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Definition

The upper half-space $H^n$ in $\Bbb R^n$ is the set of those $x\in\Bbb R^n$ such that $x_n\ge 0$.

So I ask if it is true that any not empty and open set $U$ in $H^n$ has interior (in $\Bbb R^n$) not empty. Probably this is a consequence that the interior of open set in a convex set is not empty? Is my this last statement true? So could someone help me, please?

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Antonio Maria Di Mauro
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    The empty set is open and its interior is empty. – Vercassivelaunos Sep 17 '20 at 07:02
  • Okay, and if $U$ is not empty? – Antonio Maria Di Mauro Sep 17 '20 at 07:03
  • Then it's true, but you assertion that $X\subseteq Y$ convex, $U$ open in $X$ $\Rightarrow$ interior of $U$ in $Y$ not empty, is wrong. For instance, take the $x$-axis as a convex subset of $\mathbb R^2$. Any of its subsets, including open ones, have empty interior as subsets of $\mathbb R^2$. The interior of the convex set should also be non-empty for your assertion to hold. – Vercassivelaunos Sep 17 '20 at 07:11

2 Answers2

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$\newcommand{gae}[1]{\newcommand{#1}{\operatorname{#1}}}\gae{cl}\gae{int}$ The topological property you are looking for is that $H^n$ is contained in the closure of its interior. In fact, this property is necessary and sufficient for your condition to hold. Let $X$ be a topological space and $S\subseteq X$. If $\cl_X\int_X S\supseteq S$, then every open set $U$ such that $U\cap S\ne\emptyset$ satisfies $U\cap\int_X S\ne\emptyset$ as well. Since $U\cap \int_X S$ is open in $X$ and $U\cap \int_X S\subseteq U\cap S$, we have that $U\cap \int_X S\subseteq \int_X(U\cap S)$. Vice versa, suppose that $S\setminus \cl_X \int_X S\ne \emptyset$. Then, $S\setminus\cl_X\int_X S$ is a non-empty subset of $S$ which is open in $S$ and which cannot contain open subsets of $X$ (because $(S\setminus \cl_X\int_X S)\cap\int_X S=\emptyset$). Therefore $\int_X(S\setminus\cl_X\int_X S)=\emptyset$.

  • Well, I read you wrote. So if $S\subseteq\text{cl}_X\big(\text{int}(S)\big)$ then you state that if $U\cap S\neq\emptyset$ for a open set $U$ in $\Bbb R^n$ then $U\cap\text{int}_X(S)\neq\emptyset$ but I don't understand completely your proof. Could you write it better? – Antonio Maria Di Mauro Sep 17 '20 at 07:42
  • I say that for all topological spaces $X$ and for all $S\subseteq X$, the interior of $S$ is dense in $S$ if and only if every non-empty subset which is open in $S$ has non-empty interior in $X$. –  Sep 17 '20 at 07:45
  • Okay, perhaps I understand!!! if $S\subseteq\text{cl}_X\big(\text{int}_X(S)\big)$ then any $x\in S$ is an adherent poin of $\text{int}(S)$ so if $U$ is an open set containing $x$ then by definition of adherent point there must necessarly be $U\cap\text{int}(S)\neq\emptyset$, right? – Antonio Maria Di Mauro Sep 17 '20 at 07:47
  • Yes.${}{}{}{}{}$ –  Sep 17 '20 at 07:48
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By definition of the relative topology, if $U$ is open in $H^n$ it exists an open subset $V$ of $\mathbb R^n$ such that $U = V \cap H^n$. If $U$ is not empty, it exists $x \in U$. Therefore $x \in V$ and it exists an open ball $B(x,r)$ centered on $x$ with $B(x,r) \subseteq V$.

If $x=(x_1, \dots, x_{n-1}, 0)$ then $B(\bar x, r/4) \subseteq B(x,r) \subseteq U $ where $\bar x = (x_1,\dots,x_{n-1}, r/2)$. And if $x=(x_1, \dots, x_{n-1}, x_n)$ with $x_n >0$ then $B(x, \bar r) \subseteq B(x,r) \subseteq U $ where $\bar r = \min(r, x_n/2)$. Proving that the interior of $U$ is not empty.

Here, the main argument is that for an open ball $B(x,r) \subseteq \mathbb R^n$ we have $B(x,r) \cap H^n = B(x,r)$ for $r$ small enough and $x_n >0$.

This is not related to the fact that $H^n$ is convex. For example $L= \{(x,0) \mid x \in \mathbb R\}$ is a convex subset of the plane $\mathbb R^2$. $I= \{(x,0) \mid x \in (0,1)\}$ is an open subset of $L$. However, the interior of $I$ is empty in $\mathbb R^2$.

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