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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 $\mathcal I^*(\mathcal A)=P(\omega)\setminus \mathcal I(\mathcal A)$. Question is: Suppose $X \in \mathcal I^*(\mathcal A)$. Is there a mad family $\mathcal M\supset \mathcal A$ such that $X \in \mathcal I^*(\mathcal M)$?

I tried to apply Zorn's lemma to define such an $\mathcal M$ but I think it didn't work.

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    Let $\mathcal{F}$ be an ad family containing $\mathcal{A}$ which is maximal among all ad families $\mathcal{G} \supseteq \mathcal{A}$ satisfying: No finite subcollection of $\mathcal{G}$ covers $X$ (mod finite). If $\mathcal{F}$ is not mad then there is some infinite $Y$ which is almost disjoint with every member of $\mathcal{F}$ and some $\{Y_k : k < n\} \subseteq \mathcal{F}$ covers $X \setminus Y$ (mod finite). Let $Z$ be an infinite subset of $X \cap Y$ such that $X \setminus Z$ cannot be covered by finitely many members of $\mathcal{F}$ (mod finite): Contradiction. – 2017-02-07

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