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As the title suggest, I have the following question:

For which $n\in\mathbb{N}$ does there exists a subset $S\subset\mathbb{R}^n$ so that $S\cong\mathbb{S}^{n}=\{ x\in\mathbb{R}^{n+1}:\|x\|=1 \}$

I have a feeling that the answer should be: For none. For example, for the case $n=1$ is quite easy to show given the property that all connected sets of $\mathbb{R}$ are intervals. However the cases $n\geq2$ are significantly harder.

I think there should be same topological property that could solve the problem but so far I couldn't find any.

Any ideas? Any help is appreciated :)

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    $\begingroup$ +1: The answer (based on what you've written) is: all of them. Note that $$\Bbb S^n=\bigl\{x\in\Bbb R^n:\lVert x\rVert=1\bigr\}\subset\Bbb R^n.$$ However, you should instead have $$\Bbb S^n:=\bigl\{x\in\Bbb R^{n+1}:\lVert x\rVert=1\bigr\},$$ which makes the question much more interesting! $\endgroup$ Jul 9 '19 at 23:47
  • $\begingroup$ @CameronBuie Of course you're right, it was a typo. Thank you very much in pointing it out $\endgroup$ Jul 9 '19 at 23:50
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If by contradiction you could find $S\subset \Bbb R^n$ and a homeomorphism $\phi:S\to \Bbb S^n$, then the composition of $\phi^{-1}$ with $i:S\to \Bbb R^n$ would give a continuous injection $\Bbb S^n\to \Bbb R^n$. This is in contradiction with the Borsuk–Ulam theorem, which says in particular that any continuous map $\Bbb S^n\to \Bbb R^n$ can't be injective.

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    $\begingroup$ Nice! I wasn't aware of the statement of the Borsuk-Ulam theorem $\endgroup$ Jul 10 '19 at 0:23
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    $\begingroup$ I think you mean $i:S\to\Bbb R^n,$ and "continuous injection." $\endgroup$ Jul 13 '19 at 13:02
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Here's a generalization: if $M$ is any compact $n$-manifold then there is no subset of $\mathbb R^n$ homeomorphic to $M$. For suppose there were, and let $f : M \to \mathbb R^n$ be a homeomorphism onto its image $f(M)$. By the invariance of domain theorem, $f(M)$ is an open subset of $\mathbb R^n$. Since $f(M)$ is compact, it is also a closed subset of $\mathbb R^n$. Since $\mathbb R^n$ is connected, it follows that $f(M)=\mathbb R^n$, but $\mathbb R^n$ is not compact.

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