Proving a set is closed; Metric Spaces

Question

Let $A=\{(w,z)\in \mathbb C^{2}: |w|+|z|+Im(z) \leq 1\}$

Prove that $A$ is closed in $\mathbb C^{2}$

My ideas: We have just started metric spaces so I am aware that I need to prove that $A^C$ is open in $\mathbb C^{2}$, however in the all the examples before we have only had to prove a set being closed in $\mathbb R$ or $\mathbb C$. Now we have two dimensions, so I'd suggest starting:

Let $(x,y) \in A^{C}$ which in turn means that $|x|+|y|+Im(y) > 1$ . In order to prove that it is an open set I need to find a "radius" $r$ $\in \mathbb C^{2}$ such that for $(x^{'},y^{'}) \in B_{r}((x,y))$, it follows that $(x^{'},y^{'}) \in A^{C}$.

Is the basic outline of my proof correct here for two dimensions? How do I solve this?

Answer 1

What you have can be treated in a more general framework.

Assume you have a continuous function $f$ on $\mathbb{C}$ (or on some other nice space), and consider the set $$ A = \{z \in \mathbb{C}: \ |f(z)| \leq a \}, $$ where $a > 0$ if fixed. Then, it's straightforward to see that $A$ is closed, in view of continuity of $f$. Now, chose $f$ as you need, and the claim follows.

Answer 2

Jut use the fact that if you define $f\colon\mathbb{C}^2\longrightarrow\mathbb R$ by $f(z,w)=|z|+|w|+\operatorname{Im}z$, then $f$ is continuous and $A=f^{-1}\bigl((-\infty,0]\bigr)$.