"The zero set $$Z(f) = \{ z \in U \vert f(z) = 0 \}$$ where $$U \subset \mathbb{C}^n \ \text{is open and the map} \ f: U \to \mathbb{C} \ \text{is holomorphic}$$ cannot be contained in a compact subset $K \subset U$ unless it is empty."
I have been trying to prove this statement by using Hartogs' extension theorem, but I'm stuck. Any suggestions?
Best Answer
Suppose toward a contradiction that $Z(f) \subset K$ for a non-empty compact set $K \subset U$. Take a neighborhood $V \subset U$ of $K$ such that $V \setminus K$ is connected. Then the function $1/f: V \setminus K \rightarrow \mathbb C$ is holomorphic and can thus, by the extension theorem, be extended to all of $V$. Now $f$ agrees with this extension on $V$, so since $V$ is open we have that they agree on all of $U$. But this is a contradiction since dividing by something cannot give zero.