# complement of a finite subset of a path-connected space is path-connected

Given a smooth connected manifold $X$ of dim $\geq 2$, I need to show that $X\setminus Y$ is connected, for some $Y\subseteq X$ finite.

The claim is intuitively obvious to me, but is not finding the right argument to prove.

For smooth manifolds, I know that path-connectedness and connectedness are equivalent. To prove that $X\setminus Y$ is path-connected, I take $x,y\in X\setminus Y$. I now need to show that there exists a cont's path $\gamma:[0,1]\longrightarrow X\setminus Y$ such that $\gamma(0)=x, \gamma(1)=y$. Along the way, I need to use the finiteness of $Y$, and also the fact that $X$ is also path-connected. I don't know how?

I haven't done fundamental group yet, so hint that does not involve it would be helpful.

• Whereever you meet one of the points in $Y$ you just make a little "detour" around it. Its easy to show this explicitly in $\mathbb{R}^n$. Then use local charts to transfer to manifolds. – J.R. Feb 28 '14 at 18:00

Since $X$ is path-connected, there is a path $\gamma : [0,1] \longrightarrow X$ from $x$ to $y$ in $X$ for every $x,y \in X$. Now if $p$ is a point on the path but is also in $Y$, there is a coordinate chart ($U, \phi$) of $p$ such that $Y\cap U = \emptyset$ (you can do this because $Y$ is finite). Map $\gamma([0,1])\cap U$ to $\mathbb{R}^n$, make a detour around $\phi(p)$ in $\mathbb{R}^n$ like TooOldForMath suggested and map it back to $X$. For each $p \in Y$ that the original path intersects, do this and you'll have pieced together a path in $X - Y$.

Remark: You can see why you need the dimension to be greater than 1 because then you can't make the 'detour'.

The idea mentioned by TooOldForMath in the comments to your question works fine of course. Let me suggest a more formal proof "just for fun".

I'm going to use the following characterization of connectedness, which is straightforward to prove and sometimes useful:

Fact. Let $X$ be a topological space and $Z$ be a discrete topological space with at least two points. Then $X$ is connected iff every continuous map $f: X \to Z$ is constant.

Now coming back to your problem. Let $Z$ be a discrete topological space with at least two points (e.g. $Z = \{0,1\}$) and $f : X \setminus Y \to Z$ a continuous map. The goal is to show that $f$ is constant. But $f$ extends continuously at points of $Y$ (see below), so it extends as a continuous map $X \to Z$ which must be constant by connectedness of $X$.

Why does $f$ extend continuously at points of $Y$? Let $y \in Y$, then there exists a neighborhood $U$ of $y$ in $X$ such that $U \setminus \{y\}$ is connected: this is easy to see using a local chart (also, you can choose $U$ small enough so that it only meets $Y$ at $y$). So in restriction to $U \setminus \{y\}$, $f$ must be constantly equal to some $z \in Z$, just extend it by setting $f(y) = z$.