half space is not homeomorphic to euclidean space

Solution 1:

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.

Solution 2:

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.