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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 $\bf B$.

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  • $\begingroup$ I want to say that $S$ intersects $C$ but does not totally contain $C$ -> will correct . Thanks for spotting my error. $\endgroup$ Nov 19, 2022 at 15:55
  • $\begingroup$ Set $\mathcal{C}=\{C_\alpha\mid\alpha<\kappa\}$. Construct two sequences $(x_\alpha)_{\alpha<\kappa},(y_\alpha)_{\alpha<\kappa}$ by letting $x_\alpha,y_\alpha\in C_\alpha$ but where $\{x_\alpha\mid \alpha<\kappa\}\cap\{y_\alpha\mid\alpha<\kappa\}=\emptyset$, and set $S=\{x_\alpha\mid\alpha<\kappa\}$. Then $S$ splits $\mathcal{C}$. We do not need the condition that $|C\cap D|<\kappa$ for $C,D\in\mathcal{C},C\neq D$. $\endgroup$ Nov 19, 2022 at 16:10
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    $\begingroup$ I agree that if $|{\cal C}| = \kappa$ then ${\cal C}$ has property $\bf B$. But what if $|{\cal C}| > \kappa$? $\endgroup$ Nov 19, 2022 at 20:24

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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.

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  • $\begingroup$ Thanks very much! I would assume that the natural example that you constructed above the original answer (which used CH) can be extended to any cardinal $\kappa > \aleph_0$? $\endgroup$ Nov 20, 2022 at 11:04
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To confirm Andreas' suspicion: Balcar and Vojtáš proved in Almost Disjoint Refinement of families of subsets of $\mathbb{N}$ that every ultrafilter on $\mathbb{N}$ has an almost disjoint refinement. That is: if $u$ is an ultrafilter on $\mathbb{N}$ then there is a family $\{C_U:U\in u\}$ of infinite sets such that $C_U\subseteq U$ for all $U$, and $C_U\cap C_V$ is finite when $U\neq V$. So the earlier example can be done without $\mathsf{CH}$.

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