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 family if $a, b$ are almost disjoint for all $a \neq b \in A$. A standard application of Zorn's Lemma shows that every almost disjoint family is contained in a maximal almost disjoint family (MAD family) (maximal with respect to $\subseteq$).
A diagonalization argument shows that all infinite MAD families have uncountable cardinality. By ${\frak a}$ we denote the minimum cardinality that a MAD family can have. It is consistent that ${\frak a} < {\frak c} = 2^{\aleph_0}$.
Question. Is it consistent that
- ${\frak a} < {\frak c}$,
- there is a MAD family $A\subseteq [\omega]^\omega$ with $|A| = {\frak c}$, and
- there is a cardinal ${\frak g}$ with ${\frak a} \in {\frak g} \in {\frak c}$ such that there is no MAD family with cardinality ${\frak g}$?
Best Answer
Yes, this is consistent.
Suppose we force to add $\kappa$ mutually generic Cohen reals to a model of $\mathsf{CH}$, where $\kappa$ is some cardinal with uncountable cofinality. In the extension, there are MAD families of cardinality $\aleph_1$ and cardinality $\kappa = \mathfrak{c}$, but there are no MAD families of any intermediate cardinality.
The proof of this is a basic instance of what's known as an "isomorphism of names" argument. I think this result appears as an exercise in Kunen's 1980 book, although I think it's originally due to Arnie Miller. The argument is greatly extended in
They prove that the set of all cardinals $\kappa$ such that there is a MAD family of size $\kappa$ can be almost completely arbitrary. For example, given any $A \subseteq \omega \setminus \{0\}$, there is a forcing extension in which $A = \{ n < \omega :\, \text{there is a MAD family of size } \aleph_n\}$.