Can we strengthen this exercise in Forster's book on Riemann surfaces?

Question

Exercise 2.5 in Otto Forster's Lectures on Riemann Surfaces states

Suppose $p_1,\ldots,p_n$ are points on the compact Riemann surface $X$ and $X':=X\setminus\{p_1,\ldots,p_n\}$. Suppose $$f:X'\to\mathbb{C}$$ is a non-constant holomorphic function. Show that the image of $f$ comes arbitrarily close to every $c\in\mathbb{C}$.

There are two ways in which I was wondering if this exercise can be strengthened. The first is to shrink the domain by replacing $\{p_1,\ldots,p_n\}$ with a closed countable set. That is, if $C$ is any closed countable subset of $X$, $X':=X\setminus C$, and $f:X'\to\mathbb{C}$ is a non-constant holomorphic function, must the image of $f$ still be dense?

Secondly, I was wondering if there is a bound on the number of points missing from the image. Specifically, I want to know if we can show that the image of $f:X\setminus\{p_1,\ldots,p_n\}\to\mathbb{C}$ contains all but finitely many points of $\mathbb{C}$, and if so, I want to know if there is an upper bound on the number of points not in the image in terms of $n$.

For example, when $n=1$, we can show that the image of $f$ contains all but at most one point of $\mathbb{C}$. To see this, suppose $f:X\setminus\{p\}\to\mathbb{C}$ is holomorphic. If $p$ is an essential singularity, then Picard's great theorem implies that the image of $f$ contains all but at most two points of $\mathbb{P}^1_\mathbb{C}$ and thus all but at most one point of $\mathbb{C}$. In the case that $p$ is a pole, $f$ extends to a holomorphic map $X\to\mathbb{P}_\mathbb{C}^1$. But such a map is surjective since $X$ is compact, so $f(X\setminus\{p\})=\mathbb{C}$. Thus, the upper bound when $n=1$ is just $1$.

Edit: The argument I gave for $n=1$ easily generalizes to $n\geq 2$ which somehow didn't occur to me (Thank you AndyPutman for pointing this out). I have added the proof to my answer.

Answer 1

The answer to your first question is positive. First of all, it suffices to consider the case when every point of your subset $C\subset S$ is a non-removable singularity of $f: S\to \mathbb C P^1$. (If not, then work with the subset $C'\subset C$ of nonremovable singularities. If $f$ extends to the entire $C$ then $f(S-C)$ will be dense as a complement to a countable subset.) Next, as a closed countable metrizable space, $C$ has at least one isolated point $c$. Since $f$ has nonremovable singularity at $c$, we can apply the Big Picard Theorem to a suitable punctured disk neighborhood $U$ of $c$. Then $f(U)$ will be dense in $\mathbb C P^1$.

Answer 2

After some thinking, I came up with a solution to the first strengthening by mimicking the idea for the finite case and doing some transfinite tomfoolery. Let $C$ be a closed set with cardinality less than the continuum and let $f:X\setminus C\to\mathbb{C}$ be a holomorphic function and suppose the image is not dense. If $c\in\mathbb{C}$ is not in the closure of the image, then $\frac{1}{f(x)-c}$ is a bounded holomorphic function. So, by Riemann's removable singularities theorem, we can remove all of the isolated singularities.

Mimicking the proof of the Cantor-Bendixson Theorem, we can keep removing the subset of isolated points from our set of singularities repeatedly. After apply Riemann's removable singularities theorem some transfinite number of times, we are left with a perfect set $P\subseteq C$ and a holomorphic function $g:X\setminus P\to\mathbb{C}$ extending $\frac{1}{f(x)-c}$. However, because $X$ is a Polish space, every nonempty perfect set has cardinality equal to the continuum. Since $\vert P\vert\leq\vert C\vert<\vert\mathbb{R}\vert$, $P$ must be empty.

Thus, we have a holomorphic extension $g:X\to\mathbb{C}$. Since $X$ is compact, $g$ must be constant, implying that $\frac{1}{f(x)-c}$ is constant, and thus, $f$ is constant. So, if $f:X\setminus C\to\mathbb{C}$ is a non-constant holomorphic map, then its image must be dense.

As for the second strengthening, I claim that for $n\geq 2$, the image of a non-constant holomorphic function $f:X\setminus\{p_1,\ldots,p_n\}\to\mathbb{C}$ contains all but at most $n-1$ complex numbers. To see this, if at least one of $p_1,\ldots,p_n$ is an essential singularity, then Picard's great theorem implies the image contains all but at most one point in $\mathbb{C}$. So suppose none of $p_1,\ldots,p_n$ are essential singularities. Then all of them are poles or removable singularities (unlike the case of $n=1$, we can have a removable singularity). Thus, $f$ extends to a surjective holomorphic map $\tilde{f}:X\to\mathbb{P}_\mathbb{C}^1$, and therefore $$f(X\setminus\{p_1,\ldots,p_n\})\supseteq\mathbb{P}_\mathbb{C}^1\setminus\{\tilde{f}(p_1),\ldots,\tilde{f}(p_n)\}.$$ Now, at least one of $\tilde{f}(p_1),\ldots,\tilde{f}(p_n)$ must be $\infty$, but the rest could belong to $\mathbb{C}$. Hence, the image of $f$ must contain all but at most $n-1$ points of $\mathbb{C}$, as desired.

To see that these upper bounds are attained, let $X=\mathbb{P}_\mathbb{C}^1$, $p_1=\infty$ and let $p_2,\ldots,p_n$ be any distinct points of $\mathbb{C}$. If $n\geq 2$, let $f$ be the inclusion into $\mathbb{C}$; then the image is $\mathbb{C}\setminus\{p_2,\ldots,p_n\}$ which attains the upper bound of $n-1$. If $n=1$, let $f(z)=e^z$. Then the image of $f$ contains every point except $0$, so the upper bound of $1$ is obtained.