# When is every closed set (/singleton) the zero set of complex valued functions?

We know some things for real valued functions. Let $X$ be a topological space, and call $Z \subset X$ a zero set if there exists a continuous function $f:X \rightarrow \mathbb{R}$ s.t. $f^{-1}(0) = Z$.

We know:

1. Every closed is a zero set iff $X$ is perfectly normal

2. Every closed is an intersection of zero sets iff $X$ is Tychonoff

3. If $X$ is compact Hausdorff, then every singleton is a zero-set iff $X$ is first countable.

So my question is: Do the same results hold true if we consider zero sets of complex valued functions $f:X \rightarrow \mathbb{C}$? Or what does change?

The motivation is, that for a compact Hausdorff space $X$, I would like to understand which closed sets are zero sets of complex continuous functions $f:X \rightarrow \mathbb{C}$. Or under what hypotheses every closed (/singleton) is such a zero set.

• $g(x,y) = |f(x+iy)|$ ? – reuns Sep 30 '16 at 10:10
• The zero set of $f \colon X \to \mathbb{C}$ is the intersection of the zero sets of $\operatorname{Re} f$ and $\operatorname{Im} f$. – Daniel Fischer Sep 30 '16 at 10:26
• True, but does this fully answer the question? What is the condition on $X$ such that every closed is the zero set of a complex function? This only shows that if every closed is a complex zero set, then every closed is the intersection of two real zero sets and hence Tychonoff. But what about a converse? – Niki Sep 30 '16 at 11:16
• The usual terminology for "zero-set" is "functionally closed." – DanielWainfleet Oct 1 '16 at 4:20

## 1 Answer

Nothing changes: for any space $X$, the zero-sets of continuous complex-valued functions on $X$ are precisely the zero-sets of continuous real-valued functions on $X$. Call the former complex zero-sets and the latter real zero-sets for short.

If $Z\subseteq X$ is a real zero-set, then certainly $Z$ is a complex zero-set. Now suppose that $Z$ is a complex zero-set, i.e., that $Z=f^{-1}[\{0\}]$ for some continuous $f:X\to\Bbb C$. Let

$$g:X\to\Bbb R:x\mapsto\big(\operatorname{Re}f(x)\big)^2+\big(\operatorname{Im}f(x)\big)^2\;.$$

then $g$ is continuous, and $g(x)=0$ iff $\operatorname{Re}f(x)=\operatorname{Im}f(x)=0$ iff $f(x)=0$, so $Z$ is a real zero-set.