# Is any countable set in a bounded domain of $\mathbb{C}$ an analytic set

complex-analysiscomplex-geometryseveral-complex-variables

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.

(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 Theory by Remmert.