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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    $\begingroup$ The empty set is open and its interior is empty. $\endgroup$ – Vercassivelaunos Sep 17 at 7:02
  • $\begingroup$ Okay, and if $U$ is not empty? $\endgroup$ – Antonio Maria Di Mauro Sep 17 at 7:03
  • $\begingroup$ 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. $\endgroup$ – Vercassivelaunos Sep 17 at 7:11

$\newcommand{gae}[1]{\newcommand{#1}{\operatorname{#1}}}\gae{cl}\gae{int}$ 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 $\int_X(S\setminus\cl_X\int_X S)=\emptyset$.

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  • $\begingroup$ 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? $\endgroup$ – Antonio Maria Di Mauro Sep 17 at 7:42
  • $\begingroup$ 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 $X$. $\endgroup$ – Gae. S. Sep 17 at 7:45
  • $\begingroup$ 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? $\endgroup$ – Antonio Maria Di Mauro Sep 17 at 7:47
  • $\begingroup$ Yes.${}{}{}{}{}$ $\endgroup$ – Gae. S. Sep 17 at 7:48

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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  • $\begingroup$ Excuse me but I don't fully understand: if $U$ is open in $H^n$ it is not necessarly open in $\Bbb R^n$ so why for any $x\in U$ there exist $r>0$ such that $B(x,r)\subseteq H^n$? Could you explain better, please? $\endgroup$ – Antonio Maria Di Mauro Sep 17 at 7:16
  • $\begingroup$ For example what happens if $x\in U\cap\text{Bd}(H^n)$? In this case it looks to me that there not exist $r>0$ such that $B(x,r)\subseteq H^n$!!! $\endgroup$ – Antonio Maria Di Mauro Sep 17 at 7:20
  • $\begingroup$ @AntonioMariaDiMauro I added additional elements in my answer. Let me know if this clarify. $\endgroup$ – mathcounterexamples.net Sep 17 at 7:20
  • $\begingroup$ Unfortunately what you write don't say to me what happens when $x\in\text{Bd}(H^n)$. $\endgroup$ – Antonio Maria Di Mauro Sep 17 at 7:22
  • $\begingroup$ Perhaps you implicitely state that $U\cap\text{int}(H^n)$ is not empty??? $\endgroup$ – Antonio Maria Di Mauro Sep 17 at 7:23

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