It is well-known that
In a normal topological space, every closed $G_\delta$-set is a zero-set.
The proof goes as follows:
Suppose $A$ is a closed $G_\delta$-set in $X$, a normal topological space, i.e., $A = \bigcap_{i=1}^\infty G_i$ where $\{G_i\}$ is a countable family of open sets in $X$. For each $i$, there is $f_i: X\to [0,1]$ such that $f_i = 1$ on $G_i^c$ and $f_i = 0$ on $A$. Consider $$f(x) = \sum_{i=1}^\infty \frac{f_i(x)}{2^i}$$ for all $x\in X$. Clearly, $f$ is continuous, since $\sum_{i=1}^\infty \frac{f_i(x)}{2^i}$ converges uniformly. Observe that $f = 0$ on $A$, and for $x\notin A$, $x\in G_i^c$ for some $i$, so that $f_i(x)\ne 0$, implying $f(x)\ne 0$. Therefore, $A = Z(f)$.
Question: Is the converse true, i.e., if every closed $G_\delta$-set in a topological space $X$ is a zero-set of some continuous real-valued function, is $X$ a normal topological space?
Thank you for any insights you may have to offer!
Notation and Terminologies:
- We say that $Z$ is a zero-set in $X$ if there exists a continuous function $f: X\to \Bbb R$ such that $Z = \{x \in X: f(x) = 0\}$.
Consider the Tikhonov plane $X=(\omega_1+1)\times(\omega+1)\setminus\{(\omega_1,\omega)\}$. Then $X$ is not normal, but any closed $G_{\delta}$ is zero-set.