Exercise 16, § 2

An ordinal $\rho > 0$ is said to be indecomposable if there exists no pair of ordinals $\xi$, $\eta$ such that $\xi < \rho$, $\eta < \rho$, and $\xi + \eta = \rho$.

(a) An ordinal $\rho$ is indecomposable if and only if $\xi + \rho = \rho$ for every ordinal $\xi$ such that $\xi < \rho$.

(b) If $\rho > 1$ is an indecomposable ordinal and if $\alpha$ is any ordinal $> 0$, then $\alpha\rho$ is indecomposable, and conversely (use Exercise 15 (f)).

(c) If $\rho$ is indecomposable and if $0 < \alpha < \rho$, then $\rho = \alpha\xi$, where $\xi$ is an indecomposable ordinal (use Exercise 15 (f)).

(d) Let $\alpha$ be an ordinal $> 0$. Show that there exists a greatest indecomposable ordinal among the indecomposable ordinals $\leqslant \alpha$ (consider the decompositions $\alpha = \rho + \xi$, where $\rho$ is indecomposable).

(e) If E is any set of indecomposable ordinals, deduce from (d) that the least upper bound of E (Exercise 14 (d)) is an indecomposable ordinal.