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I am having trouble on understanding Michael Hrusak's proof on "Recent Progress in General Topology III", page 605. The theorem states the following:

Given an AD family $\mathcal A$ and a decreasing sequence $(X_n: n \in \omega)$ of elements of $\mathcal I^*(\mathcal A)$, there is $X \in I^*(\mathcal A)$ such that $X\subset^*X_n$ for all $n$.

I understood the whole proof but one thing: In the first line, the author states that it suffices to suppose that $\mathcal A$ is mad, since we can extend it to a mad family so that each $X_n$ remain positive. I tried to apply Zorn's lemma but I don't know how to prove that the maximal element given by Zorn's lemma is a mad family.

The definitions are the same as in this other question I made today, and to be honest, the question is very similar. I thought that I could try to make an argument similar to hot_queen's comment but I think it didn't work.:

An almost disjoint family is an infinite collection $\mathcal A$ of infinite subsets of $\omega$ such that for all $A, B \in \mathcal A$, the intersection $A \cap B$ is finite. A mad family is a maximal almost disjoint family.

The free ideal generated by $\mathcal A$ is the collection $\mathcal I(\mathcal A)=\{X \subset \omega: \exists B_1,\dots, B_n \in \mathcal A, m \in \omega(X \subset B_1\cup \dots \cup B_n\cup m)\}$. It's easy to see that $\mathcal I(\mathcal A)$ is a proper ideal on $\omega$, that is: $\emptyset \in \mathcal I(\mathcal A)$, $\omega \notin \mathcal I(\mathcal A)$ and $\mathcal I(\mathcal A)$ is closed under finite uniions and is closed downwards. This ideal is also free, that is, all finite subsets of $\omega$ belongs to it.

The positive subsets of $\mathcal A$ are defined as the elements of $\mathcal I^*(\mathcal A)=P(\omega)\setminus \mathcal I(\mathcal A)$.

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    Let $\mathcal{F} \supseteq \mathcal{A}$ be an ad family which is maximal w.r.t. not covering any member (mod finite) of $\{X_n : n < \omega\}$ via finitely many sets. Suppose $\mathcal{F}$ is not mad and fix $Z_{-1}$ witnessing this. Choose a decreasing sequence of infinite sets $\{Z_n : n < \omega\}$ such that for each $n$, if $X_n \setminus Z_{n-1}$ can be covered by finitely many members of $\mathcal{F}$, then $Z_n \subseteq X_n \cap Z_{n-1}$ and $X \setminus Z_n$ can't be covered by finitely many sets in $\mathcal{F}$. Let $Z \subset^{\star} Z_n$ for all $n$: Contradiction. – 2017-02-08

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