Exercise 16, § 6

Every totally ordered set has a well-ordered cofinal subset (§2, Exercise 2). The least of the ordinals Ord (M) of the well-ordered cofinal subsets M of E is called the final character of E.

(a) An ordinal $\xi$ is said to be regular if it is equal to its final character, and singular otherwise. Show that every infinite regular ordinal is an initial ordinal $\omega_\alpha$ (Exercise 10). Conversely, every initial ordinal $\omega_\alpha$, whose index $\alpha$ is either 0 or has a predecessor, is a regular ordinal. An initial ordinal $\omega_\alpha$ whose index $\alpha$ has no predecessor is singular if $0 < \alpha < \omega_\alpha$; in particular, $\omega_\omega$ is the least infinite singular initial ordinal.

(b) An initial ordinal $\omega_\alpha$ is said to be inaccessible if it is regular and its index $\alpha$ has no predecessor. Show that, if $\alpha = 0$, then $\omega_\alpha = \alpha$; in other words, $\alpha$ is a critical ordinal with respect to the normal functional symbol $\omega_\eta$ (Exercise 10 (d) and 13 (c)). Let $\varkappa$ be the least critical ordinal with respect to this functional symbol. Show that $\omega_\varkappa$ is singular, with final character $\omega$ (cf. Exercise 13 (d)). In other words, there exists no inaccessible ordinal $\omega_\alpha$ such that $0 < \alpha \leqslant \varkappa$.[^1]

(c) Show that there exists only one regular ordinal which is cofinal in a given totally ordered set E; this ordinal is equal to the final character of E, and if E is not empty and has no greatest element, it is an initial ordinal. If $\omega_{\bar{\alpha}}$ is the final character of $\omega_\alpha$, then $\bar{\alpha} \leqslant \alpha$; and $\omega_\alpha$ is regular if and only if $\alpha = \bar{\alpha}$.

(d) Let $\omega_\alpha$ be a regular ordinal and let I be a well-ordered set such that Ord (I) $< \omega_\alpha$. Show that, for each family $(\xi_\iota)_{\iota \in I}$ of ordinals such that $\xi_\iota < \omega_\alpha$ for all $\iota \in I$, we have $\sum_{\iota \in I} \xi_\iota < \omega_\alpha$.

[^1]: At present it is not known whether or not there exist inaccessible ordinals other than $\omega$.