I'm studying set theory from Kunen's book, and I came across this result:
Assume MA($\kappa$). Let $\mathcal{A},\mathcal{C}\subseteq \mathcal{P}(\omega)$, where $|\mathcal{A}|\le\kappa,|\mathcal{C}|\le\kappa$, and assume that for all $y \in \mathcal{C}$ and all finite $F\subseteq\mathcal{A}$, $|y\setminus\bigcup F| = \omega$; then there is $d\subseteq\omega$ such that $\forall x \in \mathcal{A} (|d \cap x|<\omega)$ and $\forall y \in \mathcal{C} (|d \cap y| = \omega)$
What this theorem says intuitively is that given two $\omega$-subset families of cardinality $\kappa$ there exists a subset $d \subseteq \omega$ that is almost disjoint from all elements of $\mathcal{A}$ and is not from any element of $\mathcal{C}$.
The book gives the following definition of almost disjoint family for an infinite cardinal $\lambda$
An a.d. family is an $\mathcal{A}\subseteq \mathcal{P}(\lambda)$ s.t. $\forall x \in \mathcal{A}(|x| = \lambda)$ and any two distinct elements of $\mathcal{A}$ are almost disjoint.
It seems to me that a consequence of the above theorem is that under MA($\kappa$) it doesn't exist an almost disjoint family for $\omega$ of cardinality $\kappa$ (with $\omega\le\kappa<2^{\omega}$).
If it existed then we could pick $\mathcal{A} = \mathcal{C}$ as the a.d. family and satisfy all hypotheses of the theorem, resulting in an evident contradiction.
But this seems rather strange for a number of reasons. What is the issue here? Thanks
Best Answer
You cannot pick $\mathcal{A}=\mathcal{C}$ in the first result: We pick any $y \in \mathcal{C}$ and define $\mathcal{F}=\{y\} \subseteq \mathcal{A}$ and see that $|y\setminus \bigcup \mathcal{F}|=|\emptyset|=0$ and not $\omega$. So the conditions are never met when $\mathcal{A}$ equals $\mathcal{C}$: these families should be sort of "independent": you cannot almost cover a set from one with finitely many from the other.