Why is the support of a divisor on a compact Riemann Surface is finite?

Question

I know that a closed and discrete subset of a compact is finite. But I am not sure that the support of a divisor is closed. How can I assure that?

Answer 1

Just to remove this question from unaswered and for future reference, since I had the same problem, I will expand a bit @Lee Mosher comment.

Forster defines a divisor on a (not necessarily compact) Riemann surface $X$ as a mapping $$D:X \to Z$$ such that for any compact subset $K\subset X$, $D(x)\ne 0$ only for finitely many points $x \in K$. This is easily seen to imply both that $\mathrm{supp }\, D$ is always discrete and finite in the case $X$ is compact.

This seems to a me a better definition than Miranda's. Probably Miranda uses another definition of "discrete", this is not the only passage of his book where this is evident.