Only finitely many singularities contribute to the residue sum of an admissible cycle

Let $U:=\{z\in \mathbb{C}\setminus {\mathrm{\Gamma }}^{\ast }:n(\mathrm{\Gamma },z)\mathrm{\ne }0\}\mathrm{.}$ By the local constancy part of [L1], $U$ is open in $\mathbb{C}\setminus {\mathrm{\Gamma }}^{\ast }$. The same local constancy also shows that if $z\mathrm{\notin }{\mathrm{\Gamma }}^{\ast }\cup U$, then a whole neighbourhood of $z$ lies outside $U$, so every limit point of $U$ outside the trace already lies in $U$. Therefore $\bar{U}\subseteq U\cup {\mathrm{\Gamma }}^{\ast }$.

By the far-from-the-trace clause of [L1], the set $U$ is bounded. Since the trace ${\mathrm{\Gamma }}^{\ast }$ is compact, step 1.1 makes $\bar{U}$ a bounded closed subset of $\mathbb{C}$, hence compact.

Because $\mathrm{\Gamma }$ is admissible, its trace avoids the pole set. Thus [given, step 2.1, L2] ∎ $S\cap \bar{U}=S\cap U=\{a\in S:n(\mathrm{\Gamma },a)\mathrm{\ne }0\}\mathrm{.}$ The right-hand set is a closed discrete subset of the compact set $\bar{U}$ by, and every closed discrete subset of a compact metric space is finite. So only finitely many poles have nonzero index.