Let $X$ be a compact connected Riemann surface and let $\{U_\alpha\}$ be an open covering. Suppose that on any $U_\alpha$ we have a meromorphic function $g_\alpha$ such that $\frac{g_\alpha}{g_\beta}$ is holomorphic and invertible on $U_\alpha\cap U_\beta$.

For any point $x\in X$ we can define the order at $x$ of the collection $g=\{g_\alpha\}$ simply as: $$ord_x(g):= ord_x(g_{\alpha})\quad \text{if }\; x\in U_\alpha$$

This number is well defined thanks to the property on intersections.

I don't understand why there are only finitely many points such that $ord_x(g)\neq 0$. The open sets $U_{\alpha}$ are not compact, so $g_{\alpha}$ can have infinitely many points of nonzero order on $U_\alpha$.


Every zero must have a neighborhood around it that contains no other zeros; otherwise the zero set would contain a limit point, which is not possible for meromorphic functions. Taking all these sets together with the complement of the zero set ( which is also open ), we get an open cover of the compact Riemann surface. This cover must have a finite subcover, which shows there are only finitely many zeroes


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