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A have a (non-simply) bounded path-connected open set $S\subset\mathbb{R}^2$.

Given $x\in S$, there are paths in $S$ from $x$ to any point in the boundary $\partial S$. However, can all these paths be constructed so that they don't cross?.

Note: If $S$ is simply connected, then by the Riemann mapping theorem, we can map $S$ to the unit disk, and connect $x$ to points in the boundary by (shifted) wheel spokes. In the non-simply connected case, the paths have to somehow "avoid" the holes.. Perhaps there's some differential equation with whose solution produces such paths? (maybe something like a laplace equation solution with boundary value f=1 at x and at the holes, and f=0 at the boundary, and paths following the gradient of f..?)

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    $\begingroup$ Your argument using the Riemann mapping theorem is flawed. The Riemann map is not guaranteed to extend continuously to the boundary. See Carathéodory's theorem. $\endgroup$ – Douglas Zare Jul 1 '15 at 16:44
  • $\begingroup$ Understood. I believe in my case the boundary is always a Jordan curve so Carathéodory's theorem may apply. Many thanks. $\endgroup$ – user2059893 Jul 2 '15 at 19:57
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Consider a double-sided topologist's comb, with teeth $\{1/n\} \times I$ for $n \in \mathbb{Z}, n \ne 0$ together with $[-1,1] \times {0}$ and ${0} \times I$. This is closed. The complement is path-connected. There is no path in the complement to $(0,1/2)$.

While this set is not simply connected, that's not important. The intersection with the upper half plane is simply connected and has the same property. Your argument using the Riemann mapping theorem assumed that there is a continuous extension of the Riemann map to the boundary, but that might not be possible when the complement is not locally connected.

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  • $\begingroup$ I have a pretty well behaved domain, so I can assume it is open, regular, bounded, path-connected, and simply connected, I will post a question asking if its boundary is a Jordan curve (saw a previous question on this for possibly unbounded sets). Thanks again. $\endgroup$ – user2059893 Jul 2 '15 at 20:18
  • $\begingroup$ @user2059893 there were two answers to that question already, showing the boundary need not be a Jordan curve. I just added a third (more specific or perhaps more elementary) answer, I believe that the "inside" of the Warsow circle would be an example. I realize it is pretty much the same as what is described in the present answer by Douglas Zare which I just read. $\endgroup$ – Mirko Jul 26 '15 at 1:52
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(let's assume the boundary is rectifiable). By a theorem of Peter Jones (see TREE-LIKE DECOMPOSITIONS OF SIMPLY CONNECTED DOMAINS, by Chris Bishop, which generalizes it), this can be decomposed into Lipschitz disks. This reduces the problem to the simply connected case (I believe that in the rectifiable case, there is no problem with boundary extension...)

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