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Let $\newcommand{\oo}{[\omega]^\omega}\oo$ denote the collection of all infinite subsets of the set of nonnegative integers $\omega$. We say that $\newcommand{\ss}{{\cal S}}\S\subseteq \oo$ hasproperty $\newcommand{\B}{\mathbf{B}}\B$ if there is $A_0\subseteq \omega$ such that $$(A_0\cap S) \neq \emptyset \neq (S\setminus A_0)$$ for all $S\in \ss$.

Moreover we call $\ss$ almost disjoint if $S\cap S'$ is finite whenever $S\neq S'\in \ss$.

Zorn's Lemma shows that every almost disjoint family is contained in a maximal almost disjoint (MAD) family, and a diagonal argument establishes that every MAD family is uncountable.

Question. Is there an almost disjoint family $\ss$ with property $\B$ and a MAD family $\cal M$ such that $|\ss|=|{\cal M}|$?

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One of the standard examples of an almost disjoint family of cardinality $\mathfrak c$ is the set of paths through the complete binary tree $2^{<\omega}$ (identified with $\omega$ via your favorite bijection). Let $\mathcal S$ be this family. It has property B, witnessed by the set $A_0$ of nodes of even height, because every path contains (infinitely many) nodes of even height and (infinitely many) of odd height. Any extension of $\mathcal S$ to a maximal almost disjoint family has cardinality $\mathfrak c=|\mathcal S|$.

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    $\begingroup$ Alternatively, if $\{A_i:i\in I\}$ is a maximal almost disjoint family of subsets of $\{2n:n\in\omega\}$ and $\{B_i:i\in I\}$ is a maximal almost disjoint family of subsets of $\{2n+1:n\in\omega\}$, then $\{A_i\cup B_i:i\in I\}$ is a maximal almost disjoint family of subsets of $\omega$ and has property B. $\endgroup$ Commented Jul 17, 2024 at 4:32

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