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

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.

• It might be easier to think in terms of path-connectedness. If your connected manifold has dimension>1, then any two points which are not p can be joined by a path avoiding p. – user45861 Apr 7 '16 at 21:41

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.
• Oh thank you very much. But it remains the case $\dim M = 1$. What would you suggest in this case? – JonSK Apr 7 '16 at 21:51
• If $M$ is 1-dimensional and has no boundary, then $M$ without a point has at most two components. It cannot have more. – Peter Franek Apr 7 '16 at 22:22