Definition. Let X be an infinite set and $\mathcal{F}\subseteq \mathcal{P}(X)$. We will say that $\mathcal{F}$ has the property of the strong finite intersection if
\begin{align*}
\forall \sigma\in[\mathcal{F}]^{<\omega}:\left|\bigcap\sigma\right|\geq\omega
\end{align*}
If $\mathcal{F}$ has the property of the strong finite intersection it is clear that the filter generated by $\mathcal{F}$ also has that property.
If $\mathcal{U}$ is any ultrafilter that extends to $\mathcal{F}$, does $\mathcal{U}$ also have such property?
Best Answer
Unfortunately no.
Suppose that $X=\omega\cup\{\omega\}$ and $\cal F$ is the filter generated by cofinite subsets of $\omega$ on $X$. Then $\cal F$ has an extension to a principal ultrafilter by adding $\{\omega\}$ to $\cal F$.
And clearly, a principal ultrafilter does not have the strong finite intersection property.