I am trying to solve exercises from Pugh's Real Mathematical Analysis 2nd ed. and there is a question related to the connectedness of a disc. The question statement is as follows:
Is the disc minus a countable set of points connected? Path-connected? What about the sphere or the torus instead of the disc?
Intuitively, I think the disc is connected and also path-connected after removing those points. However, I would like to see proof of this. Can someone help me with that?
Given two points in the disk, there exist an uncountable family of paths connecting them, pairwise disjoint (except the endpoints). Subtracting a countable set will still leave at least one of these paths undisturbed.
Let $A$ be a countable set and $a,b\in\Bbb D\setminus A$. A line that passes between $a$ and $b$ intersects $\Bbb D$ in continuum-many points and each such point $p$ gives rise to a path $apb$ in $\Bbb D$, consiusting of two line segments. Any distinct two such paths have only $a,b$ in common and only countably many of them pass through a point $\in A$. Hence there are still uncountably many such paths left that are all insider $\Bbb D\setminus A$.
The same works for any connected manifold of dimension $\ge 2$, where you can start with uncountably many disjoint (up to ends) paths between given endpoints
Any path-connected space $X$ such that for every $x \neq y \in X$ there are uncountably many paths $p_i: [0,1] \to X, i \in I$ from $x$ to $y$ such that $p_i[(0,1)] \cap p_{j}[(0,1)] = \emptyset$ for all $i \neq j \in I$ (so the paths only intersect at end points $x$ and $y$) has the property that $X\setminus D$ is still path-connected when $D$ is countable. (Because at least one of the uncountably many paths must miss $D$).
This holds for spaces like $R^n, (n >1)$ and the disk too.
