A question on a fat $\Psi$ space Let $\kappa > \omega$ an infinite cardinal and $\mathcal E \subset [\kappa]^\omega$ is a maximal almost disjoint family, where $[\kappa]^\omega = \{ A \subset \kappa: |A| = \omega \}$. Let $\Psi(\mathcal E)$ denote the topological space whose point-set is $\kappa \cup \mathcal E$, with the topology generated by isolating each $\alpha \in \kappa$, and the basic nbhds about $E \in \mathcal E$ are all sets of the form $\{E\}\cup (E\setminus F)$, where $F \in [E]^{< \omega}$. This we can called fat $\Psi$ space.
My question is this: does such space always have a $G_\delta$ diagonal for any $\kappa$?
Thanks ahead.
 A: $X=\Psi(\mathscr{E})$ has a $G_\delta$ diagonal iff it has a family $\{\mathscr{U}_n:n\in\omega\}$ of open covers such that 
$$\bigcap_{n\in\omega}\operatorname{st}(x,\mathscr{U}_n)=\{x\}\tag{1}$$
for each $x\in X$. Clearly each $\mathscr{U}_n$ may be replaced by a refinement, so we may assume that $\mathscr{U}_n$ has the following form.

For each $E\in\mathscr{E}$ let $E=\{\xi^E(n):n\in\omega\}$ be an enumeration of $E$. For $n\in\omega$ and $E\in\mathscr{E}$ let $B_n(E)=\{E\}\cup\left\{\xi^E(k):k\ge n\right\}$. Then for each $E\in\mathscr{E}$ there is an increasing function $\varphi_E:\omega\to\omega$ such that $$\mathscr{U}_n=\left\{B_{\varphi_E(n)}(E):E\in\mathscr{E}\right\}\cup\big\{\{\xi\}:\xi<\kappa\big\}$$ for each $n\in\omega$.

For $E\in\mathscr{E}$ and $n\in\omega$ we have $\operatorname{st}(E,\mathscr{U}_n)=B_{\varphi_E(n)}(E)$, so $(1)$ holds for each $x\in\mathscr{E}$. If $\eta<\kappa$ and $E\in\mathscr{E}$, then $E\in\operatorname{st}(\eta,\mathscr{U}_n)$ iff $\eta=\xi^E(k)$ for some $k\ge\varphi_E(n)$, so for each $E\in\mathscr{E}$ there is an $n\in\omega$ such that $E\notin\operatorname{st}(\eta,\mathscr{U}_n)$, and therefore 
$$\bigcap_{n\in\omega}\operatorname{st}(\eta,\mathscr{U}_n)\subseteq\kappa\;.$$
Suppose that $\bigcap_{n\in\omega}\operatorname{st}(\eta,\mathscr{U}_n)=\{\eta\}$. Then for each $\xi\in\kappa\setminus\{\eta\}$ there is an $n\in\omega$ such that $\xi\notin\operatorname{st}(\eta,\mathscr{U}_n)$, i.e., there is no $E\in\mathscr{E}$ such that $\{\xi,\eta\}\subseteq B_{\varphi_E(n)}(E)$. Let $$m\big(\{\xi,\eta\}\big)=\min\left\{n\in\omega:\forall E\in\mathscr{E}\big(\{\xi,\eta\}\nsubseteq B_{\varphi_E(n)}(E)\big)\right\}\;;$$ then $m:[\kappa]^2\to\omega$.
Now suppose that $\kappa=\left(2^\omega\right)^+$. The $\lambda=\omega$ case of the partition relation $(2^\lambda)^+\to(\lambda^+)^2_\lambda$ says that there are an $n\in\omega$ and an uncountable $S\subseteq\kappa$ such that $m\big(\{\xi,\eta\}\big)=n$ for all $\{\xi,\eta\}\in[S]^2$. $\mathscr{E}$ is a MAD family, so there is an $E\in\mathscr{E}$ such that $|E\cap S|=\omega$. But then there are $k,\ell\in\omega$ such that $\xi^E(k),\xi^E(\ell)\in S$ and $k,\ell\ge\varphi_E(n)$, contradicting the definition of $m$. Thus, for $\kappa=\left(2^\omega\right)^+$ the space $X$ does not have a $G_\delta$-diagonal.
I’ve not been able to settle the question for $\omega_1\le\kappa\le 2^\omega$.
