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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?

user1101010
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    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 $x=(0;2/3)$. –  Jul 05 '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 05 '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. –  Jul 05 '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. –  Jul 05 '18 at 18:24
  • Assume E is regular open. – William Elliot Jul 05 '18 at 20:51
  • @WilliamElliot: Thanks. May I ask in the case ($E$ is regular open): would $E \cap D$ be regular open? I have seen counterexample to show union of regular open sets is not necessarily open, but the counterexample is about two open sets with empty intersection. – user1101010 Jul 05 '18 at 21:14
  • Yes, intersection of two regular open sets is regular open. For regular open U, boundary U = boundary closure U. – William Elliot Jul 06 '18 at 09:22

1 Answers1

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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.

William Elliot
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  • 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 06 '18 at 20:22
  • @iris2017. Did you ignore the question mark? Regular open, which got rid of some counter examples, did not get rid of all of them. So I added another premise, that got rid of the new counfer example I presented. Does it get rid of all of them? I do not know. Will the math muse inspire me again tonight? – William Elliot Jul 07 '18 at 04:11