There is this statement in my topology notes, where it states that:
Denote $\mathbb{R}^n_{+}$ as the upper half surface in $\mathbb{R}^n,$ denoted by $\{(u_1,...,u_n)\in \mathbb{R}^n:u_n\ge0\}$. Then by the subspace topology, $\mathbb{R}^n_{+}$ is open if there exists a set $V^*$ such that $V=V^* \cap \mathbb{R}^n_{+}$, where $V^*\subset \mathbb{R}^n$. It is clear that if $V\subset \mathbb{R}^n_+,$ is a disjoint open subset from the subspace $\{u_n=0\},$ then $V$ is open $\mathbb{R}^n_+,$ iff V is open in $\mathbb{R}^n$
What I have trouble seeing here is I get the subspace topology part as it is directly following from the definition of subspace topology. However, I dont understand the last part where it mentions the iff part.
Could someone please explain why that is the case?
EDIT: Also what is the significance of the set being disjoint to $\{u_n=0\}$ in this case?
Remember that in $\mathbb R^n$ a set is open if an only if it contains an $n$-dimensional open ball about each of its points. Let's look at this in $\mathbb R^2$ and take $\mathbb R^2_+$ to be $\{(x,y)\in \mathbb R^2\mid y\ge0\},$ the upper half-plane. Consider the square $S=\{(x,y)| 2>y>0, -1<x<1\}$ This is open, whether considered as a subset of $\mathbb R^2$ or as a subset of $\mathbb R^2_+.$
Now consider the square $\overline S=\{(x,y)| 2>y\ge0, -1<x<1\}.$ This is not an open subset of $\mathbb R^2,$ for it contains no neighborhoods of the points on the $x-$axis. It is an open subset of $\mathbb R^2_+,$ however, because it is the intersection of $\mathbb R^2_+$ with $\{(x,y)| 2>y>-1, -1<x<1\}$ for instance.