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.

  • $\begingroup$ $g(x,y) = |f(x+iy)|$ ? $\endgroup$ – reuns Sep 30 '16 at 10:10
  • 2
    $\begingroup$ The zero set of $f \colon X \to \mathbb{C}$ is the intersection of the zero sets of $\operatorname{Re} f$ and $\operatorname{Im} f$. $\endgroup$ – Daniel Fischer Sep 30 '16 at 10:26
  • $\begingroup$ 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? $\endgroup$ – Niki Sep 30 '16 at 11:16
  • $\begingroup$ The usual terminology for "zero-set" is "functionally closed." $\endgroup$ – DanielWainfleet Oct 1 '16 at 4:20

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.