Basic bounds for p and t

Let $⟨C_n:n\in \mathbb{N}⟩$ be a sequence of infinite subsets of $\omega$ all of whose finite intersections are infinite, and set $B_n=C_0\cap \cdots \cap C_n$; then each $B_n$ is infinite, and $B_0\supseteq B_1\supseteq \cdots$.

$p\le t$: by [F2] there is a strictly decreasing tower $⟨A_{\alpha }:\alpha <t⟩$; its member family $\mathcal{F}$ has cardinality $t$, since $\alpha \mapsto A_{\alpha }$ is injective. The family has SFIP: for ${\beta }_0<\cdots <{\beta }_n<t$ the intersection contains $A_{{\beta }_n}$ minus the finitely many finite sets $A_{{\beta }_n}\setminus A_{{\beta }_i}$. It has no pseudointersection, because it is a tower. Hence some SFIP family without pseudointersection has size $t$, and $p\le t$.

Recursively choose $x_n$ to be the least element of $B_n\setminus \{x_0,\dots ,x_{n-1}\}$; this is legitimate because $B_n$ is infinite and only finitely many elements have been removed, so the set is nonempty, and it has a least element by [F3]. Then $x_n\in B_n$ and the $x_n$ are pairwise distinct, since $x_n\mathrm{\notin }\{x_0,\dots ,x_{n-1}\}$.

$X=\{x_n:n\in \mathbb{N}\}$ is infinite by step 2.1, and $X{\subseteq }^{\ast }{C}_k$ for every $k$: indeed $x_n\in B_n\subseteq C_k$ whenever $n\ge k$, so $X\setminus C_k\subseteq \{x_0,\dots ,x_{k-1}\}$ is finite. Hence $X$ is a pseudointersection of $\{C_n:n\in \mathbb{N}\}$.

Now let $\mathcal{F}\subseteq [\omega {]}^{\omega }$ be countable and have the strong finite intersection property. If $\mathcal{F}=\mathrm{\varnothing }$, then $\omega$ is a pseudointersection and the claim follows. Otherwise list $\mathcal{F}$ as a sequence $⟨A_n:n\in \mathbb{N}⟩$, repeating one member if $\mathcal{F}$ is finite; that is possible by [F3] and [F4], and the finite intersections of the $A_n$ are still infinite. Put $B_n=A_0\cap \cdots \cap A_n$; each $B_n$ is infinite by SFIP, and the sequence $⟨B_n⟩$ has all finite intersections infinite, since $B_0\cap \cdots \cap B_m=B_m$. By steps 1.1, 2.1 and 3.1 applied to $C_n:=B_n$ there is $X\in [\omega {]}^{\omega }$ with $X{\subseteq }^{\ast }{B}_n$ for every $n$.

Every countable descending family $⟨A_n:n\in \mathbb{N}⟩$ of infinite sets has a pseudointersection: reaching $A_n$ from earlier members removes only finitely many points, so $B_n=A_0\cap \cdots \cap A_n$ satisfies $A_n\setminus B_n\subseteq {\bigcup }_{i<n}(A_n\setminus A_i)$, a finite set, and $B_n$ is infinite because $A_n$ is; the family $\{B_n:n\in \mathbb{N}\}$ therefore consists of infinite sets with all finite intersections infinite, and steps 1.1, 2.1 and 3.1 applied to it give $X\in [\omega {]}^{\omega }$ with $X{\subseteq }^{\ast }{B}_n\subseteq A_n$ for every $n$.

For each $n$ the pseudointersection $X$ of step 4.1 is almost contained in $B_n\subseteq A_n$, hence $\mathcal{F}$ has a pseudointersection.

$t\ge {\mathrm{\aleph }}_1$: a tower of length ${\mathrm{\aleph }}_0$ would be a countable descending family of infinite sets, so by step 4.2 it would have a pseudointersection, which a tower cannot have. Since $t$ is a cardinal, $t\mathrm{\ne }{\mathrm{\aleph }}_0$ is excluded, so $t\ge {\mathrm{\aleph }}_1$.

${\mathrm{\aleph }}_1\le p$: if $\mathcal{F}$ has size below ${\mathrm{\aleph }}_1$, then $\mathcal{F}$ is finite or countably infinite and, when nonempty, can be listed as a sequence with repetitions if finite; if $\mathcal{F}$ has SFIP then step 5.1 supplies a pseudointersection. Hence no family of size below ${\mathrm{\aleph }}_1$ has SFIP and lacks a pseudointersection, and since $p$ is a cardinal with an attained minimum, $p\ge {\mathrm{\aleph }}_1$.

$t\le \mathfrak{c}$ is clause [F2]. Combining steps 6.1, 1.2, 5.2 and 7.1 gives ${\mathrm{\aleph }}_1\le p\le t\le \mathfrak{c}=2^{{\mathrm{\aleph }}_0}$, and steps 5.1 and 4.2 are the two countable pseudointersection assertions. This is the statement. ∎