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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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It means that the sequence $\{ Z_{\nu} \}_{\nu \in \mathbb{N}}$ converges to $0$, the origin of $\mathbb{C}^{n}$. – rla Oct 12 '12 at 18:38
It is identically zero, because the zero set of a non-zero holomorphic function is discrete. – rla Oct 12 '12 at 18:49
So do you see which $K_{\nu}$ will work in this case? – Davide Giraudo Oct 12 '12 at 18:50
I was trying to take a dense sequence $\{ Z_{\nu}' \}$ in the unit ball, and define $\{ Z_{\nu} = Z_{\nu}' / \nu \}$. I should prove that all the derivatives of all orders are zero in $0$, but I did'nt get any success. – rla Oct 12 '12 at 22:32
Davide: I think the idea here is that in $n = 1$ case, if a holomorphic $f: \mathbb{C}\rightarrow \mathbb{C}$ vanishes on any infinite set with a limit point, it must vanish identically. The same does not hold in $\mathbb{C}^n$. Consider, for example, $f(x,y) = x - y$. – William Oct 13 '12 at 1:49
up vote 3 down vote accepted

Let $\{ Z_{\nu}' \}_{\nu \in \mathbb{N}}$ be a dense sequence in the unit ball $B(0;1)$. Define a new sequence $\{ Z_{\nu} \}_{\nu \in \mathbb{N}}$ by

$$Z_{1}', \frac{Z_{1}'}{2}, \frac{Z_{2}'}{3}, \frac{Z_{1}'}{4}, \frac{Z_{2}'}{5}, \frac{Z_{3}'}{6}, \frac{Z_{1}'}{7}, \frac{Z_{2}'}{8}, \frac{Z_{3}'}{9}, \frac{Z_{4}'}{10} \cdots$$

Then for each fixed $Z' \in \{ Z_{\nu}' \}$, we have $\nu Z_{\nu} = Z'$ for infinite number of $\nu \in \mathbb{N}$. It means that $Z_{\nu} = \frac{Z'}{\nu} \rightarrow 0$. Then, any holomorphic function in $\mathbb{C}^{n}$ that vanishes in the sequence $\{ Z_{\nu} \}$ has, in particular, a subsequence of zeros accumulating to the origin that belong to the complex line generated by any $Z'$. 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.

Thanks to Pietro Majer.

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Hello rla, where did you get this exercise from? Because my professor gave me the same and others that you have asked. Thanks – Miguemate Sep 1 '14 at 2:53
Hi @Miguemate, I got the exercicise from my professor, here in Brazil. I did not find it in any book. – rla Sep 5 '14 at 13:47

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