Topological manifold in $\mathbb R^n$, of dimension at least 2, minus a countable set

Question

Let $A \subseteq \mathbb R^n$ be a topological manifold i.e. there is $k \in \mathbb N$ such that locally $A$ (with subspace topology from $\mathbb R^n$) is homeomorphic with $\mathbb R^k$. In this case $k$ is uniquely determined and is called the dimension of $A$. Now let $A \subseteq \mathbb R^n$ be a topological manifold of dimension at least $2$ and let $B$ be a countable subset of $A$. Then is the topological space $A\setminus B$ connected ? Path connected ?

Answer 1

Certainly if $A$ is disconnected, then so is $A\setminus B$. So we need to restrict our attention to connected $A$. Below is certainly not the fastest proof, but I like the concrete geometric intuition it provides (at least, to me):

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We first show that the result is true "locally:"

Lemma 1: $\mathbb{R}^k\setminus C$ is path connected whenever $C$ is countable.

Proof: One way to do this is by constructing "lots of disjoint paths" between any two points. For example, given point $p,q,r\in\mathbb{R}^n$, let $l_r(p,q)$ be the path from $p$ to $q$ gotten by going from $p$ to $r$ in a straight line and then $r$ to $q$ in a straight line. If $r_1, r_2$ are each equidistant between $p$ and $q$, then $l_{r_1}(p,q)\cap l_{r_2}(p,q)=\{p,q\}$, and there are continuum many such points, so if $p,q\in\mathbb{R}^n\setminus C$ then there is some $r$ with $l_r(p,q)\subseteq\mathbb{R}^n\setminus C$. $\quad\Box$

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We next show that this is enough:

Lemma 2: Suppose $A$ is a connected manifold of dimension $k$ and $p,q\in A$. Then there is an open subset $U$ of $A$ containing $p$ and $q$ and homeomorphic to $\mathbb{R}^k$.

Proof: Since $A$ is path connected, let $l$ be a path connecting $p$ and $q$. We can "thicken" $l$ by considering the set $$l[\epsilon]:=\{x\in A: d(x,l)<\epsilon\}$$ for some appropriate positive $\epsilon$. As long as $l$ is "weakly-non-self-intersecting" - that is, it never leaves a point and then comes back to that point later (we allow it, however, to "linger" at a point for a while - hence the "weakly") - we can find some small enough $\epsilon$ such that $l[\epsilon]$ is homeomorphic to $\mathbb{R}^k$ (exercise). However, what if $l$ is not weakly-non-self-intersecting?

It turns out that we can "remove redundancies" from paths: any path has a weakly-non-self-intersecting "subpath." To prove this, let's remember first what a path is:

Definition: A path is a continuous function from $[a,b]$, for some finite $a,b$.

(Often we restrict attention to $a=0,b=1$, but this makes no difference.)

Now suppose $l:[a,b]\rightarrow A$ is a path. Say that $x\in [a,b]$ is bad if there are $y,z$ with $a\le y<x<z\le b$ such that $l(y)=l(z)$; that is, $x$ "lies between two points of self-intersection." Let $S\subseteq[a,b]$ be the set of non-bad points in $[a, b]$.

Now we can show (exercise) that the path $$\hat{l}: [a,b]\rightarrow A: x\mapsto l(\sup\{y\le x: y\in S\})$$ is weakly non-self-intersecting.