About Mrowka spaces. I'm studying some properties of Mrowka spaces and I'm stuck proving the next things. First, consider $\mathcal{M}$ a maximal almost disjoint family such that $|\mathcal{M}|>\aleph_0$. Let be $\psi=\omega\cup\mathcal{M}$ and let $$B=\{\{n\}\mid n\in\omega \}\cup\{\{M\}\cup S\mid M\in\mathcal{M} \ \text{and $S$ is a cofinite subset of $M$ } \}$$an open basis for $\psi$. I want to prove 


*

*$\mathcal{M}$ is a zero set of $\psi$, i.e., there exist $f\in C(\psi)$ such that $f^{-1}[\{0\}]=\mathcal{M}$. 

*Each subset of $\psi$ is $G_\delta$.


For the first one consider the next function.
\begin{equation}
f(x)=\left\{
\begin{array}{ccc}
\frac{1}{x} & \text{if} & x\in\omega \\\\
0 & \text{if} & x\in\mathcal{M}
\end{array}
\right.
\end{equation}
We need to prove that $f\in C(\psi)$. Let $U$ be an open set of $\mathbb{R}$. If $0\in U$ then $U$ contains infinitely many points of the form $1/n$ and only finitely many are outside $U$. Thus, $f^{-1}[U]=\mathcal{M}\cup\{n\in\omega\mid 1/n\in U \}$. This set is open because if $x\in f^{-1}[U]$ then we have two cases: if $1/x\in U$ then $\{ x\}$ is an open neighborhood of $x$ contained in $f^{-1}[U]$. If $x\in \mathcal{M}$ then, by the fact that $\omega\setminus \{n\in\omega\mid 1/n\in U \}$ is finite, there exist $S\subseteq \{n\in\omega\mid 1/n\in U \}$ such that $\{ x\}\cup S$ is an open neighborhood of $x$ contained in $f^{-1}[U]$. 
If $0\notin U$ then $f^{-1}[U]\subseteq\omega$ and by the fact that every element of $\omega$ is open in $\psi$ then $f^{-1}[U]$ is open. It is correct?
For the second, I don't know how to proceed. Any hint? I really appreciate any help you can provide me.  
 A: Your $f$ is fine (maybe better use $\frac{1}{x+1}$, as $0 \in \omega$ too..). 
And if $O \subseteq \Bbb R$ does not contain $0$, $f^{-1}[O] \subseteq \omega$ so is open, and if $0 \in O$, cofinally many $f(n)$ are contained in $O$, say we miss only $f(n), n \in F$, where $F \subseteq \omega$ finite. Then for every $A \in \mathcal{M}$, $\{A\} \cup (A\setminus F) \subseteq f^{-1}[O]$ so all points of $\psi$ are interior points of $f^{-1}[O]$, and $f^{-1}[O]$ is open.
Any zero-set like $\mathcal{M}$ is a closed $G_\delta$. And $\mathcal{M}$ is a discrete subspace, so any $B \subseteq \mathcal{M}$ is open in $\mathcal{M}$, so also a $G_\delta$ in $\psi$ (as $(\psi\setminus B) \cap \mathcal{M}$, the intersection of a $G_\delta$ and an open set). It follows that any subset of $\psi$ is $G_\delta$: $A = (A \cap \mathcal{M}) \cup (A \cap \omega)$ is a union of a $G_\delta$ and an open set, so $G_\delta$. Or note that all subsets of $\mathcal{M}$ are closed in $\psi$ (closed in closed is closed) and adding countably many points from $\omega$ keeps it an $F_\sigma$. So all subsets are $F_\sigma$ too.
Note that if $\mathcal{M}$ has size $\kappa$ such that $2^\kappa > 2^\omega$ then Jones' lemma implies $\psi$ is not normal. It is Tychonoff, being Hausdorff and locally compact, or because it's zero-dimensional (it has a clopen base).
