Is the space $\psi$ (described in problem **5I** of *L. Gillman and M. Jerison, Rings of continuous functions, Springer Verlag, 1976*) a F-Z-space (i.e, space with $cl(X-Z(f))$ is a zero set for every $f$ in $C(X)$)? 


$\textbf{Clarification}$ A collection $\mathcal{A}$ of infinite subsets of $\mathbb{N}$ is said to be an almost disjoint family if $R\cap S$ is finite whenever $R,S\in\mathcal{A}$ and $R\neq S$.


Let $\mathcal{E}$ be a maximal almost disjoint family of subsets of $\mathbb{N}$. Let $D=(w\_{E})\_{E\in\mathcal{E}}$ be a set of new distinct points. Let $\psi=\mathbb{N}\cup D$. We shall give $\psi$ the topology as follows. The points in $\mathbb{N}$ are isolated. The neighborhood filter around a point $w\_{E}$ is generated by sets of the form $w\_{E}\cup R$ where $R\subseteq E$ and $E\setminus R$ is a finite set. In this space, $w\_{E}\cup E$ is the one-point-compactification of the space $E$.


The problem in the book states that the space $\psi$ is completely regular, pseudocompact, but not countably compact. In particular, $\psi$ is not realcompact and $\psi$ is not normal. Furthermore, every subspace of $\psi$ is a $G\_{\delta}$-set.



The space $\psi$
As I commented above, I think the answer is that a $\psi$-space need not be an FZ-space, and that a counterexample may be constructed from a Luzin gap. Here are the details, which did not fit into the comment. 


We first construct the MAD family, which will give the counterexample. Start by splitting ${\mathbb N}$ into two infinite sets $X\_0,X\_1$. Then let ${\mathcal A}\_0=\{A^0\_\alpha:\alpha<\omega\_1\}$ be an arbitrary AD family on $X\_0$ and ${\mathcal A\_1}=\{A^1\_\alpha:\alpha<\omega\_1\}$ a Luzin gap on $X\_1$ (i.e. an AD family which, in particular, has the property that no uncountable pair of subfamilies can be separated; here ${\mathcal B},{\mathcal C}\subseteq{\mathcal A}\_1$ can be separated if there is a set $S\subseteq X\_1$ which almost-contains all sets from $B$ and is almost disjoint from all sets in $C$.). 


We shall fuse these two families into a single new AD family which we will extend to a MAD family. Proceed as follows: for $\alpha<\omega\_1$ let
$A\_\alpha=A^0\_\alpha\cup A^1\_{\alpha+1}$ and let $\{B\_\alpha:\alpha<\omega\_1\}$ be the enumeration of $\{A^1\_\beta:\beta\in Lim(\omega\_1)\}$. Let ${\mathcal A}$ be an arbitrary MAD family extending the family 
$\{A\_\alpha,B\_\alpha:\alpha<\omega\_1\}$. 


**Proposition** The space $\psi({\mathcal A})$ is not an FZ-space.


*Note* In the following we identify the points of $\psi({\mathcal A})={\mathbb N}\cup D$ which are in $D$ with the elements of the MAD family ${\mathcal A}$.


*Proof*.
Define $f:\psi({\mathcal A})\to{\mathbb R}$ as follows: $f(n)=1/n$ for $n\in X\_0$ 
and $f(x)=0$ otherwise. It is clear that $f$ is continuous. Let $G=\psi(\mathcal{A})\setminus f^{-1}(0) = X\_0$. To show that $\psi({\mathcal A})$ is not an FZ-space it will be sufficient to show that $F=cl(G)$ is not a zero set. Notice first that for each $\alpha<\omega\_1$ we have $A\_\alpha\in F$ and $B\_\alpha\not\in F$. Suppose now that $g:\psi({\mathcal A})\to{\mathbb R}$ is a continuous function such that $g^{-1}(0)=F$. Since $B\_\alpha\not\in F$
for each $\alpha<\omega\_1$, we can fix $n<\omega$ such that ${\mathcal B}=\{B\_\alpha:g(B\_\alpha)>1/n\}\subseteq{\mathcal A}\_1$ is uncountable. Now let $S=\{k<\omega:g(k)>1/n\}$. Then $S$ almost contains all sets from ${\mathcal B}$ while, by continuity of $g$, being almost disjoint from every set in $\{A\_\alpha:\alpha<\omega\_1\}$ hence also from every set in ${\mathcal C}=\{A^1\_{\alpha+1}:\alpha<\omega\_1\}\subseteq{\mathcal A}\_1$. This contradicts the fact that ${\mathcal A}\_1$ is a Luzin gap, since $S$ separates the two uncountable subfamilies ${\mathcal B,C}\subseteq{\mathcal A}\_1$. $\square$


