Is any countable set in a bounded domain of $\mathbb{C}$ an analytic set? 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.
 A: (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.
A: If $A\subset \Omega$ is a set without accumulation points and $n \colon A \to \mathbb{N}$ is a function, then there exists a holomorphic function on $\Omega$ such that for every $a \in A$, $f$ has a zero of multiplicity $n_a$ at $a$.
Sketch of proof: consider an open cover $\mathcal{U}$ of $\Omega$ such that every $U \in \mathcal {U}$ contains at most one element of $A$.  Now, on every $U$ consider a function $\phi_U$ having a zero of prescribed order $n_a$ at $a$, it it contains an element $a$ of $A$. Now, we want to piece together these $\phi_U$ to get a global function $\phi$ on $\Omega$. Suppose that we can divide $\phi_U$ by some $c_U$ a holomorphic function on $U$ without zeroes, such that $\phi_U /c_U = \phi_V /c_V$ on $U\cap V$. We would now get a global function on $\Omega$. Now, is this possible? Note that this is equivalent to
$$\frac{\phi_U}{\phi_V} = \frac{c_U}{c_V}$$
This is possible, at least on a refinement of the cover. It has to do with sheaf cohomology. Basically we want to show that $H^1(\Omega, \mathcal{O}^*_{\Omega}) = 0$.
Sketch of proof the above fact: we have an exact sequence of sheaves
$$0\to \mathbb{Z}\to  \mathcal{O}_{\Omega} \to \mathcal{O}^*_{\Omega} \to 1$$
Now pass to the long exact sequence in cohomology and use that $H^1(\Omega, \mathcal{O}_{\Omega} ) = 0$ ( another fundamental fact)  and $H^2(\Omega, \mathbb{Z}) = 0$.
Note: in Rudin, Real and Complex Analysis, there exists another simpler proof.
