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?
On
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
On
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.
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.