Take an infinite rectangular array with $\kappa$ columns and $2^\kappa$ rows where $\kappa$ is some infinite cardinal. Can you fill it up with at most $\kappa$ different elements in such a way that for every finite subset of the rows you have at least one column that has pairwise different elements in each of the chosen positions?
In other words, and a bit more formally, given a set $A$ with $|A| = \kappa$, can you construct a set of sequences/vectors $\{a_\alpha : \alpha < 2^\kappa\} \subseteq {}^\kappa A$ in such a way that for every $\beta_1,...,\beta_n < 2^\kappa$ there exists an $i < \kappa$ such that $\beta_1(i),...,\beta_n(i)$ are pairwise different?
Actually the answer is supposed to be yes, but I'm quite stumped as to how to actually show it. I know how to do it for an array with $\kappa$ rows instead of $2^\kappa$ - in that case you simply have a bijection from the finite subsets of rows to the columns and using that the problem is easy to solve. Answers or hints would be appreciated.
Assuming Choice, there are exactly $\kappa$ finite subsets of $\kappa$, so label the columns with all finite subsets of $A$ instead of elements of $A$. Each row should of course be labeled with a unqie finite or infinte subset of $A$.
As the array elements put the intersections of the row label and the column label. As before, there are only $\kappa$ many possibilities for this.
Now if you have a finite set of $\beta_i$s, for each $i\ne j$ you can pick an $a$ in the symmetric difference of $\beta_i$ and $\beta_j$. Then take the set of all these (finitely many) picks, and go to the column marked with that subset.