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We define, half space $H^n$ = $\{(x_1,x_2,...,x_n) | x_n \geq 0\}$. Can anyone suggest, how to prove that $H^n$ is not homeomorphic to the euclidean space $R^n$.

user93620
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    $\mathbb H^n$ contains a hyperplane of dimension $n-1$ (the "boundary") that can be removed without disconnecting the space. $\mathbb R^n$ does not. I'm thinking specifically about $\mathbb H^2$ and $\mathbb H^3$, but I'm pretty sure it holds in all higher dimensions as well. – MPW Apr 04 '14 at 20:08
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    The Alexandrov compactification of a half-space is a ball, the Alexandrov compactification of $\mathbb{R}^n$ is the sphere $S^n$. – Daniel Fischer Apr 04 '14 at 20:11
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    Along the same line of the comments above: if there exists an homeomorphism from $ H^n $ to $ R^n $ then removing $ {x: x_n =0 } $ from the domain you get a homeomorphism from a connected space to a disconnected one. This holds because by Jordan Brouwer separation theorem we know that a closed set of $ R^n $ homeomorphic to $ R^{n-1} $ disconnects the space into two connected components. –  Apr 04 '14 at 20:16
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    Well, I would consider the Alexandrov compactification as still somewhat elementary. – A.P. Apr 04 '14 at 20:44
  • Thanks to all for the answers. – user93620 Apr 05 '14 at 06:56
  • @DanielFischer Could you please post your comments as answers, so that the OP can accept one? – A.P. Apr 05 '14 at 10:49
  • I meant to notify @user55449, too, but apparently I can put only one @ mention in one comment... – A.P. Apr 05 '14 at 10:50
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    @user55449: Can you please elaborate on why Jordan Brouwer separation theorem implies that a closed subset of $\mathbb{R}^n$ homeomorphic to $\mathbb{R}^{n-1}$ disconnects the space into two connected components? I only found references for the theorem where the separating set was a sphere, not a half-space? – Asaf Shachar Apr 26 '16 at 16:46

2 Answers2

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Here is a different proof from the ones in the comments.

Let $O=(0,\dotsc,0)$. If $\mathbb{R^n}$ and $H^n$ were homeomorphic, then let $P$ be the image of $O$ in this homeomorphsim. Then $\mathbb{R^n}\setminus\{P\}$ and $H^n\setminus\{O\}$ would be homeomorphic, too. In particular, they would be homotopically equivalent. But $\mathbb{R}^n\setminus\{P\}$ is homotopically equivalent to the $n-1$-sphere $S^{n-1}$ while $H^n\setminus\{O\}$ is contractible.

A.P.
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If there exists an homeomorphism from $ H^n $ to $ R^n $, then removing $ \{x:x_n=0\} $ from the domain you get a homeomorphism from a connected space to a disconnected one. This follows from Jordan Brouwer separation theorem: if a closed set of $ R^n $ is homeomorphic to $ R^{n−1} $ then it disconnects $ R^n $ into two connected components.

  • This looks like an erroneous quotation of Jordan-Brouwer (https://topospaces.subwiki.org/wiki/Jordan-Brouwer_separation_theorem). R^2 has a closed subset C such that C is homeomorphic to R and R^2\C has three components. Namely, let C be the graph of the function R_{>0} -> R : x |-> sin(1/x)/x – Maurice Dekker Oct 31 '22 at 15:10