# Intersection of two simply connected sets in $\mathbb C$

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?

• Take an open square $(0,1)^2$ and delete the segments $\{(1/n;t):\text{ for }t\in[1/2,1)\}$, for $n=1,2,3,...$. Then look at $x=(0;2/3)$. – user569098 Jul 5 '18 at 17:41
• @LB_O: Thanks for your example. May I ask whether there is some sufficient condition to guarantee this? Does path-connectedness of the boundary suffice? – user1101010 Jul 5 '18 at 17:50
• The boundary of that example is path-connected. Locally path-connected should do it, since then you can map the closure of $E$ to the closed unit disc by a continuous function that is conformal in the interior. – user569098 Jul 5 '18 at 18:09
• You can complete it to an answer and post it yourself. Find the source to the result I mentioned above. Maybe Ahlfors' book on complex analysis has it. I don't know. Show that the example is actually an example. – user569098 Jul 5 '18 at 18:24
• Assume E is regular open. – William Elliot Jul 5 '18 at 20:51

• Forgive my lack of knowledge, I could not prove the statement assuming $E$ is regular open and has simply connected closure. Could you give more hints on this part? Thanks. – user1101010 Jul 6 '18 at 20:22