Let $D$ be the open unit disk, and $J$ a Jordan arc (that is, a homeomorphic copy of $[0, 1]$) that lies in $D$, except $J(0)$ lies on the boundary of $D$, say $J(0)=1$. I would like to see that $D\smallsetminus J([0, 1])$ is a path-connected topological space. Please help, if you can. Thanks!



A Jordan arc in the unit disk
It's certainly the case that $\mathbb{R}^2\setminus J$ is path connected.
So any two points in $D\setminus J$ are joined by a path in $\mathbb{R}^2$
missing $J$. If this path isn't in $D$ it hits the boundary of $J$
but then replace part of the path by an arc of the boundary of $D$.
Then one can replace this part of the arc by an arc just inside the boundary
(there is $\epsilon>0$ such that any closed arc on the boundary not containing
$J(0)$ has distance $>\epsilon$ from $J$).


