Riemann mapping under which uncountably many boundary points correspond to a single point

Question

I am interested in the following question, which is 10.4 from this list:

Give an example of a domain conformally equivalent to the disc where uncountably many points on the unit circle correspond to a single point on the boundary.

I take this to mean something like, "Find a point $p$ on the boundary of the domain and uncountably many points $e^{i\theta_\alpha}$ such that $f(z)$ tends to each $e^{i\theta_\alpha}$ depending on how $z$ approaches $p$."

I was thinking of doing something like the following. Fix a finite number of angles $\theta_i$. Consider the unit disk, and excise line segments of length $1$ with center at the origin and angle $\theta_i$. Then excise a slit from origin to the boundary of the disk to make the region simply connected. It seems like the origin would then correspond to finitely many different points on the boundary of the disk (depending on which part of the middle "star" the sequence of points approaches the origin in the original region). However, cutting out uncountably many line segments no longer leaves an open set, so I don't think this method will work. Is there another way?

Answer 1

The following domain has this property: $$\Omega=\mathbb{C}\setminus\bigcup_{n=1}^\infty\bigcup_{k=1}^{2^n} \left\{r\exp(2\pi ik/2^n): r\ge n\right\} $$ Let me first describe this in an algorithm:

1. Begin with the complex plane $\mathbb{C}$.
2. At $n$th step ($n=1,2,\dots$) remove $2^n$ half-lines ($n$th generation) that are symmetrically placed around the origin, beginning at distance $n$ from the origin, and going off into infinity at equally spaced angles. (Half of them were already out of the domain, which is fine).

Let $f$ be a conformal map from the unit disk $\mathbb{D}$ onto $\Omega$; it is convenient to think that $f(0)=0$. Key points:

1. The preimage of each half-line under $f$ is an open subarc of $\partial \mathbb{D}$. Indeed, as one travels along the halfline, coming from infinity, turning around the tip, and going back to infinity, the corresponding point on the unit circle keeps moving.

2. These arcs are disjoint, obviously.

3. Between any two arcs there is another one, because between any two halflines there is another one.

Therefore, the complement of the union of these arcs is a nonempty, perfect set without isolated points; i.e., a Cantor set. This is $f^{-1}(\infty)$.