Let $(A\_\alpha)\_{\alpha\in\mathfrak c}$ be an almost disjoint family of infinite subsets of $\omega$. The almost disjointness of the family means that $A\_\alpha\cap A\_\beta$ is finite for any ordinals $\alpha,\beta\in\mathfrak c$. This almost disjoint family generates a filter $$\mathcal F=\{F\subseteq \omega:|\{\alpha\in\mathfrak c:A\_\alpha\not\subseteq^\* F\}|<\mathfrak c\}.$$
Let $\mathcal U$ be any ultrafilter on $\omega$ containing the filter $\mathcal F$.



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> **Question.** Can $\mathcal U$ have a base of cardinality strictly less than $\mathfrak c$?
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What is the smallest cardinality of a base of an ultrafilter on $\omega$ related to an almost disjoint family of cardinality $\mathfrak c$?
Lyubomyr Zdomskyy informed me that the answer to my question is affirmative and follows from a recent (still unpublished) result of [Osvaldo Guzman and Damjan Kalajdzievski](https://lc2019.cz/static/public/slides/Friday/room-0/guzman/Logic-Colloquim.pdf?b8f311b8201af564aceb) who proved the consistency of $\mathfrak u<\mathfrak a=\mathfrak c$. 


Take any ultrafilter $\mathcal U$ on $\omega$ that has a base $\mathcal B\subseteq\mathcal U$ of cardinality $|\mathcal B|=\mathfrak u$. Next, choose a maximal almost disjoint family of sets $\mathcal A\subseteq [\omega]^\omega\setminus\mathcal U$. Consider the filter $$\mathcal F=\{F\subseteq \omega:|\{A\in\mathcal A:A\not\subseteq^\* F\}|<\mathfrak c\}.$$ We claim that $\mathcal F\subseteq\mathcal U$ under $\mathfrak a=\mathfrak c$. Assuming that $\mathcal F\not\subseteq\mathcal U$, we can find a set $F\in\mathcal F\setminus\mathcal U$. For the set $F$ the family $\mathcal A'=\{A\in\mathcal A:A\not\subseteq^\* F\}$ has cardinality $|\mathcal A'|<\mathfrak c=\mathfrak a$. Then the family $\{A\setminus F\}\_{A\in\mathcal A'}$ is not maximal almost disjoint in $\omega\setminus F\in\mathcal U$. So, we can choose an infinite set $B\subseteq\omega\setminus F$ such that $B\notin \mathcal A$ and the family $\{B\}\cup\{A\setminus F\}\_{A\in\mathcal A'}$ is almost disjoint. Replacing $B$ by a smaller infinite subset we can additionally assume that $B\notin\mathcal U$. Then $\{B\}\cup\mathcal A$ is an almost disjoint family in $[\omega]^\omega\setminus\mathcal U$, which is strictly larger than $\mathcal A$. But this contradicts the maximality of $\mathcal A$. This contradiction shows that $\mathcal F\subseteq\mathcal U$.


