This is a follow-up to my previous question. I now make my question more precise to avoid the counter-example given there.
Let $U \subset \mathbb R^2$ be an open, regular (meaning $U$ is the interior of its closure), bounded and simply connected set. In that situation, $U$ has a connected boundary but it is not necessarily given by a simple closed curve. Let $\gamma : [0,1] \rightarrow \partial U$ be a simple curve such that $\gamma(0) \neq \gamma(1)$ and assume that $\gamma((0,1)) \subset \partial U$ is open in the subspace topology.
Is $\partial U \setminus \gamma((0,1))$ connected ?
One can note that $K = \partial U \setminus \gamma((0,1))$ is compact and $V = \partial U \setminus \gamma([0,1]) \subset K$ is open. Let $C$ be a connected component of $K$. According to the lemma in that post, if $C \subset V$, then $C$ is a connected component of $X = \partial U$, i.e. $C = \partial U$ which is impossible.
Hence $C \nsubseteq V$ and either $\gamma(0) \in C$, $\gamma(1) \in C$ or $\{\gamma(0),\gamma(1)\} \subset C$. In the latter case, $C = K$ and $K$ is connected. Otherwise, $K$ is made of two compact, disjoint, connected components $C_0$ and $C_1$, $\gamma(i) \in C_i$, $K = C_0 \cup C_1$.
In particular, $\partial U = C_0 \cup C_1 \cup \gamma((0,1))$. But is such a boundary possible for a regular simply connected set ?
I was thinking of using a separation theorem (such as an hyperplane or Zoretti's theorem, cited in my previous post). Another idea was to use the Riemann Mapping Theorem, i.e. use the biholomorphic map $f : U \rightarrow \{|z| < 1\}$ and try to understand the boundary behaviour (a useful reference is Serge Lang, Complex Analysis, Ch. X.4 : The Riemann Mapping Theorem).
In any case, I think the regularity of $U$ (which is not needed to prove that $\partial U$ is connected) should play a crucial role. Otherwise, it's easy to find a counterexample : take $U = \{z \in \mathbb C : |z| < 1|\} \setminus [-1,0]$ and remove any simple curve $\gamma$ for which $\gamma([0,1]) \subset (-1,0)$.
Thanks in advance.
