I am trying to understand the stated above Theorem. $\mathfrak p$ is the smallest cardinality of any family $\mathcal F \subseteq [\omega]^\omega$, which has the strong finite intersection property, but does not have a pseudo intersection. The definitions are:
A family $\mathcal F \subseteq [\omega]^\omega$, has the strong finite intersection property if every finite subfamily has infinite intersection. Also, a pseudo intersection of a family $\mathcal F \subseteq [\omega]^\omega$, is an infinite subset of $\omega$ which is almost contained in every member of $\mathcal F$. Where $A$ is Almost contained in $B$ means that for all except maybe a finite number of elements in $A$, $a \in A \Rightarrow a \in B$. Also, $[\omega]^\omega$ is the space of all infinite subsets of $\omega$.
I have a Proof of the theorem (see below), which I somehow don't manage to understand. I don't understand who is $a_0$? Is $X_0$ an infinite sequence from $[\omega]^\omega$?
Thank you, Shir
The following proof is quoted from Halbeisen's book Combinatorial Set Theory, page 181. The book is freely available on author's website.
Theorem 8.1. $\omega_1\le\mathfrak p$.
Proof. Let $\mathscr E=\{X_n\in[\omega]^\omega : n\in\omega\}$ be a countable family which has the sfip. We construct a pseudo-intersection of $\mathscr E$ as follows: Let $a_0:=\bigcap X_0$ and for positive $n$ let $$a_n=\bigcap \left(\bigcap \{X_i; i\in n\}\setminus \{a_i; i\in n\}\right).$$ Further, let $Y=\{a_n : n\in\omega\}$; then for every $n\in\omega$, $Y\setminus \{a_i; i\in n\} \subseteq X_n$ which shows that $Y\subseteq^* X_n$, hence, $Y$ is a pseudointersection of $\mathscr E$.
To show that $\omega_1\leq\frak p$ it suffices to show that $\omega\neq\frak p$. That is, given a countable family with strong finite intersection property, it must have a pseudo-intersection.
So suppose that $\{X_n\in[\omega]^\omega\mid n\in\omega\}$ is a family with the strong finite intersection property, we define $a_0=\min X_0$, which for ordinals is the same as $\bigcap X_0$. Next we define $$a_n = \min\left\{m\in\omega\mathrel{}\middle|\mathrel{} m\in\bigcap_{k<n}X_n\setminus\{a_k\mid k<n\}\right\}$$
Finally we claim that $Y=\{a_n\mid n\in\omega\}$ is a pseudo intersection of our $X_n$'s because for every $n\in\omega$, $Y\setminus\{a_k\mid k<n\}$ is a subset of $X_n$ (recall how $a_n$ was chosen).