Vtyjkyuk

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 $Asetminus B$ connected ? Path connected ?

Certainly if $A$ is disconnected, then so is $Asetminus B$. 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):

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

Lemma 1: $mathbbR^ksetminus C$ 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 $p,q,rinmathbbR^n$, 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 $p,qinmathbbR^nsetminus C$ then there is some $r$ with $l_r(p,q)subseteqmathbbR^nsetminus C$. $quadBox$

We next show that this is enough:

Lemma 2: Suppose $A$ is a connected manifold of dimension $k$ and $p,qin A$. Then there is an open subset $U$ of $A$ containing $p$ and $q$ and homeomorphic to $mathbbR^k$.

Proof: Since $A$ is path connected, let $l$ be a path connecting $p$ and $q$. We can "thicken" $l$ by considering the set $$l[epsilon]:=xin A: d(x,l)<epsilon$$ 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 $mathbbR^k$ (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 $xin [a,b]$ is bad if there are $y,z$ with $ale y<x<zle b$ such that $l(y)=l(z)$; that is, $x$ "lies between two points of self-intersection." Let $Ssubseteq[a,b]$ be the set of non-bad points in $[a, b]$.

Now we can show (exercise) that the path $$hatl: [a,b]rightarrow A: xmapsto l(supyle x: yin S)$$ is weakly non-self-intersecting.