zeroes of holomorphic function

Solution 1:

As others have said, it comes down to the nonexistence of an uncountable discrete subset $Z$ of an open subset $U$ of the complex plane $\mathbb{C}$. The point of this answer is to record a proof of this that I find simplest. Namely, the space $\mathbb{C}$ is second countable: there is a countable base for the topology (take, e.g., open balls with rational radii centered at points $x+yi$ with $x,y \in \mathbb{Q})$). But every subspace of a second countable space is second countable: just restrict a countable base to the subspace. In particular $U$ is second countable and so is the putative uncountable discrete subset $Z$ of $U$. But this is absurd: since the only base of a discrete space consists of all the singleton sets, an uncountable discrete space is not second-countable!

Solution 2:

Existence of such a function would imply that its zero locus were an uncountable closed discrete subset of the complex plane, contradicting that the complex plane is separable and normal.

EDIT in response to question: (For simplicity's sake $f$ is assumed to be defined on all of $\mathbb{C}$ — this isn't really a problem because $\mathbb{C}$ is hereditarily normal, so if $f$ is defined on an open set $U$ we also have that $U$ is separable and normal)

The zero locus of $f$ means $f^{-1}(\{0\})$ (i.e. the set of points where $f$ is $0$ — call it $Z$) and it's the pre-image of a closed set under a continuous function so it's closed. Assume that $Z$ is uncountable and discrete. Then every function from $Z$ to $\mathbb{R}$ is continuous and thus there are $2^{2^{\aleph_0}}$ many continuous functions from $Z$ to $\mathbb{R}$. Since $\mathbb{Q}(i)$ is a countable dense subset of $\mathbb{C}$ any continuous real-valued function on $\mathbb{C}$ is determined by its values on $\mathbb{Q}(i)$, i.e. there are $2^{\aleph_0}$ many continuous real-valued functions on $\mathbb{C}$. Hence there is some continuous real-valued function on $Z$ that cannot be extended to the whole space, but $\mathbb{C}$ is normal so any continuous function on a closed subspace is extendable to a continuous function on the whole space by the Tietze extension theorem; contradicting our assumption of $Z$ being uncountable and discrete.

This is essentially the same argument as why the Moore plane isn't normal.

Solution 3:

A holomorphic function on a connected open set that is not identically zero cannot have uncountably many zeros. Open subsets of $\mathbb{C}$ are $\sigma$-compact, so if $G$ is the domain, then there is a sequence $K_1,K_2,\ldots$ of compact subsets of $G$ such that $G=K_1\cup K_2\cup\cdots$. (It is not hard to construct $K_n$; e.g., if $G$ is the whole plane, you can take $K_n=\{z:|z|\leq n\}$. Otherwise you can take $\{z\in G:|z|\leq n\text{ and }d(z,\partial G)\geq \frac{1}{n}\}$.) An uncountable subset of $G$ must have uncountable intersection with one of the $K_n$s, because a countable union of countable sets is countable. An infinite subset of $K_n$ has a limit point in $K_n$ by compactness. The rest follows from the result you mentioned that a holomorphic function that is not identically zero cannot have a limit point of its zero set in the connected open set on which it is defined.