Holomorphic separability Let $X$ be a compact Riemann surface, $A$ is a finite subset of $X$, $X^* = X\backslash A$, show that for any $p, q\in X^*$, $p\ne q$, there is a holomorphic function $f$ on $X^*$ such that $f(p)=0, f(q)=1$.
Could you please give me any hints? Thanks!
Edit: "for any $p, q\in X$" $\to$ "for any $p, q\in X^*$".
 A: This can be done nicely with Riemann-Roch. We need for $A \neq \emptyset$, or else this is impossible as there are no nonconstant holomorphic functions defined on a compact Riemann surface.
So let $z_0 \in A$. Let $D$ and $D'$ be the divisors $p - (2g +1)z_0$ and $p + q - (2g+1)z_0$. The degrees of $D$ and $D'$ are $-2g$ and $-2g + 1$, so that the degrees of $D^{-1}$ and $D'^{-1}$ are $2g$ and $2g-1$. In particular, both $D^{-1}$ and $D'^{-1}$ have degree greater than $2g-2$, hence there are no nonzero meromorphic differentials whose divisors are multiples of $D^{-1}$ or $D'^{-1}$. So Riemann-Roch then tells us:
$$\dim\{ f \text{ meromorphic on $X$} \mid D \text{ divides } f\} = 2g - g + 1 = g +1,$$
$$\dim\{ f \text{ meromorphic on $X$} \mid D' \text{ divides } f\} = 2g - 1 - g + 1 = g.$$ As the first dimension exceeds the second, let $f$ be a meromorphic function on $X$ so that $D | f$ but $D'$ does not divide $f$. As $D$ divides $f$, $f(p) = 0$ and the principal part of $f$ consists of a pole of order at most $2g+1$ at $z_0$. We can thus rule out the possibility that $f(q) = 0$, as otherwise we would have $D' | f$. So $f$ is the desired function.
