general? I think this is false, but cannot find a counterexample. What is a counterexample that $D$ is not open?
Here's what I have tried:
Fix $z\in D$.
To show that $D$ is open, it suffices to prove that $z$ is not a limit point of $G\setminus D$. So let's suppose that $z$ is a limit point of $G\setminus D$.
Then, there exists a sequence of isolated singularities $\{a_n\}$ converging to $z$ such that $f$ is NOT complex-differentiable at $a_n$, by the definition of $D$.
Since $f$ is differentiable at $z$, $f$ is continuous at $z$. Hence for some $N\in\mathbb{N}$, $n\geq N$ implies that $a_n$ is a removable singularity of $f$. This situation seems possible so this cannot lead to a contradiction. So I'm conjecturing that $D$ need not be open, but what would be a countetexample?
 A: Since you make no continuity assumptions, an example with non-open $D$ is easily constructed. Let us call $S = G\setminus D$ the set of singularities. What we want is a set $S$ that is discrete in the subspace topology (so each point of $S$ is isolated), but not closed. $S = \bigl\{ \frac{1}{n} : n \in \mathbb{N}\setminus \{0\}\bigr\}$ looks like a nice candidate. So we want a function that is complex differentiable at $0$, but has singularities exactly at the points of $S$. Taking the simplest thing one can think of is always a good plan, so let's say $f\lvert_{\mathbb{C}\setminus S} \equiv 0$. It remains to pick $f$ so that the points of $S$ are singularities - with what we have that means $f(1/n) \neq 0$ - but so that $f$ is complex differentiable at $0$. We can for example take $f(1/n) = 1/n^2$. Any choice $f(1/n) = a_n$ with $n\cdot a_n \to 0$ works.
If we require $f$ to be "as continuous as possible", that is, $f(z_0) = \lim\limits_{\substack{z\to z_0\\z\neq z_0}} f(z)$ if the limit exists in $\mathbb{C}$, then clearly all points of $S$ must be non-removable singularities, and $D$ open.
