$\langle \mathfrak{c},\mathfrak{c}\rangle$-Independent Matrix

Question

Given cardinals $\lambda,\kappa$, an $\langle \lambda,\kappa\rangle$-independent matrix on $X$ is a colection $\mathcal{A} = \{A_{\alpha}^{\beta}:\alpha<\lambda\wedge \beta<\kappa\}$ sattisfying

1. Given $\alpha <\lambda$ and $\beta_0,\beta_1<\kappa$, with $\beta_0\neq\beta_1$, $\left|A_{\alpha}^{\beta_0}\cap A_{\alpha}^{\beta_1}\right|<\omega$;
2. given $\alpha_0,\dots,\alpha_{k-1}<\lambda$ distinct, for all $\beta_0,\dots,\beta_{k-1}$, $\left|\bigcap_{i<k}A_{\alpha_i}^{\beta_i}\right| = |X|$.

How to construct an $\langle \mathfrak{c},\mathfrak{c}\rangle$-independent matrix in $\omega$?

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I'm folowing Kunen's tip: Let

$$I = \left\{\langle n,f\rangle:n<\omega\wedge f\in\,^{\wp(n)}\wp(n)\right\}$$

Then $|I|=\omega$. For each $X,Y\subseteq\omega$, define $$A_{X}^{Y} = \left\{\langle n,f\rangle\in I:f[X\cap n] = Y\cap n\right\}.$$

It's easy to see that, for $X,Y_0,Y_1\subseteq\omega$, with $Y_0\neq Y_1$, one has $\left|A_{X}^{Y_0}\cap A_{X}^{Y_1}\right|<\omega$. But I could not prove that, for $X_0,\dots X_{k-1}\subseteq\omega$ distinct and $Y_0,\dots,Y_{k-1}\subseteq\omega$, one has $\left|\bigcap_{i<k}A_{X_i}^{Y_i}\right| = \omega$.

Answer 1

Since the $X_i$ are distinct, there is an $m\in\omega$ such that the sets $X_i\cap m$ are distinct. It follows that the $k$ sets $X_i\cap n$ are distinct for each $n\ge m$. Thus, for each $n\ge m$ there is an $f_n\in{}^{\wp(n)}\wp(n)$ such that $f_n[X_i\cap n]=Y_i\cap n$ for each $i<k$, and $\{\langle n,f_n\rangle:n\ge m\}$ is an infinite subset of $\bigcap_{i<k}A_{X_i}^{Y_i}$.