# Why is open connected minus countable set is connected?

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?

• @Ian $E$ might still be path connected. Say if $G=\mathbb{C}$ then we can connect any two points in $E$ by a path made out of a horizontal and vertical segment. Oct 8 '15 at 21:29
• @Ian Why is that a counterexample? What would be $G$? If $G=\mathbb{C}$, then it is true.. Oct 8 '15 at 21:29
• Sorry, I forgot about the "the plane is not a countable union of lines" argument.
– Ian
Oct 8 '15 at 21:31
• where you are using Hahn-Mazurkiewicz theorem? Jan 10 '18 at 18:44

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.
• We expect that interior points of at least some segments lie in $E$. But then we deform the straight line segment into a circular arc, with the same end points. Since the only points common to two such circular arcs are the endpoints of the segment, only countably many of the arcs can intersect $E$. So we find such an arc that lies entirely in $G\setminus E$. Oct 8 '15 at 21:38
• I wonder if there is a softer way of proving this. The Baire category theorem comes to mind. Given points $x,y\in G\setminus E$ and $p\in E$ the set of paths avoiding $p$ is open dense in the space of paths connecting $x$ and $y$... Oct 8 '15 at 21:48
• Let $u,v$ be the endpoints of the segment. The centre of the circles is then $$\frac{u+v}{2} + it\frac{v-u}{2}$$ for some $t\in \mathbb{R}$. When $\lvert t\rvert$ is small, the arc may leave $G$, so assume $\lvert t\rvert$ large (the straight line segment is the limit of the arcs as $\lvert t\rvert\to \infty$) to have all arcs contained in $G$. Of course it can be very difficult to determine which arcs meet $E$ and which don't. For a given $E$, there are usually easier ways to find explicit paths. Oct 8 '15 at 21:49
• @MihaHabič I think that would not work since $G$ need not be complete hence $C([0,1],G)$ is not complete.. Oct 8 '15 at 22:06