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Let $H \subset \mathbb{R}^n$ be a closed subset homeomorphic to $\mathbb{R}^{n - 1}$. Could you give me an idea of how to prove that $\mathbb{R}^n - H$ has exactly two path components. I am expecting an answer using singular homology.

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    I'm suspecting this will follow from an analoguous theorem for spheres, after applying the one point compactification to $\Bbb R^n$. It's easy to see that the one closure of $H$ will be a sphere of dimension $n-1$, so the question will be to show that $\Bbb S^n\setminus \hat{H}$ has two components, where $\hat{H}\simeq\Bbb S^{n-1}$. I believe this is a well known theorem.2012-12-12
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    I cant see why $H$ together with the extra point(required for one point compactification $\mathbb{R}^n$) is closure of $H$ in $S^n$.2012-12-12
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    AFter applying the one point compactification, as @Olivier suggests, and if $n=2$, you get the statement of the Jordan curve theorem.2012-12-12
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    Because $H$ cannot be bounded, for otherwise it would be compact (since it is already closed in $\Bbb R^n$). Therefore, $\infty$ is adherent to $H\subset\widehat{\Bbb R^n}\simeq\Bbb S^n$. So $\overline{H}=H\cup\lbrace\infty\rbrace$ is compact, has $H\simeq\Bbb R^{n-1}$ as a dense subset, so is isomorphic to its one point compactification, that is $\Bbb S^{n-1}$.2012-12-12
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    You might be interested in Proposition 2B.1 on p.169 of Hatcher's book http://www.math.cornell.edu/~hatcher/AT/AT.pdf. This is the theorem that Olivier refers to in his first comment.2012-12-13
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    This theorem in Hatcher is a special case of Alexander duality.2012-12-14

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