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Let $\newcommand{\o}{[\omega]^\omega}\o$ denote the collection of infinite subsets of the set of nonnegative integers $\omega$. We say ${\cal A}\subseteq \o$ is almost disjoint if $A\cap B$ is finite whenever $A\neq B\in {\cal A}$.

Let $\frak A$ denote the collection of all almost disjoint families. Zorn's Lemma easily shows that every element of $\frak A$ is contained in a maximal element of $\frak A$ with respect to $\subseteq$. These maximal elements are referred to as maximal almost disjoint (MAD) families.

Question. If ${\cal A}\subseteq \o$ is a MAD family, is it true that there is $M\in \omega$ such that for every $n\in\omega$ with $n>M$ there are $A,B\in{\cal A}$ such that $|A\cap B| = n$?

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No. Trivially, any finite MAD family is a counterexample. To get an infinite counterexample, let $\{X_n:n\in\omega\}$ be a partition of $\omega$ into finite sets $X_n$ with $|X_n|=n!$. Let $\mathcal A$ be any infinite MAD family and let $\mathcal A'=\{A':A\in\mathcal A\}$ where $A'=\bigcup_{n\in A}X_n$. Then $\mathcal A'$ is an infinite MAD family and $\{|A\cap B|:A,B\in\mathcal A,\,A\ne B\}$ has density zero.

P.S. A commenter has asked why $\mathcal A'$ is maximal. I have to show that any infinite set $B\subseteq\omega$ has infinite intersection with some element of $\mathcal A'$. Let $C=\{n\in\omega:B_n\cap X_n\ne\varnothing\}$. Since each $X_n$ is finite, $C$ is infinite. Since $\mathcal A$ is MAD, there is a set $A\in\mathcal A$ such that $C\cap A$ is infinite. Then $A'\in\mathcal A'$, and $B\cap A'$ is infinite since there are infinitely many $n$ such that $B_n\cap X_n\ne\varnothing$ and $X_n\subseteq A'$.

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    $\begingroup$ If $\mathcal{A}$ is an uncountable almost disjoint family what can we say about $\{|A\cap B|:A,B\in \mathcal{A},A\neq B\}$? Must it be infinite? $\endgroup$ Commented Oct 8, 2024 at 0:27
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    $\begingroup$ @n901 Yes, it must be infinite. If $|A\cap B|\lt n$ whenever $A,B\in\mathcal A$, $A\ne B$, then each $n$-element subset of $\omega$ is contained in at most one element of $\mathcal A$, so $\mathcal A$ is countable. $\endgroup$ Commented Oct 8, 2024 at 3:56
  • $\begingroup$ @bof I don't see why $\mathcal{A}'$ is maximal. $\endgroup$ Commented Oct 8, 2024 at 19:36
  • $\begingroup$ @bof Thank you! $\endgroup$ Commented Oct 8, 2024 at 21:48

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