projects to a set $B^c_n(r)\subset S_n$. Since $r>e^{\frac{a}{2}}$, $ B^c(i,r)$ contains a segment of $I$ which projects to $\alpha_0.$ Than for $ M_r\leq n< m\leq \infty$ there is an isometric bijection $\varphi_{n,m}:B^c_n(r)\rightarrow B^c_m(r),$ which maps $\alpha^n_0$ to $\alpha^m_0$ and $p_n$ to $p_m$. We suppose that the finite complementary components $U_{n,j}$ and $U_{m,j}$ share the corresponding boundaries on $B^c_n(r)$ and $B^c_m(r),$ respectively. We shall prove that there is a $N_r\geq M_r$ so that for any $ n\geq N_r,$ the process of creating $S_{n+1}$ from $S_n$ by excising $E_{n+1}$ and gluing in $F_{n+1}^*$, does not change the subsurface $S_n(r)$.

Suppose $n>M_r$. Since $B^c_n(r)$ is compact, by the Collar Lemma there must exist a value $R>0$ and an integer $J_r\geq M_r$ so that for $j\geq J_r,\,  B^c_j(r)\cap  E_j\subset C_{E_j}(I_j,R).$ Consequently, the region $E_j\setminus C_{E_j}(I_j,R)$ must lie in the infinite component in the complement of $B^c_j(r)$.

Also, by the Collar Lemma we may choose $N_r\geq J_r$ so that for $j\geq N_r$ the collar $C_{F_j^*}(\alpha_{j,0},R)$ on $F_j^*$ is isometrically equivalent to $C_{E_j}(I_j,R).$ Thus, for $n\geq N_r$, $S_{n+1}$ can be constructed from $S_n$ by replacing $ E_{n+1}\setminus C_{E_{n+1}}(I_{n+1},R) $ by $F_{n+1}^*\setminus C_{F^*_{n+1}}(\alpha_{n+1,0},R).$ Consequently, $S_{n+1}$ differs from $S_n$ only in the infinite component in the complement of 