K^{\#}_j$ for each $j$. The $K_j^*$ correspond to the sets on $S_i\setminus E_{i+1},$ also called $K_j^*,$ which give a compact exhaustion of $S_i\setminus\ E_{i+1}.$ Similarly, the $ K^{\#}_j$ correspond to sets on $F_{i+1}^*,$ also called $ K^{\#}_j,$ which induce a compact exhaustion of $F_{i+1}^*$. Then $K_j=K^{\#}_j\cup K^*_j$ is a compact exhaustion of $S_{i+1}$. Let ${\sf E}_j$ be a nested sequence of complementary components of the sequence $\{K_j\}.$ Then one of the following three possibilities holds: for $j$ sufficiently large, either ${\sf E}_j\subset S_i \setminus E_{i+1}$ or ${\sf E}_j\subset F_{i+1}^*$, or ${\sf E}_j$ has non-empty intersection with both $S_i \setminus E_{i+1}$ and $F_{i+1}^*$ for all $j$. In the first two cases the corresponding end must be finite. In the last case it is possible that the sets ${\sf E}_j$ are all cylinders lying to the left of $\alpha_0$ and consequently, the corresponding end is finite. The remaining case occurs when each ${\sf E}_j$ is a union of open sets on $S_i \setminus E_{i+1}$ and $F_{i+1}^*$ belonging to the infinite end of each of those surfaces. In only this last case is it possible for the corresponding end to be infinite. Thus $S_{i+1}$ has a single infinite end and $S_n$ is a 