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

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 $$X$$?

• Your notation is very confused. Commented Dec 26, 2023 at 23:14
• I fell into the recursion $X= (P,X)$ and may never exit. Commented Dec 27, 2023 at 3:18
• $S_o$ can be an open set. Your example with d(x,y) going to infinity is equivalent to the intersection, if you like, of all open sets $S_o$ associated with x and y pertaining to a single line. Infinite intersections of open sets can be closed; there is no problem with that. Commented Dec 27, 2023 at 11:01
• @copper.hat Yes... I was actually unsure about that one, but it isn't the first time i see a topologic space defined like that. I'm kind'a self-taught math-learner. I'm open to notation suggestions, if you would care to provide them. Commented Dec 27, 2023 at 13:19
• In your edit, you seem to mix up line segments and lines. E.g. if we use as our "line to avoid" the $x$-axis, and take $o=(0,1)$, then the resulting $S_o$ is not the horizontal line through $o$ but rather the entire upper half plane, which is open. Commented Dec 29, 2023 at 5:51

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 $$S$$.

• Nice answer, I find this really interesting, the operation that assigns to each line segment and point the region that would be "lit up" by a light at that point, minding the line segment as an obstruction, behaves almost like the compliment in that an open segment creates a closed region, and a closed region creates an open region Commented Dec 28, 2023 at 12:37
• Moreover, if you take the union of such regions for all points, you retrieve the normal compliment I believe Commented Dec 28, 2023 at 12:40
• Yes, the union of the open regions $S_o$ for all $o \not\in \overline{xy}$ does indeed comprise the complement of $\overline{xy}$: it is clearly contained in the complement because $\overline{xy} \cap S_o= \emptyset$, and the compliment contains it because for any $z \not\in \overline{xy}$, $z \in S_z$. I don't think a closed segment creates a closed region under this construction. Commented Dec 28, 2023 at 21:08
• I am assuming your last statement was meant to be "...an open segment creates a closed segment..." and if so, I agree, but I believe i n $\mathbb{R}^n$ that an open all creates a closed set Commented Dec 28, 2023 at 21:20
• Apologies, I meant what you said and not what I typed, but you won't get a closed set from an open segment. E.g., assume $o$, $x$ and $y$ are not collinear, let $p = (x + y)/2$ (the midpoint of $\overline{xy}$) and consider a sequence of points $p_i$ in the interior of the segment $\overline{op}$ that converge to $p$. The $p_i$ are all in the analogue of the OP's $S_o$, but they converge to $p$, which is not. Commented Dec 28, 2023 at 21:39

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

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.