# Question about compactness of the zero set of an analytic function of several variables

Hartogs's Theorem Let $f$ be a holomorphic function on a set $G \setminus K$, where $G$ is an open subset of $\mathbb{C}^n$ ($n \ge 2$) and $K$ is a compact subset of $G$. If the complement $G\setminus K$ is connected, then $f$ can be extended to a unique to a unique holomorphic function on $G$.

This theorem can be used to show the following result about the zeros of analytic functions of several variables.

Suppose that $f$ is an analytic function on some open set $U$ and that $f$ is not identically zero on $U \subset \mathbb{C}^n$ with $n \ge 2$. Then, the set of zeros of $f$ (i.e. $\Lambda(f)=\{ z: f(z)=0\}$) is not compact.

Since $\Lambda(f)$ is not compact we can have the following three possibilities:

1. $\Lambda(f)$ is closed but is not bound
2. $\Lambda(f)$ is not closed but bounded
3. $\Lambda(f)$ is not closed and not bounded

My question is the following: Can we come up with examples of $f$ for each of the three cases?

Here is an example of the function that satisfies the first case. Let $f_1(z_1,z_2)=z_1 \cdot z_2$ then \begin{align} \Lambda(f_1)= \{ (z_1,z_2) : z_1=0 \} \cup \{ (z_1,z_2) : z_2=0 \}. \end{align} where $\Lambda(f_1)$ is closed but not bounded.

The cases 2 and 3 cannot exist by continuity. Since $f\colon U \rightarrow \mathbb{C}$ is holomorphic, it is continuous. Hence $\{0\}$ being a closed subset implies that $f^{-1}(0) = \Lambda(f)$ is closed in $U$.
• Closed and unbounded in $U$. – Alan Muniz Aug 15 '18 at 23:25
• Can you remind me what unbounded in $U$ means? There exists no ball $B$ that contains $\Lambda(f)$ such that $B \subset U$? – Boby Aug 16 '18 at 0:01
• It means that the set $\Lambda(f)$ is not contained in any compact subset of $U$. I want to emphasise that because $U$ may be bounded in $\mathbb{C}^n$. For example, let $U = B(0,1)$ the unitary ball centered at the origin. Then being unbounded in $U$ means that the closure of $\Lambda(f)$ needs to meet the boundary of $U$. However $\Lambda(f)$ is bounded as a subset of $\mathbb{C}^n$. – Alan Muniz Aug 16 '18 at 0:09