Let $\kappa\geq\aleph\_0$ be an infinite cardinal, and suppose that ${\cal A}$ is a collection of subsets of $\kappa$ such that for all $A\in {\cal A}$ we have $|A| = \kappa$ and for $A,B\in {\cal A}$ with $A\neq B$ we have $|A\cap B|<\kappa$. Is there $D\subseteq \kappa$ such that for all $A\in {\cal A}$ we have


1. $A\cap D \neq \emptyset$, and
2. $A \not\subseteq D$


?



Existence of a "diagonal" set in certain set systems
This is false in general when $\kappa=\omega$. Let $\mathcal{A}=\langle A\_\alpha:\alpha<\mathfrak{c}\rangle$ be an almost disjoint family of size continuum, and let $\langle D\_\alpha:\alpha<\mathfrak{c}\rangle$ list all subsets of $\omega$.


For each $\alpha<\mathfrak{c}$, one of $A\_\alpha\cap D\_\alpha$ and $A\_\alpha\setminus D\_\alpha$ is infinite, so we can shrink $A\_\alpha$ to an infinite subset $B\_\alpha$ such that either $B\_\alpha\subseteq D\_\alpha$ or $B\_\alpha\cap D\_\alpha=\emptyset$. The family $\langle B\_\alpha:\alpha<\mathfrak{c}\rangle$ is still almost disjoint, and for any subset $D$ of $\omega$ there is an $\alpha$ such that either $B\_\alpha\subseteq D$ or $B\_\alpha\cap D=\emptyset$.


