On Consistency of an Existence

Question

Let $\omega \leq \kappa <2^{\omega}$ , $\omega \leq\lambda \leq \kappa$ and $D(\kappa, \lambda)$ be the statement:

For all $ \mathfrak{B} \subseteq \mathbf{P}(\omega)$ with $|\mathfrak{B}|=\kappa$ that is an almost disjoint family, there exists an $\mathfrak{A}\subseteq\mathfrak{B}$ with $|\mathfrak{A}|=\lambda$, such that for some $d\subseteq \omega $ we have:

1) $\forall x \in \mathfrak{A} ~~~~~~~~~~ |x\cap d |< \omega$

2) $\forall x \in \mathfrak{B} \smallsetminus\mathfrak{A} ~~ |x \smallsetminus d|< \omega$.

Now I have three questions:

Question 1) Does $MA(\kappa) \longrightarrow D(\kappa, \lambda) ?$

Or even:

Question 2) Is it true that $$Con(ZFC+ \neg CH+MA(\kappa)) \longrightarrow Con(ZFC+ D(\kappa , \lambda)) ?$$

Remarks: Clearly $ZFC \vdash D(\omega, \omega)$ and also $MA(\kappa) \longrightarrow D(\kappa, \omega) \wedge D(\kappa, \kappa).$ On the other hand what happens in the case $\lambda=\kappa$, when we add the condition $|\mathfrak{B} \smallsetminus \mathfrak{A}|=\kappa$ into the defination of $D(\kappa, \kappa)$? I mean, let $D^{+}(\kappa, \kappa)$ be $D(\kappa, \kappa)$ with this additional condition.

Question 3) Does $MA(\kappa) \longrightarrow D^{+}(\kappa, \kappa) ?$

Thanks

Answer 1

There is a counterexample to $D^+(\omega_1, \omega_1)$ in ZFC, so the answer to question 3 is no in general. A $d$ satisfying your conditions (1) and (2) is said to separate $\mathfrak{A}$ and $\mathfrak{B} \setminus \mathfrak{A}$. Luzin showed there is an almost disjoint family $\mathfrak{B}$ of size $\aleph_1$ such that for any two uncountable disjoint subfamilies $\mathfrak{A}, \mathfrak{A}' \subseteq \mathfrak{B}$ there is no $d$ separating them. This is a slightly stronger property than just the negation of $D^+(\omega_1, \omega_1)$ for $\mathfrak{B}$. Such a family is called a Luzin gap.

The construction of a Luzin gap $\mathfrak{B}=\langle B_{\alpha}: \alpha <\omega_1 \rangle$ is almost the same as the usual inductive construction of an AD family of size $\aleph_1$, with one modification. At stage $\alpha$ one chooses $B_{\alpha}$ almost disjoint from each of the previous $B_{\beta}$ with the additional property that for every $n \in \omega$, $\{\beta < \alpha: \textrm{max}(B_{\alpha} \cap B_{\beta}) < n \}$ is finite (where max$(\emptyset)$ is defined to be $-1$). No care is needed at finite stages, so let $\langle B_n: n < \omega \rangle$ be arbitrary pairwise disjoint infinite subsets of $\omega$. Now suppose we have succeeded in building $\langle B_{\beta}: \beta<\alpha \rangle$. Enumerate this set in order type $\omega$ as $\langle A_n: n <\omega \rangle$. Let $x_n$ be any element in $A_n \setminus A_0 \cup \ldots \cup A_{n-1}$ greater than $n$, and let $B_{\alpha}=\{x_n: n\in\omega\}$. Then $B_{\alpha}$ is almost disjoint from every $B_{\beta}$, $\beta < \alpha$, and furthermore has the desired property.

The AD family $\mathfrak{B}$ is a Luzin gap. To see this, fix two disjoint and uncountable subfamilies $\mathfrak{A}, \mathfrak{A}' \subseteq \mathfrak{B}$, and suppose toward a contradiction there is a $d \subseteq \omega$ separating them, that is, a $d$ that is almost disjoint from every $B \in \mathfrak{A}$ and that almost contains every $B \in \mathfrak{A}'$. For every $B \in \mathfrak{A}$, let $n_B = \textrm{max}(d \cap B)$. There must be some $m$ such that for uncountably many $B \in \mathfrak{A}$ we have $n_B = m$. Pick some $B_{\alpha} \in \mathfrak{A}'$ of large enough index so that $\{\beta < \alpha: B_{\beta} \in \mathfrak{A} \, \textrm{and} \, n_{B_{\beta}}=m\}$ is infinite. We chose $B_{\alpha}$ so that $\{\beta < \alpha: \textrm{max}(B_{\alpha} \cap B_{\beta}) \leq m \}$ is finite. Since $d$ almost contains $B_{\alpha}$, it must be that $\{\beta < \alpha: \textrm{max}(d \cap B_{\beta}) \leq m \}$ is also finite, an immediate contradiction.