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




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**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).



What is the smallest cardinality of a maximal ultrafamily of infinite subsets of $\omega$?
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 $|\mathcal U|\ge|\{U\_s\}\_{s\in 2^\omega}|=|2^\omega|=\mathfrak c$. $\quad\square$.


