# Two paths to the boundary with no holes in between

Let $$X\subset \mathbb{R}^2$$ be open connected (and let's say bounded), let $$x\in X$$ and $$y\in\partial X$$ be such that there is a Jordan curve $$\gamma:[0,1]\to X\cup\{y\}$$ such that $$\gamma(0)=x$$ and $$\gamma(1)=y$$.

Does there always exist a Jordan curve $$\delta:[0,1]\to X\backslash\gamma(0,1)\cup\{y\}$$ such that $$\delta(0)=x$$ and $$\delta(1)=y$$ and there is no holes in between $$\gamma$$ and $$\delta$$?

By "no holes in between" I mean that $$\gamma$$ and $$\delta$$ are homotopic in $$X\cup\{y\}$$, or alternatively, if $$G$$ is the Jordan domain defined by the union of $$\gamma$$ and $$\delta$$, then $$G\subset X$$.

I think, another way of stating this question is: is it true that the set $$C([0,1], X\cup\{y\})$$ is locally path connected?

• "Jordan curve" means that $\gamma(0)=\gamma(1)$. You want to say just "curve". Nov 7, 2019 at 15:40
• @SamZbarsky I think "simple curve" would be more appropriate. Nov 8, 2019 at 16:06

Yes, this is true because every topological copy of $$[0,1]$$ in the plane is unknotted and can be transformed by a homeomorphism of the plane into the straight arc $$[0,1]\times\{0\}$$. In the latter case the existence of the curve $$\delta$$ is more-or-less obvious.