How to proof M is a compact connected n-manifold,if for any point $p\in M,M\backslash\{p\}\cong R^n$ then M is homeomorphic to $S^n$.

Question

M is a compact connected n-manifold,if for any point $p\in M,M\backslash\{p\}\cong R^n$ then M is homeomorphic to $S^n$.

I have the guess from https://mathoverflow.net/questions/117457/manifolds-with-two-coordinate-charts, but I don't know how to proof it.Beside if there are any grammatical mistake,I apologize for my poor english.

Answer 1

The open sets around $p$ are the complement of closed set of $M$ that does not contain $p$. Since closed subset of compact set is compact, it's the complement of compact sets of $M$ that does not contain $p$. This means that $M$ is a one-point compactification of $\mathbb{R}^n$, which is unique up to homeomorphism. Since $S^n$ is one such compact manifold, $M$ must be homeomorphic to $S^n$.