# (Maximal) almost disjoint families of true cardinality ${\frak c}$

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 $${\frak c}$$?

## 1 Answer

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.

• 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 $\mathfrak{a} = \mathfrak{c}$. May 25, 2022 at 22:08
• Thanks @Will, that makes sense! May 25, 2022 at 22:14