Why is this set open in a proof of a manifold with boundary?

Question

I try to understand the prove given here. There we need to find an open neighbourhood for all points on $\partial \overline{\Bbb B^n}= \Bbb S^{n-1}$ which is homeomorphic with an open subset in $\Bbb H^n$. He gives us this homeomorphism later as $\varphi: U_i^+ \cap \overline{\Bbb B^n} \to \Bbb H^n$. Now I don't understand, why $U_i^+ \cap \overline{\Bbb B^n}$ is an open neighbourhood of $x \in \partial \overline{\Bbb B^n}= \Bbb S^{n-1}$. Can you help me out?

Answer 1

The claim is that every $x\in\mathbb S^{n-1}$ has an open neighborhood of the form $V_i^\pm:=U_i^\pm\cap\overline{\mathbb B^n}$, where $U_i^\pm=\{x\in\mathbb R^n;\pm x_i>0\}$. (For details on notation, see the question linked in the question.) Therefore we have to check two things:

1. Every $x\in\mathbb S^{n-1}$ is contained in one of these sets.

2. These sets are open subsets of $\overline{\mathbb B^{n}}$.

Part 1: If $x\in\mathbb S^{n-1}$, then $x\neq0$ and thus $x_i\neq0$ for some $i$. If $x_i>0$, then $x_i\in V_i^+$. If $x_i<0$, then $x_i\in V_i^-$.

Part 2: These sets should be open in the relative topology of $\overline{\mathbb B^{n}}$. This is the topology inherited from the ambient space $\mathbb R^n$. Open sets in the relative topology are the sets of the form $U\cap\overline{\mathbb B^{n}}$, where $U\subset\mathbb R^n$ is open in the sense of the ambient space. (This is a definition, not a result. It might not have been stated explicitly in the linked question, but the closed ball is equipped with the relative topology, and with that topology it is a manifold.) The sets $U_i^\pm$ are open in the ambient space, so $V_i\pm$ are open in the closed ball. (They are not open subset of $\mathbb R^n$, but that is unimportant. They are open subsets of the space $\overline{\mathbb B^n}$ we are studying.) If you are studying manifolds, I hope you can convince yourself that $U_i^\pm\subset\mathbb R^n$ are indeed open.