Let $\kappa\geq \aleph\_0$ be a cardinal. If $X\neq \emptyset$ is a set, we say that a family ${\cal C}\subseteq {\cal P}(X)$ has *property ${\bf B}$* if there is $S\subseteq X$ such that for all $C\in {\cal C}$ we have $S \cap C \neq \emptyset \neq C\setminus S$. (In other words, $S$ intersects every $C$ but contains no $C$.)


**Question.** Suppose $\kappa\geq \aleph\_0$ is a cardinal and ${\cal C}\subseteq {\cal P}(\kappa)$ is a non-empty family of sets such that all members of ${\cal C}$ have cardinality $\kappa$, and $|C\cap D| <\kappa$ whenever $C\neq D\in {\cal C}$. Does this imply that ${\cal C}$ has property ${\bf B}$?


**Note.** Joseph Van Name's argument in the comment section shows that if $|{\cal C} | = \kappa$, then ${\cal C}$ has property $\bf B$.



Property ${\bf B}$ for families of large sets with small intersection
EDIT: I'll leave my previous answer up for now (at the end of this one), but here's an easier answer that doesn't need assumptions like CH that go beyond ZFC.


It's well-known that there is a family of continuum many infinite subsets of $\omega$ such that every two have finite intersection. List such a family as $\{A\_\xi:\xi<\mathfrak c\}$. Also, list all the subsets of $\omega$ as $\{S\_\xi:\xi<\mathfrak c\}$. For each $\xi$, let $C\_\xi$ be an infinite subset of $A\_\xi$ that is either included in $S\_\xi$ or disjoint from $S\_\xi$. Then $\mathcal C=\{C\_\xi:\xi<\mathfrak c\}$ serves as the required counterexample.


PREVIOUS ANSWER:
Here's a counterexample for $\kappa=\aleph\_0$ assuming the continuum hypothesis.


Fix a nonprincipal ultrafilter $\mathcal U$ on $\omega$ and, using CH, list its elements in an $\omega\_1$-sequence as $B\_\xi,\ \xi<\omega\_1$. Build an $\omega\_1$-sequence of sets $C\_\xi$ inductively so that (1) each $C\_\xi$ is an infinite subset of $B\_\xi$, (2) the intersection $C\_\xi\cap C\_\eta$ is finite for all $\eta<\xi$, and (3) $C\_\xi\notin\mathcal U$. To do this, at stage $\xi$, when you already have the countably many sets $C\_\eta$ for $\eta<\xi$, list those sets as $C'\_n$ for $n\in\omega$ and inductively choose distinct points $x\_n\in B\_\xi$ such that $x\_n\notin\bigcup\_{k<n}C'\_k$. These choices can be made because $B\_\xi$ is in $\mathcal U$ and each $C'\_k\notin\mathcal U$ by induction hypothesis (3). If we were to take $C\_\xi=\{x\_n:n\in\omega\}$, we'd satisfy requirements (1) and (2) but perhaps not (3). So split this $\{x\_n:n\in\omega\}$ into two infinite pieces; at least one of the pieces is not in $\mathcal U$, and we take that piece as $C\_\xi$.


The family $\mathcal C=\{C\_\xi:\xi<\omega\_1\}$ satisfies the cardinality requirements in the question: its members $C\_\xi$ are infinite and their pairwise intersections are finite. I claim that it does not have property B. Indeed, for any $S\subseteq\omega$, one of $S$ and $\omega-S$ is in $\mathcal U$, hence is one of the $B\_\xi$'s, and hence includes the corresponding $C\_\xi$.


Remark: CH is overkill here. Essentially the same argument works if $\mathfrak r$, the unsplitting number, is $\aleph\_1$; just enumerate a $\pi$-base for $\mathcal U$ instead of all of $\mathcal U$. Also, $\mathfrak p=\mathfrak u$ seems to be sufficient: Let $\{B\_\xi:\xi<\mathfrak u\}$ be a base for $\mathcal U$ (rather than all of $\mathcal U$) and use the assumption about $\mathfrak p$ to obtain the desired $x\_n$'s.


I wouldn't be surprised if there's a counterexample in ZFC without any special hypotheses, but I don't yet see one.


