When a $\Psi$-Space is metrizable

Question

Let $I$ a set such that $I\cap \omega = \emptyset$ and $|I|\leq 2^\omega$. Given an almost disjoint family $\mathcal A = \{A_i: i\in I \} \subset [\omega]^\omega$, the space $\Psi( \mathcal A)$ is defined by $\Psi( \mathcal A) \doteq \omega \cup I$ and the topology defined as follows:

1. $\{n\}$ is an open set, for all $n\in \omega$;
2. For all $i\in I$, $B\in cof(A_i)$, $\{i\}\cup B$ is open.

It is straight foward to check that $\Psi(\mathcal A)$ is Hausdorff, locally compact, first countable, zero-dimensional and separable.I am trying to prove the following:

$\Psi(\mathcal A)$ is a metrizable space $\iff$ $I$ is countable

I've don the first part as follows:

$(\implies)$ If $\Psi(\mathcal A)$ is metrizable, since it is separable, $\Psi(\mathcal A)$ is second countable as well. Then it follows that the basis defined above contains a countable basis $\mathcal C$. Hence, $I \subset \mathcal C$ is countable.

I couldn't prove the other implication and I need a guide to do this.

Answer 1

Every point of $\Psi(\mathcal A)$ has a countable basis. If $\mathcal A$ is countable, so is $\Psi(\mathcal A)$, therefore $\Psi(\mathcal A)$ is second countable. Since it is Hausdorff and locally compact, it is regular, and every Hausdorff regular second countable space is metrizable.

Answer 2

The set $I$ is closed and discrete (as a subspace), so $w(\Psi(\mathcal{A})) \ge w(I) = |I|$, while $d(\Psi(\mathcal{A})) = \omega$ and $d(X) = w(X)$ for metrisable spaces. This shows metrisable implies $I$ countable, and the reverse is just Urysohn's metrisation theorem, as noted by @ViniciusRodriguez.