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Is there any sequence $ \{ Z_{\nu} \}_{\nu \in \mathbb{N}}$ in $\mathbb{C}^{n}$, $Z_{\nu} \rightarrow 0$, such that any holomorphic function in $\mathbb{C}^{n}$ which vanishes in $Z_{\nu}$ for all $\nu \in \mathbb{N}$ is identically zero?

Thank you!

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Hi theStudent: how can any sequence fail? From the sequence, you can compute all derivatives of your function at 0, which in holomorphic land is enough to determine your function. It's possible I've misunderstood your question, but most likely it isn't really appropriate for this site. Please read the faq, and if you still think it's appropriate, read howtoask. – Theo Johnson-Freyd Oct 13 '12 at 18:13
Anything dense in an open ball, surely, just by analyticity – Yemon Choi Oct 13 '12 at 18:14
@Theo: in several CV your holo function might vanish on some submanifold, so if your sequence is confined to that submanifold it won't detect the non-zero nature of the original function – Yemon Choi Oct 13 '12 at 18:15
@Yemon Choi, How can a sequence which is dense in an open ball converge to $0$? – the L Oct 13 '12 at 18:18
To fine tune Yemon's suggestion, take a countable set $Z_\nu'$ dense in the ball, and consider $Z\nu=\frac{1}{\nu} Z_\nu'$. I suspect this works, but surely "the Student" ought to check. – Donu Arapura Oct 13 '12 at 18:25
up vote 4 down vote accepted

Start from a countable dense subset $S$ of the unit ball, and take a sequence $Z_\nu$ such that for all $s\in S$, one has $\nu Z_\nu=s$ infinitely often. Then, any holomorphic function in $\mathbb{C}^n$ that vanishes along the sequence $Z_\nu$ has in particular a subsequence of zeros accumulating to the origin that belong to the complex line generated by any $s\in S$. Therefore it vanishes identically on that complex line, by the principle of isolated zeros in one variable. Since the union of these lines is dense in $\mathbb{C}^n$, the function is identically zero by continuity.

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