I would like to know if my proof to this statement is true:
Statement: Let $\mathbb{H}^n\subset\mathbb{R}^n$ be $\mathbb{H}^{n}=\left\{ \left(x_{1},\dots,x_{n}\right)\in\mathbb{R}^{n}\;\lvert\;x_{n}\geq0\right\} $ and and let $p\in\partial\mathbb{H}^n$ where $\partial\mathbb{H}^{n}=\left\{ \left(x_{1},\dots,x_{n}\right)\in\mathbb{R}^{n}\;\lvert\;x_{n}=0\right\}$. prove that for every open set $U\subset\mathbb{R}^n$ containing $p$ the set $U\cap\mathbb{H}^n$ is not diffeomorphic to any open $V\subset\mathbb{R}^n$
my arguement is this:
Arguement: first i'll show that for any open subset $U\subset\mathbb{R}^n$ the set $U\cap\mathbb{H}^n$ is diffeomorphic to some open set in $\mathbb{R}^{n-1}$, and then i'll use the fact that diffeomorphism is an equivalence relation to reach a contradiction. first, notice that $\partial\mathbb{H}^ n=\operatorname{span}\left\{ e_{1}^{n},\dots,e_{n-1}^{n}\right\} $ where $\mathcal{S}=\left\{ e_{i}^{n}\right\} _{i=1}^{n}$ is the standart basis of $\mathbb{R}^n$. now, we define $f:\partial\mathbb{H}^ n\rightarrow\mathbb{R}^{n-1}$ by $f\left(\left(x_{1},\dots,x_{n-1},0\right)\right)=\left(x_{1},\dots,x_{n-1}\right)$. it is clear that $f$ is a bijection. moreover, it is a linear transformation which satisfies $\left[f\right]_{\mathcal{S}}^{\mathcal{B}}=I_{n-1}$ where $\mathcal{B}=\left\{ e_{i}^{n}\right\} _{i=1}^{n-1}$ is a basis to $\partial\mathbb{H}^n$. therefore it is easy to see that $f$ is smooth, and also $\left[f^{-1}\right]_{\mathcal{B}}^{\mathcal{S}}=\left(\left[f\right]_{\mathcal{S}}^{\mathcal{B}}\right)^{-1}=I_{n-1}$ which means $f^{-1}$ is smooth as well, and we have showed that $\mathbb{R}^n$ and $\partial\mathbb{H}^n$ are diffeomorphic. now suppose there is an open subset $U\subset\mathbb{R}^n$, an open subset $V\subset\mathbb{R}^n$ and a diffeomorphism $\varphi:U\cap\mathbb{H}^n\rightarrow V$. since $U$ is open, there exists $\delta>0$ such that $B_{\delta}^{n}\left(p\right)\cap\mathbb{H}^ n\subset U\cap\mathbb{H}^n$ where $B_{\delta}^{n}\left(p\right)=\left\{ x\in\mathbb{R}^{n}\;\lvert\;\Vert x-p\Vert<\delta\right\} $. now observe the set $W=B_{\delta}^{n}\left(p\right)\cap\partial\mathbb{H}^ n$ and notice that $f\left(W\right)=B_{\delta}^{n-1}\left(f\left(p\right)\right)$ which is an open subset of $\mathbb{R}^{n-1}$ and therefore $W$ is diffeomorphic to an open subset of $\mathbb{R}^{n-1}$. but, according to our assumption, the function $\varphi_{\lvert W}:W\rightarrow\varphi\left(W\right)$ is a diffeomorphism between $W$ and $\varphi\left(W\right)$. now, recall that diffeomorphism is transitive and therefore $f\left(W\right)$ is diffeomorphic to $\varphi\left(W\right)$. if $\varphi\left(W\right)$ is not open, it cannot be diffeomorphic (let alone homeomorphic) to an open subset of $\mathbb{R}^{n-1}$. if it is open, then we found a diffeomorphism between open subsets of $\mathbb{R}^n,\mathbb{R}^{n-1}$ which cannot be (Invariance of domain theorem)
if there is a flaw in the arguement, I'd be happy to know. also, excuse me for my english, I'm not a native speaker
