Let $X$ be a complex manifold. A closed subset $Y$ of $X$ is called an **analytic set** if for each $y \in Y$, there is an open neighborhood $U$ of $y$ so that $Y \cap U$ is the zero set of finitely many holomorphic functions $f_{1}, \cdots, f_{k} \in \mathscr{O}(U)$.

Let $\Omega$ be a bounded domain in the complex plane $\mathbb{C}$, and $A$ a countable set of $\Omega$ admitting no accumulation point.

Ques: Is $A$ necessarily an analytic set in $\Omega$?

As well known, the Weierstrass factorization theorem tell us that any countable

set of $\mathbb{C}$ tending to infinity is an analytic set.

## Best Answer

(Note that the condition that $\Omega$ is bounded is not used anywhere)

Here's a rather silly and trivial solution: If $A$ has no accumulation point, then it is discrete, so for every point $p \in A$, we can find an open neighborhood $U$ such that $A \cap U =\{p\}$. But clearly $\{p\}$ is the zero set of $z-p \in \mathscr O(U)$.

You might say that this is unsatisfying and I'd agree. The reason is that for $\Bbb C$, Weierstraß factorization tells you much more: not only is a discrete subset of $\Bbb C$ locally the zero set of holomorphic functions, it's actually the zero set of a globally defined holomorphic function!

So one might ask the same question about $A \subset \Omega$. And indeed, it is true that there some $f \in \mathscr O(U)$ such that $A$ is the zero set of $f$. One can show this via sheaf cohomology, as orangeskid explains in his answer. But it is also possible to do a more general version of Weierstraß factorization, along the same lines as the version for entire functions. For this, see chapter 4 of

Classical Topics in Complex Function Theoryby Remmert.