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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 $\omega$? (Only one question needs to be answered for acceptance.)

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    $\begingroup$ What is $e$ in the definition of bipartite? $\endgroup$ Jun 27, 2022 at 12:35
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    $\begingroup$ It's hard for me to tell, because your definition of "bipartite" doesn't quite make sense at the moment, but I think you might be defining the reaping number. (en.wikipedia.org/wiki/…) If so, the answer to both of your questions is yes. $\endgroup$
    – Will Brian
    Jun 27, 2022 at 13:35
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    $\begingroup$ I think it should say "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" $\endgroup$
    – Louis D
    Jun 27, 2022 at 14:00
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    $\begingroup$ @DominicvanderZypen: You're right! I missed that. $\endgroup$
    – Will Brian
    Jun 27, 2022 at 16:53
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    $\begingroup$ But I think your number is still $\mathfrak{r}$. If $\mathcal R$ is a reaping family, then closing $\mathcal R$ under finite modifications gives a non-bitartite family. And if $\mathcal E$ is a non-bipartite family, then it is also a reaping family. $\endgroup$
    – Will Brian
    Jun 27, 2022 at 18:44

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The cardinal $\mathfrak{nb}$ is equal to the reaping number $\mathfrak{r}$.

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, $\mathfrak{r}$.

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