Let $I$ be an index set. Given $A\subseteq I\times 2^\omega$ and $i\in I$, set $(A)\_i = \{x\in 2^\omega: (i, x)\in A\}$. 


Now let $I\times 2^\omega = \bigcup\_{n<\omega} A\_n$. How large must $I$ be to ensure that there are indices $i\neq j\in I$ and $n< \omega$ with $(A\_n)\_i \cap (A\_n)\_j$ non-meager? In particular, is it consistent that there is a partition of $\omega\_1\times 2^w$ into countably many pieces not satisfying the above?



Countable partitions of Cantor space mod meager
Let $\kappa$ be least such that whenever $\{A^i\_n : n < \omega\}$, for $i < \kappa$ are coverings of $2^{\omega}$, there are $i < j < \kappa$ and $n < \omega$ such that $A^i\_n \cap A^j\_n$ is non meager.


Claim: CH implies $\kappa > \omega\_1$.


Proof: Using an Ulam matrix.


Claim: Assume MA plus not CH. Then $\kappa = \omega\_1$.


Proof: Let $\{A^i\_n: n < \omega\}$ be coverings of $2^{\omega}$ for $i < \omega\_1$. Towards a contradiction, assume that for every $i < j < \omega\_1$ and $n < \omega$, $A^i\_n \cap A^j\_n$ is meager. Put $B^i\_n = A^i\_n \setminus \bigcup \{A^j\_n : j < i\}$ and $B^i = \bigcup \{B^i\_n : n < \omega\}$. Then each $B^i$ is comeager and so is $B = \bigcap \{B^i : i < \omega\_1\}$ (by MA plus not CH). But now if $x \in B$, then $(\forall i < \omega\_1)(\exists n < \omega)(x \in B^i\_n)$. Choose $i < j < \omega\_1$ and $n < \omega$ such that $x \in B^i\_n \cap B^j\_n$: Contradiction.


So the lower bound $\omega\_1 \leq \kappa$ cannot be improved in ZFC. I am not sure about the optimal upper bound for $\kappa$.


