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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 $\mathcal F$.

Question. Can $\mathcal U$ have a base of cardinality strictly less than $\mathfrak c$?

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Lyubomyr Zdomskyy informed me that the answer to my question is affirmative and follows from a recent (still unpublished) result of Osvaldo Guzman and Damjan Kalajdzievski who proved the consistency of $\mathfrak u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 $\mathcal F\subseteq\mathcal U$.

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