# Is this proof that the number of Weierstrass points on a compact Riemann surface $X$ is finite correct?

Let $$X$$ be a compact connected Riemann surface. Let $$\{h_1(z),\ldots,h_g(z)\}$$ be a base of the space of holomorphic forms on $$X$$ in a local map centered in $$p\in X$$. Let’s assume that we know that $$p$$ is a Weierstrass point iff the Wronskian

$$W(z):=\begin{vmatrix}h_1(z) & h_2(z) & \ldots & h_g(z) \\h_1’(z) & h_2’(z) & \ldots & h_g’(z) \\ \vdots & \vdots & & \vdots \\h_1^{(g-1)}(z) & h_2^{(g-1)}(z) & \ldots & h_g^{(g-1)}(z)\end{vmatrix}$$

verifies $$W(0)=0$$ (we can take this as the definition of a Weierstrass point for the purposes of this question) (I asked for a proof of this equivalence here).

Let’s also assume that we know that $$W(z)$$ isn’t identically null on any local chart.

I’d like to check if the following proof (that I came up with) of the fact that the number of Weierstrass points in $$X$$ is finite is correct, or if it could be made more rigorous:

Since all the $$h_i(z)$$ are holomorphic functions, then $$W(z)$$ is holomorphic too. But then the set of zeros of $$W(z)$$ must be discrete, or else it would contain an accumulation point, and by the identity principle $$W(z)$$ would be identically null. We also know it is closed since $$W(z)$$ is continuous, so it is closed and discrete. But every closed and discrete subset of a compact is finite, so the set of zeros of $$W(z)$$ is finite… and since it must contain the set of Weierstrass points of $$X$$, this set is finite too.

I think you are using the right concepts from complex analysis and topology. However, you should be careful that $$W(z)$$ is only defined locally, and cannot be a holomorphic function on all of $$X$$, because all holomorphic constant function on a compact Riemann surface are constant.