Suppose $E \subset \mathbb C$ is an open, simply connected set and the boundary $\partial E$ is not empty. Let $x \in \partial E$. Is it true: we can always find some open disk $D(x, r) = \{y \in \mathbb C: |y-x| < r\}$ such that the intersection $E \cap D(x, r)$ is simply connected? As discussed here and here, essentially we want to know whether there exists some $r$ such that $E \cap D(x, r)$ is path-connected?
As the example showed in the comment, this is not true. What sufficient condition would guarantee such open disk exists? For example, is path-connectedness of the boundary $\partial E$ sufficient?
Best Answer
The space between a circle of radius 2 centered at 0 and a
circle of radius 1 centered at 1 is simply connected, regular
open. The point of tangency is a counter example.
Is the property you'd like, E is regular open with a
simply connected closure? Here we are considering every
disk D. If you want just one disk, then if E is bounded,
any disk containing E will suffice.