Let $X$ be a connected complex manifold with dimension $n \geq 1$ and $f:X \rightarrow \mathbb{C}$ be a holomorphic function. Suppose that on a coordinate open set $U$, the function takes infinitely many zero that accumulates. If $n = 1$, this implies that $f = 0$ on the whole $X$.
If $n = 2$, this is not the case as one can consider $\mathbb{C}^{2} \rightarrow \mathbb{C}$ given by $(z, w) \mapsto z$. Now suppose that $n = 2$ and say we have parallel lines in a coordinate open set $U$ that "accumulate" to a line. Then I think this necessarily implies that $f = 0$ on the whole $X$.
Question 1. Can we say more generally about this phenomenon? If $f$ vanishes on "accumulating" hyperplanes, does it necessarily imply that $f = 0$ on the whole $X$?
̶̶Q̶u̶e̶s̶t̶i̶o̶n̶ ̶2̶.̶̶̶ ̶C̶o̶n̶s̶i̶d̶e̶r̶ ̶t̶h̶e̶ ̶c̶a̶s̶e̶ ̶$̶n̶ ̶=̶ ̶2̶$̶ ̶(̶o̶r̶ ̶h̶i̶g̶h̶e̶r̶)̶.̶ ̶I̶f̶ ̶t̶h̶e̶r̶e̶ ̶i̶s̶ ̶a̶ ̶c̶o̶o̶r̶d̶i̶n̶a̶t̶e̶ ̶o̶p̶e̶n̶ ̶s̶e̶t̶ ̶$̶U̶$̶ ̶s̶u̶c̶h̶ ̶t̶h̶a̶t̶ ̶$̶Z̶(̶f̶)̶ ̶=̶ ̶f̶^̶{̶-̶1̶}̶(̶0̶)̶$̶ ̶i̶s̶ ̶d̶e̶n̶s̶e̶ ̶i̶n̶ ̶$̶U̶$̶,̶ ̶i̶s̶ ̶$̶f̶ ̶=̶ ̶0̶$̶ ̶o̶n̶ ̶t̶h̶e̶ ̶w̶h̶o̶l̶e̶ ̶$̶X̶$̶?̶ ̶
I will be happy to know about that case $X = \mathbb{C}^{n}$. Notice that for compact $X$, this is a trivial problem.
The set $Y=\{f=0\}\subset X$ is analytic, hence it has countable collection of irreducible analytic components. Furthermore, those components form a locally finite familly (this is basic Complex Analytic Sets theory, e.g. Gunning-Rossi, Narasimhan...). Now any analytic set $H\subset Y$ of codimension $1$ in $X$ would be an irreducible component of $Y$. Hence any collection of such $H$'s inside $Y$ must form a locally finite family (whose union is closed), and so they cannot acummulate anywhere.
Conversely, if you pick any locally finite family of hyperplanes $H_i$ in $\mathbb C^n$, say, then there is an anaytic function $f:\mathbb C^n\to\mathbb C$ whith zero-set is the union of the $H_i$'s.