# Given $x$ in a path-connected open set $S$ on the plane, are there non-crossing paths from $x$ to every point in $\partial S$?

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..?)

• 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. – Douglas Zare Jul 1 '15 at 16:44
• Understood. I believe in my case the boundary is always a Jordan curve so Carathéodory's theorem may apply. Many thanks. – user2059893 Jul 2 '15 at 19:57

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 $(0,1/2)$.