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Let $\omega \leq \kappa <2^{\omega}$ 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

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

Let $\omega \leq \kappa <2^{\omega}$ 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 $MA(\kappa) \longrightarrow D(\kappa, \lambda) ?$

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

Let $\omega \leq \kappa <2^{\omega}$ 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