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Let $[\omega]^\omega$ denote the collection of infinite subsets of $\omega$. We say $a,b\in [\omega]^\omega$ are almost disjoint if $a\cap b$ is finite. A subset $A\subseteq [\omega]^\omega$ is said to be an almost disjoint (AD) family if $a, b$ are almost disjoint for all $a \neq b \in A$. Zorn's Lemma shows that every almost disjoint family is contained in a maximal almost disjoint (MAD) family (maximal with respect to $\subseteq$).

If ${\cal A}$ is an AD family, we say that ${\cal A}$ is of true cardinality ${\frak c}$ if for every $X\in [\omega]^\omega$ the set $\{A\in {\cal A}: |A\cap X| = \aleph_0\}$ is either finite or of size ${\frak c}$.

Question. Let ${\cal A}$ be an AD family, and let ${\cal M}$ be a MAD family such that ${\cal A}\subseteq {\cal M}$. If ${\cal A}$ is of true cardinality ${\frak c}$, is ${\cal M}$ necessarily of true cardinality ${\frak c}$?

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No.

Suppose that there is a MAD family of size $\aleph_1$ and $\sf CH$ fails. Let $\mathcal E=\{E_\alpha\mid\alpha<\omega_1\}$ be a MAD family on the even integers.

Suppose now that $\cal A$, your AD family, happened to be an almost disjoint family only on the odd integers. It can happen, who knows. Extend it to a MAD family on the odd integers, and take its union with $\cal E$. Call this $\cal M$.

Now, if $X$ is any infinite set of integers, its intersection with either the even or the odds is infinite, and by the maximality, it must meet at least one of the members of $\cal E$ or the extension of $\cal A$ on an infinite set. So $\cal M$ is indeed MAD.

However, taking $X$ to be the even integers is a counterexample for true cardinality $\frak c$. In fact this shows that if the statement is correct, then $\frak a=c$. The obvious follow is, does $\frak a=c$ implies that every AD family with true cardinality $\frak c$ extends to a maximal such family.

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    $\begingroup$ Nice! Regarding your last sentence, the answer to Dominic's question is yes if $\mathfrak{a} = \mathfrak{c}$. This is because if $\mathcal M$ is a MAD family on $\mathbb N$ and $A \subseteq \mathbb N$, then either $A$ only meets finitely many members of $\mathcal M$ in an infinite set, or else the restriction of $\mathcal M$ to $A$ is a MAD family on $A$, hence of cardinality $\mathfrak{c}$. So it seems the answer to Dominic's question is yes if and only if $\mathfrak{a} = \mathfrak{c}$. $\endgroup$
    – Will Brian
    May 25, 2022 at 22:08
  • $\begingroup$ Thanks @Will, that makes sense! $\endgroup$
    – Asaf Karagila
    May 25, 2022 at 22:14

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