Strong Finite Intersection Property and Pseudo-intersection (Kunen III.1.23)

Question

I have a question on the proof of a lemma in Kunen's Set Theory regarding pseudo-intersections and the cardinal $\mathfrak{p}$ (least size of a family $\mathcal{F} \subseteq [\omega]^{\omega}$ which has SFIP and no pseudo-intersection).

Lemma III.1.23 $\,\,\,\,$Fix $\mathcal{F}, \mathcal{H} \subseteq [\omega]^{\omega}$ such that $|\mathcal{F}|, |\mathcal{H}| < \mathfrak{p}$ and assume for all $H\in \mathcal{H}$, the set $\{Z \cap H: Z\in \mathcal{F}\}$ has the SFIP (strong finite intersection property). Then $\mathcal{F}$ has a pseudo-intersection $K$ such that $K\cap H$ is infinite for all $H\in \mathcal{H}$.

The following the proof given in the text.

Let $J= [\omega]^{<\omega}-\{\emptyset\}$ be the set of nonempty finite subsets of $\omega$. For each $Z\in \mathcal{F}$ define $\hat{Z} = \{s\in J: s\subseteq Z\}$. For each $H\in \mathcal{H}$ define $\tilde{H} = \{s\in J: s\cap H \neq\emptyset\}$. For each $l\in \omega$ define $T_{l} = \{s\in J: \text{min}(s) > l\}$. Then define $$\mathcal{F}^* = \{\hat{Z}:Z\in \mathcal{F}\} \cup\{\tilde{H}:H\in \mathcal{H}\}\cup \{T_l: l\in \omega\} \subseteq \mathcal{P}(J).$$ Then $\mathcal{F}^*$ has the SFIP: that is the a finite intersection $$I = \hat{Z_1}\cap\cdots\cap\hat{Z_n}\cap\tilde{H_1}\cap\cdots\cap\tilde{H_m}\cap T_l$$ is infinite by the following. By assumption of the lemma, $Z_1\cap\cdots\cap Z_n$ meets each $H_i$ in an infinite set, so there are $s\in J$ with arbitrarily large minimums so that $s \subseteq Z$ and $s$ meets each $H_i$. Since $|\mathcal{F}^*| < \mathfrak{p}$, fix an infinite $M\subseteq J$ such that $M$ is a pseudo-intersection of $\mathcal{F}^*$. Let $K = \bigcup M\subseteq \omega$. Then $K$ is infinite because $M$ is infinite. For each $Z\in \mathcal{F}$: $M\subseteq^{*}\hat{Z}$ implies that $K\subseteq^* Z$, so $K$ is a pseudo-intersection of $\mathcal{F}$.

To prove that $K\cap H$ is infinite for $H\in \mathcal{H}$, fix any $l\in \omega$. $M\subseteq^*\tilde{H}\cap T_l$, so fix an $s\in M$ with $s\in \tilde{H}\cap T_l$. Then $s\cap H \neq\emptyset$ and $\text{min}(s) > l$, so $s\cap H \subseteq\bigcup M = K$ implies that $K\cap H$ contains a number larger than $l$. Since $l$ was arbitrary, $K\cap H$ is infinite.

I understand everything in the proof pretty well. However I do have one question. When defining $\hat{Z}, \tilde{H}$ and $T_l$, why do we require them to only contain finite sets. I cannot find a part of this proof that uses the fact the these are sets of finite sets. Would this proof still hold if we defined $\hat{Z}, \tilde{H}$ and $T_l$ to contain infinite sets of numbers too? If not what part of the proof uses the fact that they are finite.

Answer 1

The reason why using $J = [\omega]^{<\omega} - \{\emptyset\}$ is necessary is because it is countable infinite. Recall that $\mathfrak{p}$ is defined such that if $\mathcal{F} \subseteq [\omega]^\omega$ with SFIP and $|\mathcal{F}|< \mathfrak{p}$ then $\mathcal{F}$ has a pseudo-intersection, i.e. it is defined for a family of countably infinite subsets of $\omega$. We define $J$ so that it is countably infinite and so we are actually allowed to use the definition of $\mathfrak{p}$. Because $|\omega| = |J|$ we may use a bijection to apply the definition of $\mathfrak{p}$ and find such a pseudo-intersection for $\mathcal{F}^*$. If we defined $J= \mathcal{P}(\omega)$ then $|J| = 2^{\aleph_0}$ and we cannot apply the definition of $\mathfrak{p}$.

Answer 2

We need only finite subsets (from $J$) to "force" the pseudo-intersection of $\mathcal{F}^\ast$ to have the right property and to have this family has the SFIP in the first place.

I'll give the proof that $I$ is infinite in even more detail, I think I see a subtle use of finiteness there:

Let $k \in \omega$ arbitrary (a pregiven lower bound introduced by me, unrelated to $l$). The sets $Z_1 \cap \ldots Z_n \cap H_1$, $Z_1 \cap \ldots Z_n \cap H_2$ up to $Z_1 \cap \ldots Z_n \cap H_m$ are infinite subsets of $\omega$ by the assumptions of the lemma. So pick $k_1 \in Z_1 \cap \ldots Z_n \cap H_1$ such that $k_1 > k$ and $k_1 > l$. Also pick $k_2 \in Z_1 \cap \ldots Z_n \cap H_2$ such that $k_2 > k$ and $k_2 > l$ and (after finitely many steps, so we avoid choice here) $k_m \in Z_1 \cap \ldots Z_n \cap H_m$ with $k_m > k$ and $k_m >l$. Then $s:=\{k_1, k_2,\ldots, k_m\} \in J$ and it's clear that: $s \in \hat{Z_i}$ for all $i=1,\ldots,n$ and $s \in \tilde{H}_j$ for all $j \in \{1,\ldots,m\}$ and $s \in T_l$, all by construction. And because we ensured that the minimum of $s$ is beyond "my" $k$ as well, we see that $I$ is infinite (as a finite number of $s$ in $I$ could never all lie beyond any arbitrary bound, consider the max of their maxima which exists when all $s$ are finite, as is the case).

I think that's the (admittedly hidden) reason why we consider the finite conditions $s \in J$ (they're much easier to reason with) instead of infinite ones only, say.