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A family $\mathcal U$ of infinite subsets of $\omega$ is called an ultrafamily if for any sets $U,V\in\mathcal U$ one of the sets $U\setminus V$, $U\cap V$ or $V\setminus U$ is finite.

By the Kuratowski-Zorn Lemma, each ultrafamily $\mathcal U\subseteq [\omega]^\omega$ can be enlarged to a maximal ultrafamily.

Let $\mathfrak{uf}$ be the smallest cardinality of a maximal ultrafamily. It can be shown that $$\max\{\mathfrak s,\mathfrak a\}\le\mathfrak {uf}\le\mathfrak c,$$ where $\mathfrak a$ is the smallest cardinality of a maximal infinite almost disjoint family of infinite sets in $\omega$ and $\mathfrak s$ is the smallest cardinality of a family $\mathcal S\subseteq[\omega]^\omega$ such that for every $X\in[\omega]^\omega$ there exists $S\in\mathcal S$ such that the sets $X\cap S$ and $X\setminus S$ are infinite.

Problem 1. Find non-trivial upper (and lower) bounds for the cardinal $\mathfrak{uf}$.

Problem 2. Is $\mathfrak{uf}$ equal to any known cardinal characteristic of the continuum?

Problem 3. Is $\mathfrak{uf}<\mathfrak c$ consistent?


Added in Edit. The diagrams of known small uncountable cardinals in the surveys of Blass and Vaughan do not show up any cardinal characteristic between $\max\{\mathfrak s,\mathfrak a\}$ and $\mathfrak c$. So the answer to Problem 2 is rather ``no'' unless $\mathfrak{uf}=\mathfrak c$ (which would be a bit surprising).

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  • $\begingroup$ A piece of heuristics: this statement has a very similar flavour to Booth's lemma (that is equivalent to Martin's axiom), so very likely $uf = c$ under MA and probably it is consistent that there is a strict inequality. $\endgroup$ – Tomasz Kania Apr 15 '20 at 15:38
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    $\begingroup$ @TomekKania That $\mathfrak{uf}=\mathfrak{c}$ under MA follows from the inequalities in the question plus the fact that MA implies $\mathfrak{s}=\mathfrak{c}$. $\endgroup$ – Andreas Blass Apr 15 '20 at 16:54
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To my surprise, I found that this my ``new'' cardinal $\mathfrak{uf}$ is equal to $\mathfrak c$.

Theorem. $\mathfrak{uf}=\mathfrak{c}$.

Proof. Fix any maximal ultrafamily $\mathcal U\subseteq[\omega]^{\omega}$. For two sets $A,B$ we write $A\subset^* B$ if $A\setminus B$ is finite but $B\setminus A$ is infinite.

A subfamily $\mathcal L$ will be called

$\bullet$ linearly ordered if for any distinct sets $A,B\in\mathcal L$ we have either $ A\subset^* B$ or $B\subset^* A$;

$\bullet$ densely linearly ordered if for any distinct sets $A,B\in\mathcal L$ with $A\subset^* B$ there exists a set $C\in\mathcal L$ such that $A\subset^* C\subset^* B$.

Claim 1. If $\mathcal U$ contains a set $U\in\mathcal U$ such that the family ${\downarrow}U=\{V\in\mathcal U:V\subset^* U\}$ is linearly ordered, then ${\downarrow}U$ is densely linearly ordered.

Proof. Assuming that ${\downarrow}U$ is not densely linearly ordered, we can find two sets $A,B\in{\downarrow}U$ such that $A\subset^* B$ and the set $\{C\in\mathcal U:A\subset^* C\subset^* B\}$ is empty. Taking into account that $B\setminus A\subset^* U$ and the family ${\downarrow}U$ is linearly ordered, we conclude that $B\setminus A\notin\mathcal U$. By the maximality of $\mathcal U$, there exists a set $W\in\mathcal U$ such that the sets $(B\setminus A)\cap W$, $(B\setminus A)\setminus W$, $W\setminus(B\setminus A)$ are infinite. Then also the sets $B\cap W$, $B\setminus W$ are infinite. Taking into account that $\mathcal U$ is an ultrafamily, we conclude that $W\subseteq^* B\subset^* U$ and hence $W\in{\downarrow}U$. Now the choice of the sets $A,B$ guarantees that $W\subseteq^*A$ and then $(B\setminus A)\cap W$ is finite, which is a desired contradiction. $\quad\square$

Claim 2. If $\mathcal U$ contains a set $U\in\mathcal U$ such that the family ${\downarrow}U=\{V\in\mathcal U:V\subset^* U\}$ is linearly ordered, then $|\mathcal U|=|{\downarrow}U|=\mathfrak c$.

Proof. By Claim 1, the family ${\downarrow}U$ is densely linearly ordered. Consider the countable set $\mathbb Q_2=\{\frac{k}{2^n}:n\in\omega,\;0\le k\le 2^n\}$ of binary rational numbers in the unit interval $[0,1]$. Using the density of the linear order on $\mathcal L$, we can inductively construct a subfamily $\{L_q\}_{q\in\mathbb Q_2}\subseteq\mathcal L$ such that for any rational numbers $r<q$ in $\mathbb Q_2$ we have $L_r\subset^* L_q$.

To see that $|{\downarrow}U|=\mathfrak c$, it remains to prove the following

Claim 3. For every $r\in[0,1]\setminus \mathbb Q_2$ there exists a set $L_r\in\mathcal U$ such that $L_p\subset^* L_r\subset^* L_q$ for every rational numbers $p,q\in\mathbb Q_2$ with $p<r<q$.

Proof. To derive a contradiction, assume that the maximal ultrafamily $\mathcal U$ does not contain such set $L_r$.

Since the poset $\mathcal P(\omega)/\mathrm{Fin}$ contains no $(\omega,\omega)$-gaps, there exists a set $\tilde L_r\subseteq \omega$ such that $L_p\subset^* \tilde L_r\subset^* L_q$ for any $p,q\in\mathbb Q_2$ with $p<r<q$. By our assumption, $\tilde L_r\notin\mathcal U$.

By the maximality of $\mathcal U$ we can find a set $L_r\in\mathcal U$ such that the sets $\tilde L_r\cap L_r$, $\tilde L_r\setminus L_r$ and $L_r\setminus \tilde L_r$ are infinite. The infiniteness of the intersections $\tilde L_r\cap L_r$ and $\tilde L_r\setminus L_r$ implies that for any $q\in\mathbb Q_2$ with $r<q$ the intersections $L_q\cap L_r$ and $L_q\setminus L_r$ are infinite. Taking that $\mathcal U$ is an ultrafamily, we conclude that $L_r\subseteq^* L_q\subset^* U$ and hence $L_r\in{\downarrow}U$. For every $p\in\mathbb Q_2$ with $p<r$ the infiniteness of the set $L_r\setminus \tilde L_r$ and the almost inclusion $L_p\subset^* \tilde L_r$ implies the infiniteness of the set $L_r\setminus L_p$. Since the family ${\downarrow}U$ is linearly ordered, we conclude that $L_p\subseteq^* L_r$. Therefore, we proved that $L_p\subset^* L_r\subseteq^* L_q$ for any rational numbers $p,q\in\mathbb Q_2$ with $p<r<q$. But the existence of the set $L_r$ contradicts our assumption. $\quad\square$

So, Claims 3 and 2 have been proved. $\quad\square$

Claim 2 reduced the proof of the theorem to the case when for every $U\in\mathcal U$ the family ${\downarrow}U$ is not linearly ordered and hence contains two sets $U_0,U_1$ such that $U_0\cap U_1$ is finite. In this case we can inductively construct a family of sets $\{U_s\}_{s\in 2^{<\omega}}\subseteq\mathcal U$ indexed by the elements of the binary tree $2^{<\omega}=\bigcup_{n\in\omega}2^n$ such that $$\mbox{$U_{s\hat{\;}0}\cup U_{s\hat{\;}1}\subseteq^* U_s$ and $U_{s\hat{\;}0}\cap U_{s\hat{\;}1}=^*\emptyset$ for any binary sequence $s\in 2^{<\omega}$}.$$

Claim 4. For every $s\in 2^\omega$ there exists a set $U_s\in\mathcal U$ such that $U_s\subseteq^* U_{s{\restriction}n}$ for every $n\in\omega$.

Proof. To derive a contradicion, assume that for some $s\in 2^\omega$ the sequence $(U_{s{\restriction}n})_{n\in\omega}$ has no pseudointersection in $\mathcal U$. Choose any infinite pseudointersection $\tilde U_s$ of the sequence $(U_{s{\restriction}n})_{n\in\omega}$. By our assumption, $\tilde U_s\notin\mathcal U$ and by the maximality of the ultrafamily $\mathcal U$, there exists a set $U_s\in\mathcal U$ such that the sets $U_s\cap\tilde U_s$, $U_s\setminus\tilde U_s$ and $\tilde U_s\setminus U_s$ are infinite. Then for every $n\in\omega$ the sets $U_{s{\restriction}n}\cap\tilde U_s$ and $U_{s{\restriction}n}\setminus U_s$ are infinite. Taking into account that $\mathcal U$ is an ultrafamily, we conclude that $U_s\subseteq^* U_{s{\restriction}n}$ for every $n\in\omega$, which contradicts our assumption. $\quad\square$

It is easy to see that the family $(U_s)_{s\in 2^\omega}$ given by Claim 4 is almost disjoint and hence $|\mathcal U|\ge|\{U_s\}_{s\in 2^\omega}|=|2^\omega|=\mathfrak c$. $\quad\square$.

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