Exercise 10, § 6

Let $\omega$ or $\omega_0$ denote the ordinal Ord (N) (§ 2, Exercise 14). The set of all integers is then a well-ordered set isomorphic to the set of all ordinals $< \omega$. For each integer $n$ we denote again by $n$ (by abuse of language) the ordinal Ord $([0, n[)$.

(a) Show that for each cardinal $\mathfrak{a}$ the relation "$\xi$ is an ordinal and Card $(\xi) < \mathfrak{a}$" is collectivizing (use Zermelo's theorem). Let $W(\mathfrak{a})$ denote the set of all ordinals $\xi$ such that Card $(\xi) < \mathfrak{a}$.

(b) For each ordinal $\alpha > 0$ define a function $f_\alpha$ on the well-ordered set $O'(\alpha)$ of ordinals $\leqslant \alpha$ by transfinite induction as follows : $f_\alpha(0) = \omega_0 = \omega$, and for each ordinal $\xi$ such that $0 < \xi \leqslant \alpha$, $f_\alpha(\xi)$ is the least upper bound (§ 2, Exercise 14 (d)) of the set of ordinals $\zeta$ such that $\mathrm{Card}(\zeta) \leqslant \mathrm{Card}(f_\alpha(\eta))$ for at least one ordinal $\eta < \xi$. Show that, if $0 \leqslant \eta < \xi \leqslant \alpha$, then $\mathrm{Card}(f_\alpha(\eta)) < \mathrm{Card}(f_\alpha(\xi))$ and that, if $\xi \leqslant \alpha \leqslant \beta$, then $f_\alpha(\xi) = f_\beta(\xi)$. Put $\omega_\alpha = f_\alpha(\alpha)$; $\omega_\alpha$ is said to be the initial ordinal with index $\alpha$. We have $\omega_\alpha \geqslant \alpha$. Put $\aleph_\alpha = \mathrm{Card}(\omega_\alpha)$; $\aleph_\alpha$ is said to be the aleph of index $\alpha$. In particular, $\aleph_0 = \mathrm{Card}\ (\mathbf{N})$.

(c) Show that for each infinite cardinal $\mathfrak{a}$ the least upper bound $\lambda$ of the set of ordinals $W(\mathfrak{a})$ is an initial ordinal $\omega_\alpha$, and that $\mathfrak{a} = \aleph_\alpha$ (consider the least ordinal $\mu$ such that $\omega_\mu \geqslant \lambda$); in other words, $\omega_\alpha$ is the least ordinal $\xi$ such that $\mathrm{Card}(\xi) = \aleph_\alpha$. For each ordinal $\alpha$ the mapping $\xi \rightarrow \aleph_\xi$, defined on $O'(\alpha)$, is an isomorphism of the well-ordered set $O'(\alpha)$ onto the well-ordered set of cardinals $\leqslant \aleph_\alpha$; in particular, $\aleph_{\alpha+1}$ is the least cardinal $> \aleph_\alpha$. Show that, if $\alpha$ has no predecessor, then for every strictly increasing mapping $\xi \rightarrow \sigma_\xi$ of an ordinal $\beta$ into $\alpha$ such that $\alpha = \sup_{\xi < \beta} \sigma_\xi$, we have

(d) Deduce from (c) that $\omega_\xi$ is a normal ordinal functional symbol (§ 2, Exercise 17).