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.

  • $\begingroup$ 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. $\endgroup$ – 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.

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

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