# Property ${\bf B}$ for families of large sets with small intersection

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 $$\bf B$$.

• I want to say that $S$ intersects $C$ but does not totally contain $C$ -> will correct . Thanks for spotting my error. Nov 19, 2022 at 15:55
• 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 $C,D\in\mathcal{C},C\neq D$. Nov 19, 2022 at 16:10
• I agree that if $|{\cal C}| = \kappa$ then ${\cal C}$ has property $\bf B$. But what if $|{\cal C}| > \kappa$? Nov 19, 2022 at 20:24

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 $$x_n\notin\bigcup_{k. These choices can be made because $$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.

• 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 $\kappa > \aleph_0$? Nov 20, 2022 at 11:04

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 $$\mathsf{CH}$$.