Suppose we are given a finite set $S$ of line segments in the plane and
- the intersection between two segments is empty or a single point in the interior of both segments
- at most two segments intersect in a single point
- all endpoints of segments and intersection points have pairwise different $x$-coordinate
Let's imagine that we sweep an imaginary vertical line $\ell$ rightwards over the plane, starting from a position left of all segments. Let
- $T$ be the set of segments $s\in S$ which intersect $\ell$ at a fixed time (i.e. position on the $x$-coordinate)
- $y_s$ be the $y$-coordinate of the intersection of $s\in T$ with $\ell$ and define $$s<t:\Leftrightarrow y_s<y_t\;\;\;\text{for }s,t\in T\tag 1$$
I want to show that if $s,t\in S$ with $s\cap t=\left\{p\right\}$, then $s,t$ eventually become neighbors (wrt to $<$) in T.
That's what I've tried (but I think it's too complicated): By assumption (2.), there is some $\varepsilon>0$ with $$B_\varepsilon(p)\cap u=\emptyset\;\;\;\text{for all }u\in S\setminus\left\{s,t\right\}\tag 2$$ and there is some point $v_u$ with $v_u.x<p.x$ and $$v_u\in\partial B_\varepsilon(p)\cap u$$ for $u\in\left\{s,t\right\}$. We can assume that $v_s.x>v_t.x$.
Now, the idea is to show that $s,t$ are neighbors in $T$ in the whole time interval $I_x:=(v_s.x,p.x)$, i.e. there is no $u\in s$ whose orthogonal projection onto the $x$-coordinate contains some $x\in I_x$ such that $$s<u<t\text{ or }t<u<s\;\;\;\text{in }T\tag 3$$ at time $x$.
What's the easiest way to prove this?
My idea was to let $$I_y:=\begin{cases}(v_s.y,v_t.y)&\text{, if }v_s.y<v_t.y\\ (v_t.y,v_s.y)&\text{, if }v_t.y<v_s.y\end{cases}$$ and show that $I_x\times I_y\subseteq B_\varepsilon(x)$. By $(2)$ it would immediately follow there can't be any $u\in S\setminus\left\{s,t\right\}$ with $(3)$.
I've tried to prove this in the following way: Let $q\in I_x\times I_y$. Then,
\begin{equation} \begin{split} \left\|p-q\right\|_2^2&<|p.x-v_s.x|^2+\max\left(\left|p.y-v_s.y\right|^2,\left|p.y-v_t.y\right|^2\right)\\ &\le\left\|p-v_s\right\|_2^2+\left\|p-v_t\right\|_2^2=2\varepsilon^2 \end{split} \end{equation}
but that estimate is to weak. So, how can we prove the statement?
I think you can simplify your proof after (2). You already know that \begin{align} B_{\epsilon}(p) \cap u = \emptyset \tag{2} \end{align} for all $u \in S \setminus \{s, t\}$. Furthermore, there must be an $x$ such that there exist points $v_s \in s \cap B_{\epsilon}(p)$ and $v_t \in t \cap B_{\epsilon}(p)$ with $v_s.x = v_t.x$. Without loss of generality let $v_s.y < v_t.y$. By (2) you know that there exists no $u \in S$ with $v_u.x = v_s.x$ and $v_s.y < v_u.y < v_t.y$ (This would imply $v_u \in B_{\epsilon}(p)$ and, thus, $B_{\epsilon}(p) \cap u \neq \emptyset$). Therefore, $s$ and $t$ must be neighbors.