Given a sequence of numbers $1 > a_1 > a_2 > \cdots$ converging to $0$, let $U \subset \mathbb{C}$ be obtained from the open unit square $(0,1)\times(0,1)$ by removing the line $[a_n, 1]\times\{a_n\}$ for each odd value of $n$, and removing $[0, 1 - a_n]\times\{an\}$ for each even value of $n$, as illustrated in Figure 9. Give $U$ the Poincare metric. Given a basepoint $Z_0$ in the open set $U$, for each unit vector $v$ in the tangent space at $Z_0$ there is a unique geodesic ray $g_v : [0, \infty ] \to U$ which starts at $Z_0$ with initial velocity vector ${g^\prime_v}(0)$. Evidently $g_v(t)$ tends towards the boundary of $U$ as $t\to \infty$ . Now consider the sequence of line segments $L_n = U \cap([0,1] \times {a_n})$, each of which cuts $U$ into two components. Let $V_n$ be the set of all unit vectors $v$ at $Z_0$ such that $g_v[0,\infty)$ intersects the line segment $L_n$.
(1) If $Z_0$ lies near the top of $U$, show that each $V_n$ contains the closure of $V_{n+1}$ and show that there exists a vector $v$ which belongs to the intersection of the $V_n$ .
(2) Show that every point on the bottom edge of the unit square is an accumulation point for the ray $g_v(t)$ as $t\to \infty$.
(3) Now consider the conformal automorphism $f : U\to U$ which maps this geodesic to itself with $f(\hat{g}_v(t)) = \hat{g}_v(t + 1)$. (Compare Problem 2-e.) Show that every point of the bottom edge is also an accumulation point for the orbit $f: Z_0 \mapsto Z_1\mapsto · · · $.
I tried to solve this problem as following:
(1) To show that $V_n$ contains the closure of $V_{n+1}$, we observe that since $L_n$ cuts $U$ into two components, any geodesic ray $g_v$ with $v\in V_n$ will eventually cross $L_n$. As we move to the next $L_{n+1}$, the geodesic rays $g_v$ with $v\in V_{n+1}$ will also cross $L_{n+1}$, since $L_{n+1}$ includes $L_n$ and goes further. Hence, any vector $v\in V_{n+1}$ is also in $V_n$, demonstrating that the closure of $V_{n+1}$ is contained in $V_n$. To show the existence of $\hat{v}$, consider an open neighborhood around $Z_0$ within $U$. Since $U$ is compact, we can choose a sequence of vectors $v_n$ with $v_n\in V_n$ within this neighborhood. Due to the containment of the closure of $V_{n+1}$ in $V_n$, this sequence converges to a vector $\hat{v}$ which belongs to the intersection of all $V_n$.
(2) For every point on the bottom edge, the translated geodesic $f(g_{\hat{v}}(t))$ follows the same trajectory as $g_{\hat{v}}(t)$ but with a translation. Since the original geodesic approaches the bottom edge, the translated geodesic will also do the same. Therefore, every point on the bottom edge is an accumulation point for the ray $g_{\hat{v}}(t)$ as $t\to \infty$.
(3) As we demonstrated in part (1), every point on the bottom edge is an accumulation point for the translated geodesic $f(g_{\hat{v}}(t))$. Considering that the conformal automorphism $f$ preserves the geodesics, each point on the bottom edge is also an accumulation point for the orbit $f: Z_0 \mapsto Z_1\mapsto · · · $.
I need from you to correct my question, please.