# The interior of open set in a convex set is not empty

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?

• The empty set is open and its interior is empty. Sep 17, 2020 at 7:02
• Okay, and if $U$ is not empty? Sep 17, 2020 at 7:03
• 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. Sep 17, 2020 at 7:11


• 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? Sep 17, 2020 at 7:42
• 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 $X$.
– user239203
Sep 17, 2020 at 7:45
• 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? Sep 17, 2020 at 7:47
• Yes.${}{}{}{}{}$
– user239203
Sep 17, 2020 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 $$\mathbb R^2$$.

• 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? Sep 17, 2020 at 7:16
• 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 $B(x,r)\subseteq H^n$!!! Sep 17, 2020 at 7:20
• Unfortunately what you write don't say to me what happens when $x\in\text{Bd}(H^n)$. Sep 17, 2020 at 7:22
• Perhaps you implicitely state that $U\cap\text{int}(H^n)$ is not empty??? Sep 17, 2020 at 7:23