Let $X$ be an infinite set, and let ${\cal E}$ be a collection of non-empty subsets of $X$. We say that ${\cal E}$ has [property $\mathbf{B}$](https://users.renyi.hu/~p_erdos/1966-07.pdf) if there is $B\subseteq X$ such that $B\cap E\neq \emptyset$ and $E\not\subseteq B$ for all $E\in{\cal E}$. (This is equivalent to saying that the hypergraph $(X,{\cal E})$ has [chromatic number](https://mathoverflow.net/questions/321987/hypergraph-colorings-with-small-fibers) $2$.)


It is possible to find a chain ${\frak E}$ (with respect to $\subseteq$) of collections with property $\mathbf{B}$ such that $\bigcup {\frak E}$ has not property ${\mathbf B}$ - but this does not imply that not every set of subsets of $X$ with property ${\mathbf B}$ is not in a maximal such set of subsets. This leads to the following:


**Question.** What is an example of a set $X$ and and set of non-empty subsets ${\cal E}$ of $X$ with property ${\mathbf B}$ such that ${\cal E}$ is not contained in a set of non-empty subsets ${\cal E'}$ of $X$ that is maximal with respect to having property ${\mathbf B}$ and set inclusion?



Maximizing set systems with property $\mathbf{B}$
Any collection $\cal E$ satisfying property ${\mathbf B}$ (further: ${\mathbf B}$-collection) is contained in an inclusion-maximal ${\mathbf B}$-collection.


Proof: consider the corresponding 2-coloring of $\cal E$ and define the over-collection $\cal E\_0$ as the collection of all sets which contain elements of both colors. It is easy to see that the 2-coloring for $\cal E\_0$ is unique (sets of size 2 from $\cal E\_0$ are enough for such a conclusion), therefore if we add any set to $\cal E\_0$, it is no longer a ${\mathbf B}$-collection.


