TITLE: On existence of subset of $\mathbb{R}^{n}$ homemorphic to $\mathbb{S}^n$
QUESTION [8 upvotes]: 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 :)

REPLY [4 votes]: 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.