About branch points of a holomorphic map

Let $$F:X \to Y$$ be a holomorphic map between Riemann surfaces. $$q \in Y$$ is a branch point if it is the image of a ramification point. How to prove that the set of branch points is a discrete subset of $$Y$$. This is from Rick Miranda's Algebraic curves and Riemann surfaces. Thank you.

• I think it is a mistake. I don't believe branch points is always discrete. However, it's discrete for compact $X$ and $Y$ supposedly. I'm about to make a question about this. In the mean time, see here: math.stackexchange.com/questions/3813927/… – John Smith Kyon Oct 31 '20 at 4:03
• Thanks a lot.@JohnSmithKyon – yuan Dec 1 '20 at 2:50
• You're welcome yuan. how about upvoting and accepting my answer then? – John Smith Kyon Dec 3 '20 at 23:30

You forgot to say $$F$$ is non-constant. Then again, I guess $$Ram(F)$$ is not defined for $$F$$ non-constant.
In general for any map $$F: X \to Y$$ of any topological spaces $$X$$ and $$Y$$ with $$X$$ compact and $$Y$$ Fréchet/T1 and for any closed discrete subspace $$A$$ of $$X$$, we have $$F(A)$$ discrete.
Proof: Closed discrete subspaces $$A$$ of compact is finite $$\implies$$ $$A$$ is finite $$\implies$$ $$F(A)$$ is finite $$\implies$$ $$F(A)$$ is discrete because finite subspaces of Fréchet/T1 are discrete. QED
Apply this to the case of $$A=Ram(F)$$ when $$F$$ is a non-constant holomorphic map between connected Riemann surfaces with $$X$$ compact (and thus $$F$$ is surjective, open, closed and proper and $$Y$$ is compact) to get $$F(A)=Branch(F)$$ is discrete.
In particular, this means we do not use that $$F$$ is proper, closed, open, surjective, non-constant or holomorphic or that $$X$$ is connected or that $$Y$$ is connected. We can relax this to $$X$$ compact (and not necessarily Riemann surface) and $$Y$$ Fréchet/T1 (and not necessarily Riemann surface, Hausdorff/T2 or compact).