Existence of a "diagonal" set in certain set systems 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


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*$A\cap D \neq \emptyset$, and

*$A \not\subseteq D$


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 A: 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 $B_\alpha\cap D=\emptyset$.
A: At least, this is true if $|\mathcal A|=\aleph_0$, and also if $|\mathcal A|<\kappa$.
In the former case ($|\mathcal A|=\aleph_0$) we can write $\mathcal A=\{A_1,A_2,\ldots\}$. Choose now elements $a_1\in A_1$, $a_2\in A_2$, $a_3\in A_3,\ldots$ so that $a_2\notin A_1$ (this is possible as $|A_2|>|A_1\cap A_2|$), then $a_3\notin A_1\cup A_2$ (possible in view of $|A_3|>|(A_1\cup A_2)\cap A_3|$), etc. Taking $D:=\cup_i\{a_i\}$, we get $\{a_i\}\subseteq D\cap A_i\subseteq\{a_1,\ldots,a_i\}\subsetneq A_i$ for each $i\ge 1$.
In the latter case ($|\mathcal A|<\kappa$), for each $A\in\mathcal A$ we have $|\cup_{B\ne A}(A\cap B)|<|A|$, and therefore there exists $a\in A$ with $a\notin \cup_{B\ne A} B$. Thus, $\mathcal A$ admits a system of distinct representatives, and taking $D$ to be the union of all these representatives, we get $D\cap A=\{a\}$, for every $A\in\mathcal A$.
