Exercise 14, § 2

An ordinal is the order-type of a well-ordered set (Exercise 13).

(a) Show that, if $(\lambda_\iota)_{\iota \in I}$ is a family of ordinals indexed by a well-ordered set $I$, then the ordinal sum $\sum_{\iota \in I} \lambda_\iota$ is an ordinal; \* and that if, moreover, $I$ is finite, the ordinal product $\mathbf{P}_{\iota \in I} \lambda_\iota$ is an ordinal (Exercise 11). \* The order-type of the empty set is denoted by 0, and that of a set with one element by 1 (by abuse of language, cf. § 3). Show that

for every ordinal $\alpha$.

(b) Show that the relation "$\lambda$ is an ordinal and $\mu$ is an ordinal and $\lambda \prec \mu$" is a well-ordering relation, denoted by $\lambda \leqslant \mu$. (Note that, if $\lambda$ and $\mu$ are ordinals, the relation $\lambda \prec \mu$ is equivalent to "$\lambda$ is equal to the order-type of a segment of $\mu$" (no. 5, Theorem 3, Corollary 3): given a family $(\lambda_\iota)_{\iota \in I}$ of ordinals, consider a well-ordering on $I$ and take the ordinal sum of the family of sets ordered by the $\lambda_\iota$; finally, use Proposition 2 of no. 1.)

(c) Let $\alpha$ be an ordinal. Show that the relation "$\xi$ is an ordinal and $\xi \leqslant \alpha$" is collectivizing in $\xi$, and that the set $O_\alpha$ of ordinals $< \alpha$ is a well-ordered set such that $\mathrm{Ord}\,(O_\alpha) = \alpha$. We shall often identify $O_\alpha$ with $\alpha$.

(d) Show that for every family of ordinals $(\xi_\iota)_{\iota \in I}$ there exists a unique ordinal $\alpha$ such that the relation "$\lambda$ is an ordinal and $\xi_\iota \leqslant \lambda$ for all $\iota \in I$" is equivalent to $\alpha \leqslant \lambda$. By abuse of language, $\alpha$ is called the least upper bound of the family of ordinals $(\xi_\iota)_{\iota \in I}$, and we write $\alpha = \sup_{\iota \in I} \xi_\iota$ (it is the greatest element of the union of $\{\alpha\}$ and the set of the $\xi_\iota$). The least upper bound of the set of ordinals $\xi < \alpha$ is either $\alpha$ or an ordinal $\beta$ such that $\alpha = \beta + 1$; in the latter case, $\beta$ is said to be the predecessor of $\alpha$.