Is the set of all non-intersecting points an open set of the standard topology in $\mathbb{R}^2$?

Question

I'm wondering if the set of all non-intersecting points on the plane $P=\mathbb{R}^2$ is an open set of the standard topology in $\mathbb{R}^2$.

To make this precise, suppose $X=(P,X)$ is the standard topology on $\mathbb{R}^2$, and that $d$ is the euclidean metric on $X$. Let $\{x,y\}\in X$ be a pair of points in the plane. We define the closed set $\overline{xy}\in X$ as: $$\overline{xy}=\{p\in X|d(x,p)+d(p,y)=d(x,y)\}$$ This is, $\overline{xy}$ is the unique set containing all points in the line segment joining $x$ and $y$. Let $o\notin\overline{xy}$ be a point in $X$. We define the set of all non-intersecting points, $S_o\subseteq X$ as: $$S_o=\{p\in X|\overline{xy}\cap\overline{op}=\emptyset\}$$ Where $\overline{op}$ is the unique set containing all points in the line segment joining $o$ and $p$. This is, $S_o$ contains all points $p$ such that the line segment $\overline{op}$ does not intersect $\overline{xy}$, for an arbitrary $o\in X$.

Given this, how can we prove that $S_o\in X$ is an open set of $X$?

Answer 1

I think you are asking whether, given $o, x, y \in \Bbb{R}^2$, where $o \not\in \overline{xy}$, is the set $S = \{p \in \Bbb{R}^2 \mid \overline{xy} \cap \overline{op} = \emptyset\}$ open. The answer is yes. To see this, we can assume w.l.o.g. that $o = 0$. Then $S = A \cup B$, where $A$ is the interior of the triangle $\triangle{0xy}$ and $B$ is the complement of the cone formed by the rays $0x\infty$ and $0y\infty$ that do pass through $\overline{xy}$. $A$ and $B$ are both open, and hence so is $S$.

Answer 2

Hint: the distance between two compact disjoint subsets, (e.g. - two closed and bounded disjoint intervals) of the Euclidean plane is positive.

Added Later:

Let $p$ be a point in the set $S_o$. The intervals $\overline{op},\overline{xy}$ both being compact, the distance between them is positive, say, $\delta>0$. Given any point $q$, the maximal distance between the intervals $\overline{op}$ and $\overline{oq}$ is $d(p,q)$. Assume $d(p,q)<\delta$,if $w\in\overline{xy}\cap\overline{oq}$, then $d(w,p)\leq d(p,q)<\delta$, contradicting the assumption that the distance between $\overline{xy}$ and $\overline{op}$ is precisely $\delta$. Hence the intersection $\overline{xy}\cap\overline{oq}$ must be empty. Since this holds for every $q\in B(p,\delta)$, this proves that $S_o$ is open.