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Consider the $n-$dimensional closed half-space $\mathbb{H^n}=\{x \in \mathbb{R^n}|x_1 \le 0 \}$ and let $\partial \mathbb{H^n}=\{x \in \mathbb{H}|x_1=0\}$ be its boundary. Show that an open subset $U \subset \mathbb{H^n}$ is open in $\mathbb{R^n}$ iff $U \cap \partial\mathbb{H^n}=\phi$

My try:

Suppose that an open subset $U \subset \mathbb{H^n}$ is open in $\mathbb{R^n}$. Then let's assume that $\vec{x} \in U \cap \partial\mathbb{H^n}$.Since $U$ is open in $\mathbb{R^n}$, there exists $r \gt 0$ such that $B_{d}(\vec{x},r) \subset U$. Let $\vec{y}=(y_1,y')\in \mathbb{H^n} \cap B_{d}(\vec{x},r) $ such that $y_1 \lt 0$. $z=(-y_1,y') \in B_{d}(\vec{x},r) $ but $z \not \in B_d(\vec{x},r) \cap \mathbb{H^n}$. I believe this should lead to a contradiction at the point $x$. For some reason I am unable to see it.

For the other side since $ U$ is open in $\mathbb{H^n}$, $U$ can be written as some $V \cap \mathbb{H^n}$, where $V$ is open in $\mathbb{R^n}$. Since $U \cap \partial\mathbb{H^n}=\phi$ , we have $V \cap \partial\mathbb{H^n}=\phi$.Thus $V \subset \mathbb{H^n}$. Thus $U=V$.

Is this part alright? Thanks for the help!!

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  • $\begingroup$ If $x\in \partial\mathbb{H}^n$, then every open set containing $x$ intersects the complement of $\mathbb{H}^n$. $\endgroup$ – Vincent Boelens Apr 6 '15 at 14:11
  • $\begingroup$ @VincentBoelens How does that lead to a contradiction?? That's what I am asking $\endgroup$ – tattwamasi amrutam Apr 6 '15 at 14:17
  • $\begingroup$ $U\subset \mathbb{H}^n$ is open in $\mathbb{R}^n$ and contains $x$. Now apply what I said before. $\endgroup$ – Vincent Boelens Apr 6 '15 at 14:20
  • $\begingroup$ @VincentBoelens thank you $\endgroup$ – tattwamasi amrutam Apr 6 '15 at 14:26
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This theorem is actually true generally.

Theorem: If $S$ is a subset of a topological space $X$, then a set $U\subseteq S$ which is open in $S$ is open in $X$ iff $U\cap\partial S=\emptyset$.

Proof: Suppose that $U\cap\partial S=\emptyset$, and let $x\in U$. Then $x$ is an interior point of $S$ because it is in $S$ and not on the boundary, so there is a neighborhood $x\in V\subseteq S$. Since $U$ is open in $S$ there is also an open set $W$ such that $U=W\cap S$; then $V\cap W$ is a neighborhood of $x$ contained in $U$, so $U$ is open.

Conversely, if $U$ is open and $x\in U\cap\partial S$, then since $U$ is a neighborhood of $x$ and $x$ is a boundary point of $S$, $U$ intersects the complement of $S$, which is a contradiction because $U\subseteq S$. Thus $U\cap\partial S=\emptyset$.

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