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?

  • 1
    $\begingroup$ 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 $x=(0;2/3)$. $\endgroup$ – user569098 Jul 5 '18 at 17:41
  • $\begingroup$ @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? $\endgroup$ – user1101010 Jul 5 '18 at 17:50
  • $\begingroup$ 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. $\endgroup$ – user569098 Jul 5 '18 at 18:09
  • $\begingroup$ 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. $\endgroup$ – user569098 Jul 5 '18 at 18:24
  • $\begingroup$ Assume E is regular open. $\endgroup$ – William Elliot Jul 5 '18 at 20:51

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.

  • $\begingroup$ 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. $\endgroup$ – user1101010 Jul 6 '18 at 20:22
  • $\begingroup$ @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? $\endgroup$ – William Elliot Jul 7 '18 at 4:11

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