$U \subset \mathbb{C}$ open, how do I construct $f$ holomorphic, such that there is no continuation of $f$ on any neighbourhood of $z \in \partial U$? Let $U \subseteq \mathbb{C}$ be open. I want to construct a holomorphic function $f: U \to \mathbb{C}$, such that for all $z \in \partial U$ and for all $\varepsilon > 0$, there is no holomorphic continuation $\tilde{f}: U \cup B_\varepsilon(z) \to \mathbb{C}$ of $f$. For simply connected $U$, I could just find a diffeomorphism to the unit circle and there, the geometric series would be an example of such a function, if I am not mistaken. But what am I supposed to do if $U$ isn't simply connected? The Laurent-series comes to mind and also is the current topic of our complex-analysis lecture, however what would I do if those holes of $U$ were not "circle-shaped"?
Thanks for any help in advance.
 A: The idea is to choose poles of first order $\frac{a_n}{z-z_n}$ at countable many $z_n \in \partial U$ such that they are dense in $\partial U$ and then to add these poles to a holomorphic function on $U$.
Let $z_n$ be a sequence of pairwise distinct complex numbers dense in $\partial U$.
Proposition 1: There exist two sequences $a_n, u_n$ of complex numbers such that


*

*for all $n \in \mathbb{N}$: $u_n \in U$ and $|u_n-z_n| < \frac{1}{n}$,

*for all $n \in \mathbb{N}$: $|a_n|\leq \frac{1}{n^2}$ and

*for all $k,n \in \mathbb{N}$ with $k \leq n$: $|f_n(u_k)| > k$,


where we define the $f_n$ through
$$f_n(z) := \sum_{k=1}^n \frac{a_k}{z-z_k}.$$
Proposition 2: These $f_n$ converge locally uniformly to a holomorphic function $f$ on $U$, and $f$ is singular at each point in $\partial U$.
Proof of Proposition 1: We prove by induction. $z_1$ is given and we choose $a_1 = 1$. Choose further $u_1 \in U$ close enough to $z_0$ (since $z_0$ lies on the boundary of $U$ we have elements in $U$ arbitrarily close to $z_0$), such that $|f_1(u_1)| = |\frac{1}{u_1-z_1}| > 1$ and $|u_1-z_1|<1$. This was the case $n=1$.
Assume the proposition holds for the elements of the sequences up to $n-1$. Again we have $z_n$ given and choose $a_n$ such that for all $k \leq n-1$:
$$|f_{n-1}(u_k)| > k \implies |f_n(u_k)| > k$$
For example, 
$$|a_n| < \min_{k=1, \dots, n-1} (|f_{n-1}(u_k)|-k)|u_k-z_n|$$
suffices. We can also ensure $|a_n| \leq \frac{1}{n^2}$. We have left to choose $u_n$: Choose it close enough to $z_n$ such that $|f_n(u_n)| > n$ and $|u_n-z_n| < \frac{1}{n}$ holds. This proves the proposition.
Proof of Proposition 2: Let $u \in U$ and $r > 0$ such that $B_{2r}(u) \subset U$. The sequence of holomorphic functions $f_n$ converges uniformly on $B_r(u)$, because each $v \in B_r(u)$ has distance further than $r$ to all $z_n \in \partial U$ and therefore, for all $n < m$:
$$|f_n(v) - f_m(v)| \leq \sum_{k=n+1}^m \frac{|a_k|}{r} \leq\frac{1}{r} \sum_{k=n+1}^m \frac{1}{k^2}$$
This shows locally uniform convergence of $f_n$ to a holomorphic function $f$.
Further, if $f$ wasn't singular in $z \in \partial U$, then $f$ would be bounded near $z$. But in every neighbourhood of $z$ there are infinitely many $z_n$ and thus also infinitely many $u_n$. Now $|f(u_n)| \geq n$ contradicts boundedness (the condition $|f_n(u_k)| > k$ becomes $|f(u_n)| \geq n$ due to uniform convergence). Q.E.D.
