Is intersection of zero set of any family of holomorphic functions an analytic set?

Question

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.

Answer 1

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.