Exercise 25, § 6

(a) Let $\mathfrak{F}$ be a covering of an infinite set E. The degree of disjointness of $\mathfrak{F}$ is the least cardinal $\mathfrak{c}$ such that $\mathfrak{c}$ is strictly greater than the cardinals Card $(X \cap Y)$ for each pair of distinct sets X, $Y \in \mathfrak{F}$. If Card $(E) = \mathfrak{a}$ and Card $(\mathfrak{F}) = \mathfrak{b}$, show that $\mathfrak{b} \leqslant \mathfrak{a}^{\mathfrak{c}}$ (note that a subset of E of cardinal $\mathfrak{c}$ is contained in at most one set of $\mathfrak{F}$).

(b) Let $\omega_\alpha$ be an initial ordinal and let F be a set such that $2 \leqslant \mathfrak{p} =$ Card $(F) < \aleph_\alpha$. Let E be the set of mappings of segments of $\omega_\alpha$, other than $\omega_\alpha$ itself, into F. Then we have Card $(E) \leqslant \mathfrak{p}^{\aleph_\alpha}$. For each mapping $f$ of $\omega_\alpha$ into F, let $K_f$ be the subset of E consisting of the restrictions of $f$ to the segments of $\omega_\alpha$ (other than $\omega_\alpha$ itself). Show that the set $\mathfrak{F}$ of subsets $K_f$ is a covering of E such that Card $(\mathfrak{F}) = \mathfrak{p}^{\aleph_\alpha}$, and that its degree of disjointness is equal to $\aleph_\alpha$.

(c) Let E be an infinite set of cardinal $\mathfrak{a}$, and let $\mathfrak{c}$, $\mathfrak{p}$ be two cardinals $> 1$ such that $\mathfrak{p} < \mathfrak{c}$, $\mathfrak{p}^{\mathfrak{m}} < \mathfrak{a}$ for all $\mathfrak{m} < \mathfrak{c}$, and $\mathfrak{a} = \sum_{\mathfrak{m} < \mathfrak{c}} \mathfrak{p}^{\mathfrak{m}}$. Deduce from (b) that there exists a covering $\mathfrak{F}$ of E consisting of sets of cardinal $\mathfrak{c}$, with degree of disjunction equal to $\mathfrak{c}$, and such that Card $(\mathfrak{F}) = \mathfrak{p}^{\mathfrak{c}}$. In particular, if E is countably infinite, there exists a covering $\mathfrak{F}$ of E by infinite sets such that Card $(\mathfrak{F}) = 2^{\aleph_0}$ and such that the intersection of any two sets of $\mathfrak{F}$ is finite.