Find a vector field on the plane such that $\omega$-limit set of a single point is two parallel lines. I came across this puzzling question in "Chaos" by K. T. Alligood. 

Sketch a vector field in the plane for which the $\omega$-limit set of a single trajectory is two (unbounded) parallel lines.

Call this point $p$, with evolution operator $t\mapsto\Phi_t$.
How can it be that for every point $z$ on two parallel lines, there exists an unbounded increasing series of times, $\{t_k\}$, such that $\|z - \Phi_{t_k}\|$ is arbitrarily small?
It seems like once the trajectory gets close enough to one of the lines it should stay near the line in order for rest of the line to be an $omega$-limit set.
Hints, anyone?
 A: Here's an example.
Start with $D \subset \mathbb{R}^2$ being the closed unit disc in the plane. On $D$ construct a vector field $\mathcal{V}$ such that the circle $\partial D$ is a closed trajectory, the origin is a zero of the vector field, and every other trajectory is repelled away from the origin and spirals around $\partial D$. Let $P,Q \in \partial D$ be a pair of antipodal points. Now take any diffeomorphism
$$f : D - \{P,Q\} \to \mathbb{R} \times [-1,+1]
$$
The pushforward vector field $f_*(\mathcal{V})$ can then be extended in any way to a vector field on all of $\mathbb{R} \times \mathbb{R}$. Any trajectory in $\mathbb{R} \times (-1,+1)$ (except the stationary trajectory at the origin) has $\omega$-limit set equal to the two lines $\bigl(\mathbb{R} \times \{-1\}\bigr) \cup \bigl(\mathbb{R} \times \{+1\}\bigr)$.
Where your intuition goes wrong is that the trajectory can go back and forth between the two lines, as long as the flow segments along which it travels back and forth themselves go off to infinity.
