Independent families of sets

Question

I'm having a difficulty understanding some exercises related to independent families of sets. Recall that $ \mathcal{A} $ is $\lambda$-independent if for any disjoint $ P, Q \in \mathcal{A} : |P|, |Q| < \lambda $ it holds that $$ \bigcap\limits_{A \in P}A \cap \bigcap\limits_{A \in Q} (X \setminus A) \neq \emptyset $$ A family is independent iff it's $ \omega $-independent and $ \sigma $-independent iff it's $ \omega_1 $-independent.

The questions are the following: 1) Suppose $ \mathcal{A} $ is $ \lambda $-independent on $ \kappa $. Show that there is a family $ \mathcal B : | \mathcal B | = | \mathcal A |$ such that $$ \left| \bigcap\limits_{A \in P}A \cap \bigcap\limits_{A \in Q} (X \setminus A) \right| = \kappa $$

2) Show that if $ \kappa = \kappa^{\aleph_0} $, then there exists a $\sigma$-independent family on $ \kappa $ of cardinality $ 2^\kappa $

I know that it holds that for $ \kappa > \omega $ there exists an independent family of such cardinality, but I'm not sure how to use this in that case

I would appreciate some hints

Answer 1

HINT: For the first part, fatten up each point of $\kappa$ into $\kappa$ points. In case that’s not enough, I’ve made it more explicit in the spoiler-protected block that follows.

Let $\mathscr{B}=\{A\times\kappa:A\in\mathscr{A}\}$. Since $|\kappa\times\kappa|=\kappa$, no harm is done by making the members of $\mathscr{B}$ subsets of $\kappa\times\kappa$.

For the second part, this answer has a proof of the theorem that if $|X|=\kappa$, there is an independent family $\mathscr{A}$ of subsets of $X$ such that $|\mathscr{A}|=2^\kappa$. Try to modify that proof to make $\mathscr{A}$ $\sigma$-independent of $\kappa^\omega=\kappa$.