# zeroes of holomorphic function

I know that zeroes of holomorphic functions are isolated,and I know that if a holomorphic function has zero set whic has a limit point then it is identically zero function,i know a holomorphic function can have countable zero set, does there exixt a holomorphic function which is not identically zero, and has uncountable number of zeroes?

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## 4 Answers

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!

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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.

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@kahen---> could you explain a little more?why if zero locus(I dont know what is zero locus means exactly) were uncountable then C would not be normal and seperable? –  elinor Mar 17 '11 at 8:45
+1: I like this. However, when you say $2^{2^\aleph_0}$, aren't you assuming that the zero set has cardinality $2^{\aleph_0}$ rather than just being uncountable? It is apparently consistent that $2^{\aleph_1}=2^{\aleph_0}$, so there seems to be something else needed. mathoverflow.net/questions/17152/… (An uncountable closed subset of $\mathbb{C}$ must have cardinality $2^{\aleph_0}$, so there isn't really a problem, but I didn't know whether you meant to skip that step.) –  Jonas Meyer Mar 17 '11 at 9:24
@ Kahen,nice I got it, thank you very much to all. –  elinor Mar 17 '11 at 9:25
@Jonas: True. If one wants to avoid that, one could go by the alternate route of using $\sigma$-compactness, $Z$ having infinitely many points in one of the compact sets, and that any infinite subset of a compact metric space has a limit point –  kahen Mar 17 '11 at 9:34
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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.

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thank you,so I can say that any uncountable subset of C has a limit point because if not then for a point in it will have a ball with no other point from the set and I will put a complex number with rational co ordinate in that ball and then i can construct an injection from this uncountable set to rational points of plane.am I correct? –  elinor Mar 17 '11 at 8:37
@elinor: That's not quite right, because from what you have described there is no guarantee that the balls will be disjoint, and therefore no guarantee that the chosen rationals are distinct. –  Jonas Meyer Mar 17 '11 at 8:42
@jonas--ok!! I got –  elinor Mar 17 '11 at 8:43
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Your question needs some clarification, as it is not internally consistent. If I understand correctly, you're asking if all holomorphic functions have uncountably many zeros? In that case, the answer is definitely no, for example $f(z) = 1$.

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@ alex-nope, I am saying that is it true--> whether a non constant holomorphic function can have uncountable number of zeroes. –  elinor Mar 17 '11 at 8:39
@elinor: Alex was responding to your original version which was later fixed by William. –  Jonas Meyer Mar 17 '11 at 8:43
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