Exercise 33, § 6

Let $a, b$ be two cardinals such that $a \geq 2, b \geq 1$, at least one of the two being infinite. Let E be a set, $\mathfrak{F}$ a subset of $\mathfrak{P}(E)$, such that $\mathrm{Card}(\mathfrak{F}) > a^b$ and $\mathrm{Card}(X) \leq b$ for every $X \in \mathfrak{F}$. We propose to show that there exists a subset $\mathfrak{G} \subset \mathfrak{F}$ such that $\mathrm{Card}(\mathfrak{G}) > a^b$ and that any two of the sets belonging to $\mathfrak{G}$ have the same intersection. One may proceed as follows: a) Let c be the smallest of the cardinals $> a^b$, and let $\Omega$ be the smallest ordinal of cardinal c. We consider an injective mapping $v \mapsto X(v)$ from $\Omega$ into $\mathfrak{F}$, and we put $M = \bigcup_{v \in \Omega} X(v)$; we may suppose that $\mathrm{Card}(M) = c$, and there is therefore a bijection $v \mapsto x_v$ of $\Omega$ onto M, ordering M. b) For every $v \in \Omega$, let $\rho_v$ be the order type of the subset $X(v)$ of M (we have $\mathrm{Card}(\rho_v) \geq b$) and let $\mu \mapsto y_\mu^{(v)}$ be the unique increasing bijective mapping from $\rho_v$ onto $X(v)$. We denote by $M_\mu$ the set of the $y_\mu^{(v)}$ when $v$ runs through $\Omega$. Now prove that there exists at least one ordinal $\mu$ such that $\mathrm{Card}(M_\mu) = c$. We designate by $\alpha$ the smallest of these ordinals; the union of the $M_\gamma$ for $\gamma < \alpha$ has a cardinal $\leq a^b < c$. c) Now prove that there exists a subset $N_0 \subset \Omega$ such that $\mathrm{Card}(N_0) = c$ and that the mapping $v \mapsto y_\alpha^{(v)}$ from $N_0$ into M is injective. Now prove, by induction on $\beta$, that there exists a subset $N_\beta \subset N_0$ of cardinal c such that the element $y_\alpha^{(v)} = z_\lambda$ is independent of $v$ for $v \in N_\beta$ and for every $\lambda \leq \beta$. Now prove that the intersection N of the $N_\beta$ for $\beta < \alpha$ has cardinal c (consider its complement). Let Q be the set of the $z_\lambda$ for $\lambda < \alpha$. d) For every $v \in \mathbf{N}$, we define by induction an ordinal $\lambda_v$ by the following condition: it is the smallest ordinal in $\mathbf{N}$ such that $y_{\alpha}^{(\lambda_v)}$ is a strict upper bound, in $M$, of the union of the $X(\lambda_\mu)$ for $\mu < v, \mu \in \mathbf{N}$. Now prove that for $\mu < v$ in $\mathbf{N}$, we have $X(\lambda_\mu) \cap X(\lambda_v) = \mathbf{Q}$.