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Define $H^n = \{(x_1, \dots, x_n)\in \mathbb R^n : x_n \ge 0\}$, $\partial H^n = \{(x_1, \dots, x_{n-1},0) : x_i \in \mathbb R\}$.

$\partial H^n$ is a manifold of dimension $n-1$: As a subspace of $H^n$ it is Hausdorff and second-countable. If $U \subseteq \partial H^n$ is open in $H^n$ with the subspace topology of $\mathbb R^n$ then $f: (x_1, \dots, x_{n-1},0) \mapsto (x_1, \dots, x_{n-1}), \partial H^n \to \mathbb R^{n-1} $ is injective and continous. Then its restriction to $U$ is a homeomorphism by invariance of domain.

I am asked to show that a nbhd $U'$ of $x \in \partial H^n$ is not homeomorphic to an open set $U \subseteq \mathbb R^n$. My try:

$H^n$ is with the subspace topology. Then a set $U' \subseteq H^n$ is open iff it is $U \cap H^n$ for some open set $U \subseteq \mathbb R^n$. $H^n$ is closed in $\mathbb R^n$. How to proof that $U \cap H^n$ can't be open? Thank you for correcting me.

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Your question is difficult to read because you keep using the adjective open without saying open in what space. – Georges Elencwajg Jan 15 '13 at 11:48
@GeorgesElencwajg I am sorry. Is it better now? – tom b. Jan 15 '13 at 12:10
Dear user58260, I'm afraid it is still not correct: it is impossible for a non-empty subset $U \subseteq \partial H^n$ to be open in $H^n$. You want $U$ to be open in $\partial H^n$ and then the rest of the section is a triviality, independent of invariance of domain. – Georges Elencwajg Jan 15 '13 at 12:25
@Georges: I understood the question to be about a set $U'$ that is open in $H^n$ but which has also a non-empty intersection with $\partial H^n$. – Marek Jan 15 '13 at 12:45
up vote 1 down vote accepted

Suppose $U'$ is open in $\mathbb R^n$ and $U' \ni x \in \partial H^n$. Then $U'$ must contain an open ball $B$ around $x$ and so there must exist a point $y \in B$ with $y_n < 0$ and therefore $U' \not \subset H^n$.

For the more general argument, suppose $\phi: U' \to \mathbb R^n$ is a homeomorphism. Then there is an open ball $B \subset \mathbb R^n$ containing $\phi(x)$ and so there is also a ball $B' \subset \phi^{-1}(B)$ that contains $x$. So we are done by the first paragraph.

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Thank you very much! – tom b. Jan 15 '13 at 13:40

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