# Property $\mathbf{B}$ for maximal linear set systems on $\omega$ with finite members

Let $$X\neq\emptyset$$ be a set. A family $${\cal S}\subseteq {\cal P}(X)$$ has property $$\mathbf{B}$$ if there is $$T\subseteq X$$ such that for all $$S\in{\cal S}$$ we have $$S\cap T\neq \emptyset$$ and $$S\not\subseteq T$$. Moreover, $${\cal S}$$ is said to be linear if $$|S_1 \cap S_2| \leq 1$$ for all $$S_1\neq S_2\in{\cal S}$$.

Let $$\text{FL}(\omega)$$ be the collection of linear families $${\cal S}$$ of $$\omega$$ such that all $$S\in{\cal S}$$ are finite and have at least $$2$$ elements. A standard application of Zorn's Lemma shows that every member of $$\text{FL}(\omega)$$ is contained in a member of $$\text{FL}(\omega)$$ that is maximal with respect to set inclusion.

Question. Is there a maximal member of $$\text{FL}(\omega)$$ which has property $${\bf B}$$?

## 1 Answer

Yes. Partition $$\omega$$ into two disjoint infinite subsets $$T_1$$ and $$T_2$$. Recursively construct a $$3$$-uniform linear hypergraph (Steiner triple system) $$\mathcal S\subseteq\binom\omega3$$ so that each element of $$\binom\omega2$$ is contained in a unique element of $$\mathcal S$$, and each element of $$\mathcal S$$ meets both $$T_1$$ and $$T_2$$. Namely, we enumerate the elements of $$\binom\omega2$$, and when we come to an element $$\{x,y\}$$ of $$\binom\omega2$$ which is not already covered, choose $$i\in\{1,2\}$$ so that $$\{x,y\}\not\subseteq T_i$$ and choose a number $$z\in T_i\setminus\{x,y\}$$ which is not in any element of $$\binom\omega3$$ which has already been put into $$\mathcal S$$, and add the triple $$\{x,y,z\}$$ to $$\mathcal S$$.