Let $B:=\{w\in \mathbb{R}^2 : \lvert w\rvert < 1\}$ be the unit circle. Suppose that $Q$ is an open, connected subset of $B$, such that a boundary point $q\in \partial Q$ lies in $\partial B$. For an arbitrary interior point $x\in Q$, I want to find a path $\Gamma$ which connects $x$ and $q$ such that $\Gamma$ lies in $Q$ (except for $q$). I may also use the boundary points of $Q$ which lie in $B$, namely $\partial Q \cap B$, in order to construct such a path.
Connecting an interior and boundary point of an open, connected subset of the unit circle using a path.
curvesgeneral-topologyreal-analysis
Best Answer
Your construction unfortunately need not be continuous at $1$. That's because middle points of each curve may converge to somewhere else than $y$, breaking the continuity.
Moreover, there is no way to fix that: the statement is not true in general. Consider
$$U_n=(1/n-\epsilon_n, 1/n+\epsilon_n)\times\mathbb{R}$$
Here I choose $\epsilon_n$ to be small enough so that $U_n\cap U_m=\emptyset$ for $n\neq m$. Something like $\epsilon_n=1/n^2$.
Then we define $B=\mathbb{R}\times(-1,1)\cup\big(\bigcup_{n=1}^\infty U_n\big)$. So these are infinitely many vertical thick lines ("convergent" to the $Y$-axis) together with a single horizontal thick line. Our $B$ is clearly connected and open. But for example $(0,2)\in\partial B$ and there is no path (Jordan or not) almost entirely contained in $B$ with $(0,2)$ as its endpoint.
Also note that the example above can be made bounded by intersecting everything with any big enough (i.e. of radius $>1$) ball around $(0,0)$.