# half space is not homeomorphic to euclidean space

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 $R^n$.

• $\mathbb H^n$ 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 4, 2014 at 20:08
• The Alexandrov compactification of a half-space is a ball, the Alexandrov compactification of $\mathbb{R}^n$ is the sphere $S^n$. Apr 4, 2014 at 20:11
• 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.
– user55449
Apr 4, 2014 at 20:16
• Well, I would consider the Alexandrov compactification as still somewhat elementary.
– A.P.
Apr 4, 2014 at 20:44
• @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? Apr 26, 2016 at 16:46

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.
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.