The following text is a quote from p.180 of Halbeisen's book Combinatorial Set Theory. This book is also available on website of a course taught by the author. (As mentioned in Asaf's comment, it is also available on the author's website.)
For two sets $x,y\subseteq\omega$ we say that $x$ is almost contained in $y$, denoted $x\subseteq^*y$, if $x\setminus y$ finite.
A pseudo-intersection of a family $\mathscr F \subseteq [\omega]^\omega$ of infinite subsets of $\omega$ is an infinite subset of $\omega$ that is almost contained in every member of $\mathscr F$.
Furthermore, a family $\mathscr F \subseteq [\omega]^\omega$ has the strong finite intersection property (sfip) if every finite subfamily has infinite intersection.
For example any filter $\mathscr F \subseteq [\omega]^\omega$ has the sfip, but no ultrafilter on $[\omega]^\omega$ has a pseudo-intersection.
- How can we show that a free ultrafilter cannot have an infinite pseudointersection?
This fact is used to show that $\mathfrak p$ is well-defined and $\mathfrak p \le \mathfrak c$, where the pseudo-intersection number $\mathfrak p$ is the smallest cardinality of any family $\mathscr F \subseteq [\omega]^\omega$ which has the sfip but which does not have a pseudo-intersection.
I will post my proof below; but I wonder whether there are different ways to show this.