Question on countability of family of holomorphic functions Let $\mathcal{F} \subset \mathcal{H}(\mathbb{C})$ a family of holomorphic functions such that for every $z \in \mathbb{C}$ the set $\{f(z): f \in \mathcal{F}\}$ is countable, my question is if $\mathcal{F}$ is necessary countble.
I don't have any idea from where to start so any tips or help would be really appreciated.
 A: Apparently, the question you ask has name and surname: it is called Wetzel's problem. To that extent, we can mention the Wikipedia page, which has a brief historical overview of the problem, as well as the article by Erdös, which proves the following:
The conclusion of the problem (i.e., there are only countably many  analytic functions in the family) is equivalent to the continuum hypothesis not holding.
Therefore, there is no simple answer to your question. The article by Erdös contains a beautiful proof of such equivalence, which I recommend that anyone who likes mathematics checks.
On the other hand, if you replace "countable" by "finite" everywhere, there is a simple answer: yes, the family must be finite, given that $\{f(z) \colon f \in \mathcal{F}\}$ is finite for all $z \in \mathbb{C}.$ The proof, which is fairly simple, is as follows:
Suppose there is an infinite sequence $\{f_i\}_{i \ge 1} \subset \mathcal{F}$ of pairwise distinct analytic functions. Let first $B_{i,j} = \{z \in \mathbb{C} \colon f_i(z) = f_j(z)\}.$ As all functions are analytic and distinct pairwise, the set $B_{i,j}$ is at most countable. Thus, the set
$$B = \cup_{i,j \ge 1, i \neq j} B_{i,j}$$
Is also countable. Now let $A_N = \{z \in \mathbb{C} \colon |\{f(z) \colon f \in \mathcal{F}\}| = N\}.$ As $\cup_{N \ge 1} A_N = \mathbb{C},$ there is $M \in \mathbb{N}$ such that $A_M$ is uncountable.
On the other hand, as $A_M$ is uncountable and $B$ is countable, there is at least a point $z_0 \in A_M \cap B^{c}.$ But this is a contradiction, as $z_0 \in B^{c}$ implies that $f_i(z_0) \neq f_j(z_0), \forall i\neq j,$ and thus $z_0 \not \in A_M.$
