follows that $\Gamma_{\infty}$ is a Fuchsian group, [@beardon]. Define $ S_{\infty}={\Bbb{H}}^2/\Gamma_{\infty}.$ Let $\pi_{\infty}\rightarrow S_{\infty}$ denote the quotient map. Set the notation $S_{\infty}=S(\alpha_0, p, C,\{F_i\})$ to emphasize the fact that the orientation on $\alpha_0,$ the point $p,$ the length $a=|\alpha_0|$, the closed set $C,$ and the sequence of untwisted flutes $F_i$ together define the surface.

As usual $B(q,r)\subset {\Bbb{H}}^2$ shall denote the set of points of distance less than r from q and we let $B^c(q,r)$ denote its closure. Let $i$ be the imaginary value $\sqrt{-1}.$ Without loss of generality, we can suppose that $i$ projects to the point $p$ on $A$. Consider the translates of the closed ball $B^c(i,r)$ by the transformations in $\Gamma_n$. Given $r>0$ there is a value $M_r\in {\Bbb{N}}$ so that for $ M_r\leq n\leq \infty$ and $g\in \Gamma_n,$ if $  g(B^c(i,r))\cap B^c(i,r)\not =\emptyset$ then $g\in \Gamma_{M_r}.$ This must be so since, in a Fuchsian group, only finitely many $\Gamma_{\infty}$-translates of the compact set $B^c(i,r)$ may intersect $B^c(i,r)$.

Thus, for $n\geq M_r$, the projections $\pi_n(B^c(i,r))$ are isometrically equivalent in the most natural way. Note $\pi_n(B^c(i,r))$ is connected and since the set of transformations that create intersections is finite, the complement of $\pi_n(B^c(i,r))$ in $S_n$ has $k<\infty$ components. 