$\newcommand{\U}{\mathcal{U}}$
$\newcommand{\F}{\mathcal{F}}$
$\newcommand{\D}{\mathcal{D}}$
$\newcommand{\C}{\mathcal{C}}$
For any infinite $X \subseteq \omega$, we define:
$$
\D\_X := \{Y \in [\omega]^\omega : Y \subseteq X \vee Y \cap X = \emptyset\}
$$
It's easy to see that $\D\_X$ is a dense open subset of $([\omega]^\omega,\subseteq)$ for all infinite $X$. Now let $\cal{C} \subseteq \lbrack\omega\rbrack^\omega$ be a collection of infinite sets. Consider the following statement:



> 
> The statement $\mathsf{U}(\C)$ asserts that: If $\F$ is a filter on $\omega$ and $\F \cap \D\_X \neq \emptyset$ for all $X \in \cal{C}$, then $\F$ is an ultrafilter.
> 
> 
> 


It's easy to verify that that $\mathsf{U}([\omega]^\omega)$ is true. Let $\mathfrak{u}'$ be the least cardinal such that there exists some $\cal{C} \subseteq \lbrack\omega\rbrack^\omega$ which $|\C| = \mathfrak{u}'$ and $\mathsf{U}(\C)$ holds. Is it consistent that $\mathfrak{u}' < \frak{c}$?



Minimum number of dense sets to make a filter an ultrafilter
No; $\mathfrak u'=\mathfrak c$.


To prove it, consider any $\mathcal C\subseteq[\omega]^\omega$ with cardinality $<\mathfrak c$. Working modulo finoite subsets of $\omega$ , and closing under (finitary) Boolean operations, we may assume that $\mathcal C$ is a Boolean subalgebra of $\mathcal P(\omega)/\text{fin}$, and of course a proper subalgebra because of the cardinality assumption. There are now two ways to finish the proof.


(1) Since the inclusion map $\mathcal C\to P(\omega)/\text{fin}$ is not surjective, its Stone dual (from $\beta\omega-\omega$ to the space of ultrafilters in $\mathcal C$) is not injective. So fix two distinct ultrafilters $U$ and $V$ on $\omega$ that have the same intersection with $\mathcal C$. Let $F=U\cap V$. For any $X\in\mathcal C$, either $X$ is in both $U$ and $V$ and therefore in $F$, or $\omega-X$ is in both $U$ and $V$ and therefore in $F$. Thus, $F$ meets $\mathcal D\_X$. But, since $U$ and $V$ are distinct, $F$ is not an ultrafilter.


(2) Fix some $A\subseteq\omega$ not in $\mathcal C$. (I'm still tacitly working mod finite.) Let
$$
I=\{X\in\mathcal C:A\cap X\in\mathcal C\},
$$
and note that $I$ is a proper ideal in $\mathcal C$. Let $F$ be an ultrafilter of $\mathcal C$ disjoint from $I$, and let $F'$ be the filter in $\mathcal P(\omega)/\text{fin}$ generated by $F$ (i.e., the upward closure of $F$ in $\mathcal P(\omega)/\text{fin}$). The fact that $F$ is ultra in $\mathcal C$ implies that $F'$ meets $\mathcal D\_X$ for each $X\in\mathcal C$. On the other hand, the fact that $F$ extends $I$ implies that every set in $F$ meets both $A$ and $\omega-A$; therefore the same holds for every set in $F'$, and thus $F'$ is not an ultrafilter.


