The exercise is the following:
Let $\mathcal{B}\subseteq \mathcal{P}(\omega)$ be an almost disjoint family of size $\kappa$, where $\omega\le\kappa<2^\omega$. Let $\mathcal{A} \subseteq \mathcal{B}$, with $|\mathcal{A}|\le\omega$. Assuming MA($\kappa$), show that there is a $d\subseteq \omega$ such that $\forall x \in \mathcal{A}(|d\cap x|<\omega)$ and $\forall x \in \mathcal{B}\setminus\mathcal{A}(|x\setminus d| < \omega)$
A more or less direct consequence of MA($\kappa$) is a similar statement, where we drop the hypothesis $|\mathcal{A}|\le\omega$ and require just that $\forall x \in \mathcal{B}\setminus\mathcal{A}(|x\cap d| = \omega)$.
I tried to use this similar statement, but I don't get much further. In particular I have a problem in using in a smart way both the hypotheses on the countability of $\mathcal{A}$ and MA. Being more specific, the coutability of $\mathcal{A}$ induces me to first use MA and then "adjust" what I have found by adding\subtracting elements to\from $d$ by iterating over $\mathcal{A}$ (for example by a diagonal argument). Could you give me a hint? Thanks
EDIT: I'm beginning to wonder whether it is solvable..
Best Answer
It seems like that similar statement is not really helpful for your problem; it is usually derived by applying $MA$ to almost disjoint coding forcing, while your problem can be solved by applying $MA$ to Hechler forcing instead as follows:
Let's write $\mathcal A$ as $\langle a_n\mid n<\omega\rangle$ (with possible repetitions if $\mathcal A$ is finite). For any $b\in\mathcal B\setminus\mathcal A$ we can choose a function $f_b:\omega\rightarrow\omega$ such that $$a_n\cap b\subseteq f_b(n)$$ for any $n$. By applying $MA$ to Hechler forcing, we find one function $f_\ast:\omega\rightarrow\omega$ that eventually dominates any such $f_b$, i.e. there is an $l_b<\omega$ s.t. for all $l_b\leq n<\omega$ we have $f_b(n)<f_\ast(n)$. Now put $$d=\bigcap_{n<\omega}\omega\setminus(a_n\setminus f_\ast(n))$$ that is $d$ is the subset of $\omega$ that has the endsegment of $a_n$ above $f_\ast(n)$ removed for all $n<\omega$. We get for free that $d\cap a_n\subseteq f_\ast(n)$ and thus is finite for any $n$. For the other property, by chasing through definitions it is easy to see that $b\setminus d\subseteq l_b$, and thus is finite, for any $b\in\mathcal B\setminus \mathcal A$.