I know that $S^{n}- p$ is homeomorphic to $R^{n}$ and the fundamental group : $\pi_{1}(R^{n} - 0) = \pi_{1}(S^{n})$ but I can't prove they aren't homeomorphic . Anybody help me ? Sorry for my bad English
Why $R^{n} - 0$ is not homeomorphic to $S^{n}$
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general-topology
1 Answers
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$\mathbb{R}^n-0$ is not compact unlike $S^n$.
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0Can you prove it ? – 2017-01-13
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1That $R^n-0$ is not compact? Obviously $a_k=(k,\ldots,k)$ does not have a convergent subsequence. – 2017-01-13
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1More generally, by Heinie-Borel, a subset of $\mathbb R^n$ is compact if and only if it is closed and bounded. – 2017-01-13
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0:( Oh sorry , I forgot Heine-Borel theorem , thank you . – 2017-01-13