Exercise 24, § 6

Let $\mathfrak{F}$ be a set of subsets of a set E such that for each $A \in \mathfrak{F}$ we have Card $(A) =$ Card $(\mathfrak{F}) = \mathfrak{a} \geqslant \aleph_0$. Show that E has a subset P such that Card $(P) = \mathfrak{a}$ and such that no set of $\mathfrak{F}$ is contained in P. (If $\mathfrak{a} = \aleph_\alpha$, define by transfinite induction two injective mappings $\xi \to f(\xi)$, $\xi \to g(\xi)$ of $\omega_\alpha$ into E such that the sets $P = f(\omega_\alpha)$ and $Q = g(\omega_\alpha)$ do not intersect and such that each of them meets every subset $A \in \mathfrak{F}$.)

(b) Suppose, moreover, that for each subset $\mathfrak{G}$ of $\mathfrak{F}$ such that Card $(\mathfrak{G}) < \mathfrak{a}$, the complement in E of the union of the sets $A \in \mathfrak{G}$ has cardinal $\geqslant \mathfrak{a}$. Show that E then has a subset P such that Card $(P) = \mathfrak{a}$ and such that, for each $A \in \mathfrak{G}$, Card $(P \cap A) < \mathfrak{a}$ (similar method).