# What is the smallest cardinality of a base of an ultrafilter on $\omega$ related to an almost disjoint family of cardinality $\mathfrak c$?

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 $$\mathfrak c$$?

## 1 Answer

Lyubomyr Zdomskyy informed me that the answer to my question is affirmative and follows from a recent (still unpublished) result of Osvaldo Guzman and Damjan Kalajdzievski who proved the consistency of $$\mathfrak u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