Accessible points of a simply connected domain

Question

We know that if $U$ is an open subset of $\mathbb{\widehat C}$ (extended complex plane), a point $v\in\partial U$ is called accessible from $U$ if there exists a curve $\gamma:[0,1)\to U$ such that $\lim_{t\to 1}\gamma(t)=v$.

Question: If $U$ is a simply connected domain, then are all points on $\partial U$ necessarily accessible from $U$ ?

Intuitively it seems to be so, as we shall always get such curves which land at boundary points of $U$. Is my argument correct ? If not, can anyone please suggest me what am I missing ? Thank you.

Answer 1

Nice references are

• Section 17 of Milnor's book "Dynamics in One Complex Variable" (old version is available in arXiv https://arxiv.org/abs/math/9201272 and it is Section 15 for this version), and
• Section 6.1 of McMullen's book "Complex dynamics and renormalizations" (PDF is available in the author's webpage: https://people.math.harvard.edu/~ctm/papers/pup.html).

For example, it is known that

• For almost all angle, the radial ray lands at some point.
• If a boundary point is accessible, then some radial ray lands at it (and vice versa of course).
• The boundary is locally connected if and only if the Riemann map can be extended continuously on the boundary. (Carathéodory's theorem). In particular, every radial ray lands if these equivalent conditions hold.

Also, interesting inaccessible examples are given by Sørensen, and he proved that there exist quadratic polynomials such that the Julia set is connected but not locally connected and the critical point is not accessible. Hence any point in the backward orbit of the critical point is inaccessible and they form a dense subset in the Julia set. (Ergod. Th. & Dyn. Sys. 18 (1998), 739-758 and The J. Geom. Anal. 10 (2000), 169–206)