If $M$ is a connected manifold, does $M\setminus\{p\}$ have finitely many components?

Question

Let $M$ be a connected manifold and $p\in M$. Is it true that $M\setminus\{p\}$ has only finitely many connected components?

(We can also suppose $M$ is compact if that helps.)

I think this is true but I can't prove it yet. This is what I thought: $M$ looks the same as some $\mathbb{R}^n$ locally. Let $U\subseteq M,V\subseteq\mathbb{R}^n$ be homeomorphic open sets with $p\in U$ and $V$ some open ball. If $M\setminus\{p\}$ has infinitely many components, would that imply that $V\setminus\{x\}$ ($x$ is the image of $p$) must also have infinitely many components? That would prove that $M\setminus \{p\}$ must have only finitely many components.

What do you think?

Thank you.

Answer 1

I think your argument is quite all right. The crucial part of it is proving the implication $M - \{p\}$ has infinitely many components $\implies$ $V - \{x\}$ has infinitely many components, and I think you should focus on making sure it you argue it convincingly.

Note though that if $\dim M \geq 2$, $M - \{p\}$ is connected whenever $M$ is -- connected manifolds are also path connected, and you can make any path omit $p$ by going down to Euclidean neighbourhood.