Let $[\omega]^\omega$ the collection of infinite subsets of $\omega$. We say that $E\subseteq [\omega]^\omega$ is *bipartite* if there is $d\subseteq \omega$ such that for all $e\in E$ the intersections $e\cap d$ and $e\cap (\omega\setminus d)$ are both non-empty. If $E\subseteq[\omega]^\omega$ is countable, then $(\omega, E)$ is bipartite. (There is an elegant argument somewhere on MathOverflow for this by user @bof; I will add it as a comment once I find it.)


Now we can define the *non-bipartiteness number* by ${\frak nb} := \min\{|E|: E\subseteq [\omega]^\omega \text{ and } E \text{ is not bipartite}\}.$


Is it consistent that ${\frak nb}<{\frak c} = 2^{\aleph\_0}$? Is it even consistent that ${\frak nb < a}$ where ${\frak a}$ is the minimum cardinality of a [maximum almost disjoint family in $\omega$](https://en.wikipedia.org/wiki/Almost_disjoint_sets)? (Only one question needs to be answered for acceptance.)



Minimal cardinality of non-bipartite sub-family of $[\omega]^\omega$
The cardinal $\mathfrak{nb}$ is equal to the [reaping number](https://en.wikipedia.org/wiki/Cardinal_characteristic_of_the_continuum#Splitting_number_%7F%27%22%60UNIQ--postMath-0000002D-QINU%60%22%27%7F_and_reaping_number_%7F%27%22%60UNIQ--postMath-0000002E-QINU%60%22%27%7F) $\mathfrak{r}$.


An *unsplit family* is a collection $\mathcal R$ of infinite subsets of $\omega$ such that there is no set $D \subseteq \omega$ with the property that $E \cap D$ and $E \setminus D$ are infinite for every $E \in \mathcal R$. The reaping number $\mathfrak{r}$ is defined to be the minimum cardinality of an unsplit family.


Note that a bipartite family (from the question) is defined in almost exactly the same way: just replace "are infinite" with "are nonempty" in the definition of an unsplit family.


The point is that $\mathcal E$ is a bipartite family then it is also an unsplit family, and if $\mathcal R$ is an unsplit family then
$$\mathcal E = \{ X \subseteq \omega :\, X \Delta E \text{ is finite for some }E \in \mathcal R\}$$
is a bipartite family. Because neither kind of family can be finite, this implies that the minimum cardinality of a bipartite family, $\mathfrak{nb}$, is equal to the minimum cardinality of an unsplit family, $\mathfrak{r}$.


