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If $n\ge 2$ then $$\Bbb S^{n-1}=\left\{\sum_{k=1}^nx_k^2=1\right\}$$ is not homeomorphic to $\Bbb R^n$ because the former is compact and the latter is not. At least I have been explained so when I studied it.

But if $E$ is an infinite dimensional Banach or Hilbert space (over $\Bbb R$ or $\Bbb C$), can we state that $$E\not\cong\{x\in E:\|x\|=1\}?$$

ajotatxe
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    I think this question will give you an answer: https://math.stackexchange.com/questions/678214/homeomorphisms-between-infinite-dimensional-banach-spaces-and-their-spheres – N.B. Nov 20 '19 at 17:22
  • @N.B. Thank you very much. I expected a proof, but perhaps the problem is too difficult to be solved here. – ajotatxe Nov 20 '19 at 17:25
  • Actually the answer provides you with two references, so you can look at the papers for the proofs. Maybe you meant an answer reasonably short, that could fit in a post? – N.B. Nov 20 '19 at 17:27
  • @N.B. Yes, I did, but when I have time, I'll check your references, of course. I see that the short answer is 'No, never', at least for Hilbert spaces. – ajotatxe Nov 20 '19 at 17:29

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