# On existence of subset of $\mathbb{R}^{n}$ homemorphic to $\mathbb{S}^n$

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 :)

• +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! Jul 9 '19 at 23:47
• @CameronBuie Of course you're right, it was a typo. Thank you very much in pointing it out Jul 9 '19 at 23:50

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.
• I think you mean $i:S\to\Bbb R^n,$ and "continuous injection." Jul 13 '19 at 13:02
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.