Definition
The upper half-space $H^k$ is the set of those $x$ of $\Bbb R^k$ whose last coordinate is not negative. The positive upper half-space $H^k_+$ is the set of those $x$ of $\Bbb R^n$ whose last coordinate is positive.
Statement
The upper hal-space is closed and its interior is the positive upper-half space.
So to show that $H^k$ is closed I try to prove that $\Bbb R^k\setminus H^k$ is open. So let be $x\in\Bbb R^k$ that is $x_k<0$ and I put
$$
\delta:=\min\{|x_i|: i=1,…,n\}
$$
and we proved to show that the open cube centered in at $x$ and radius $\delta$ is contained in $\Bbb R^k\setminus H^k$ but unfortunately it seems false: indeed if $y\in C(x,\frac\delta 2)$ then
$$
y_i-x_i<\frac\delta 2
$$
for each $i=1,…,n$ and by definition of $\delta$ this means that
$$
y_k<\delta+x_k=\frac\delta 2-|x_k|<\delta-|x_k|\le 0
$$
that is $y\in\Bbb R^k\setminus H^k$. Then if $x\in H^k$ is such that $x_k=0$ we observe that $x\in\text{Bd}(H^k)$. So for any $\delta>0$ we observe that $C(x,\delta)\cap\Bbb R^n\setminus H^k\neq\emptyset$ and $C(x,\delta)\cap H^k\neq\emptyset$ because if $\xi_k\in(-\delta,0)$ then $(x_1,…,x_{k-1},\xi_k)\in C(x,\delta)\cap\Bbb R^k\setminus H^k$ and if $\xi_k\in[0,\delta)$ then $(x_1,…,x_{k-1},\xi_k)\in C(x,\delta)\cap H^k$. So now if we prove that $\text{Bd}(H^k)\subseteq\{x\in\Bbb R^n:x_k=0\}$ then the statement holds but unfortunately I can't prove it. So could someone help me, please?
Best Answer
For $x \in \Bbb R^k\setminus H^k$ with $x_k < 0$, the (euclidean) open ball $B(x, -x_k/2)$ centered on $x$ with radius $-x_k/2$ is included in $\Bbb R^k\setminus H^k$ as for $y=(y_1, \dots, y_k) \in B(x, -x_k/2)$ you have
$$0 \le \left\vert y_k - x_k\right\vert \le \sqrt{(y_1-x_1)^2 + \dots +(y_k - x_k)^2} \lt -x_k/2$$
and therefore $$y_k \lt \frac{3}{4}x_k \lt 0$$
This proves that $\Bbb R^k\setminus H^k$ is open and therefore $H^k$ closed.
Also $H_+^k$ is open (similar proof than above using balls) and included in $H^k$ therefore $H_+^k \subseteq \text{int}(H^k)$.
Conversely if $x$ is such that $x_k >0$ then $x \in \text{int}(H^k)$: similar proof than above using balls. And if $x \in H^k$ with $x_k=0$, then any non empty open ball centered on $x$ will intersect $\Bbb R^k\setminus H^k$. This proves that $\text{int}(H^k) \subseteq H_+^k$ and finally $H_+^k = \text{int}(H^k)$.