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Let $\{f_{\alpha}\}_{\alpha\in \mathcal{I}}$ be a family of holomorphic functions on the unit ball $\mathbb{B}^{n}$ in $\mathbb{C}^{n}$. Let $$ U:=\{z\in \mathbb{B}^{n}: f_{\alpha}(z)=0,\forall \alpha\in\mathcal{I}\}. $$

Can we always conclude that $U$ is an analytic subset in $\mathbb{B}^{n}$?

That is, for any $a\in \mathbb{B}^{n}$, there exist a neighbourhood $B(a,\varepsilon)\subset \mathbb{B}^{n}$ of $a$, and a finite family of holomorphic functions $g_{1},...,g_{m}$ on $B(a,\varepsilon)$ such that

$$ B(a,\varepsilon)\cap U=\{z\in B(a,\varepsilon): g_{1}(z)=...=g_{m}(z)=0\}. $$

I think in general this is not the case. We may need to impose some structure on our family. However, I am not sure at the moment.

Thanks so much for any suggestion.

Hahn
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    Obviously $\mathcal{I}$ is not a priori finite, isn't it? I mean are you asking if, from an arbitrary family of holomorphic functions, you can chose a finite subfamily which can be used to describe the union of the zero sets of each function in the family? – Daniele Tampieri Apr 30 '19 at 10:06
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    $\mathcal{I}$ is not supposed to be finite. If it is finite then the result is trivial. I think my question is stated very clearly. – Hahn Apr 30 '19 at 10:43
  • Could you pls. say more? What is the ideal should be constructing here. Since ${f_{\alpha}}$ is already given, i don't know how to see an ideal associated with it. – Hahn Apr 30 '19 at 22:22
  • @reuns: Nope, it is not Noetherian: https://math.stackexchange.com/questions/4049/proving-that-ring-of-complex-entire-functions-is-neither-artinian-nor-noetherian – Moishe Kohan May 01 '19 at 03:13
  • @reuns: This does not matter, take a sequence going to infinity in a half-plane, then use the fact that half-plane is conformal to the unit disk. – Moishe Kohan May 01 '19 at 03:22
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    @MoisheKohan The ring of meromorphic functions analytic at $0$ is claimed to be noetherian. Thus in some neighborhood of $0$ the vanishing set is given by finitely many $f_j$, which is what the OP wants – reuns May 01 '19 at 03:57
  • @reuns Could you write some suggestions? I know that $\mathcal{O}{a}$ is Noetherian, which means that any ideal of $\mathcal{O}{a}$ is finitely generated. However, here I am not sure how to relate this fact to get the claim. The set $\mathcal{I}$ may not be countable. – Hahn May 01 '19 at 04:01
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    There is a neighborhood $V\ni 0$ and finitely many $f_j$ such that for every $\alpha, Z(f_\alpha) \supset (V\bigcap Z(f_j) )$ the latter being analytic subset of $V$. If it isn't the case you have an infinite chain of ideals $(f_1,\ldots,f_n)$. For analytic functions Z(f), for meromorphic functions I think $Z(f)$ is defined as the closure of the points where $f$ is holomorphic and vanishes, so that $Z(\frac{x}{y+1}) = { (0,y),y \in \Bbb{C}}$ – reuns May 01 '19 at 04:10

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Yes, your $U$ is an analytic subset of $\mathbb B^n$. Here is the relevant theorem:

Given an open subset $H\subset \mathbb C^n$ and an arbitrary family $(V_\alpha)_{\alpha \in I}$ of analytic subvarieties $V_\alpha \subset H$, the intersection $\bigcap_{\alpha \in I}V_\alpha\subset H$ is analytic.

This settles your case by taking $H=\mathbb B^n$ and $V_\alpha =f_\alpha^{-1}(0).$
The displayed theorem is far from trivial, but you can find a proof in Whitney's masterful Complex Analytic Varieties, Theorem 9C, page 100.

Georges Elencwajg
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