Reference: Path connectedness of the complement of countable set
Let $G$ be an open connected subset of $\mathbb{C}$.
And let $E$ be a countable subset of $G$.
How do I prove that $G\setminus E$ is path-connected?
If $E$ is discrete and has no limit point in $G$, we can apply Hahn-Mazurkiewicz theorem to prove that $G\setminus E$ is path-connected. (If $a,b\in G$, then there exists an injective path $\alpha$ connecting $a,b$ in $G$. Since $E$ has no limit point, there exist only finitely many poins of $E$ on the trace of $\alpha$. Since $\alpha$ is injective and $E$ is discrete, we can deform $\alpha$ into a path not passing through those finite intersections. Hence, the result is a path in $G\setminus E$ connecting $a,b$)
However, I saw the comment in the link saying that $G\setminus E$ is indeed path-connected even though $E$ is any countable set. My argument above for the case decrete $E$ having no limit point cannot be applied to this general one. How do I prove this?
Best Answer
Take two points $x,y \in G \setminus E$. Connect them by a path consisting of finitely many straight line segments in $G$. If an endpoint of one of the segments lies in $E$, replace that point with a point in $G\setminus E$ close to it. Thus you have a polygonal path connecting $x$ and $y$ in $G$ such that the endpoints of each line segment lie in $G \setminus E$.
Now consider one such line segment. If it lies entirely on $G\setminus E$, leave it alone. Otherwise, consider circular arcs connecting the two endpoints. Any two such circular arcs are disjoint, except for the two endpoints. There are uncountably many such arcs entirely contained in $G$, and $E$ is countable. Hence uncountably many of these arcs lie in $G\setminus E$. Replace that segment by such an arc.