$\newcommand{\U}{\mathcal{U}}$ $\newcommand{\F}{\mathcal{F}}$ $\newcommand{\D}{\mathcal{D}}$ $\newcommand{\C}{\mathcal{C}}$ 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 $\mathfrak{u}' < \frak{c}$?