Exercise 23, § 6

Let $\alpha$ be an ordinal $> 0$. A mapping $f$ of the ordinal $\alpha$ into itself is said to be divergent if for each ordinal $\lambda_0 < \alpha$ there exists an ordinal $\mu_0 < \alpha$ such that the relation $\mu_0 \leqslant \xi < \alpha$ implies $\lambda_0 \leqslant f(\xi) < \alpha$ [^1].

(a) Let $\varphi$ be a strictly increasing mapping of an ordinal $\beta$ into $\alpha$ such that

and such that

[^2]

Then there exists a divergent mapping $f$ of $\alpha$ into itself, such that $f(\xi) < \xi$ for all $\xi$ satisfying $0 < \xi < \alpha$, if and only if there exists a divergent mapping of $\beta$ into itself of the same type.

(b) Deduce from (a) that there exists a divergent mapping of $\omega_\alpha$ into itself, such that $f(\xi) < \xi$ for all $\xi$ satisfying $0 < \xi < \alpha$, if and only if the final character of $\omega_\alpha$ is $\omega_0$. (If $\omega_\alpha$ is a regular ordinal $> \omega_0$, define inductively a strictly increasing sequence $(\eta_n)$ as follows : $\eta_1 = 1$, and $\eta_{n+1}$ is the least ordinal $\zeta$ such that $f(\xi) > \eta_n$ for all $\xi \geqslant \zeta$.)

(c) Let $\omega_{\bar{\alpha}}$ be the final character of $\omega_\alpha$ (Exercise 16). Show that, if $\bar{\alpha} > 0$ and if $f$ is a mapping of $\omega_\alpha$ into itself such that $f(\xi) < \zeta$ for all $\xi$ such that $0 < \xi < \omega_\alpha$, then there exists an ordinal $\lambda_0$ such that the set of solutions of the equation $f(\xi) = \lambda_0$ has a cardinal $\geqslant \aleph_{\bar{\alpha}}$.

[^1]: If the well-ordered set $O'_\alpha$ of ordinals $\leqslant \alpha$ is endowed with the topology $\mathscr{T}_-(O'_\alpha)$ (General Topology, Chapter I, § 2, Exercise 5), this condition may be written $\lim_{\xi \to \alpha, \, \xi < \alpha} f(\xi) = \alpha$. [^2]: If we extend $\varphi$ to $O'_\beta$ by defining $\varphi(\beta) = \alpha$, the conditions above signify that $\varphi$ is continuous with respect to the topologies $\mathscr{T}_-(O'_\alpha)$ and $\mathscr{T}_-(O'_\beta)$ on $O'_\alpha$ and $O'_\beta$ respectively (loc. cit.).