Exercise 15, § 2

(a) Let $\alpha$ and $\beta$ be two ordinals. Show that the inequality $\alpha < \beta$ is equivalent to $\alpha + 1 \leqslant \beta$, and that it implies the inequalities $\xi + \alpha < \xi + \beta$, $\alpha + \xi \leqslant \beta + \xi$, $\alpha\xi \leqslant \beta\xi$ for all ordinals $\xi$, and $\xi\alpha < \xi\beta$ if $\xi > 0$.

(b) Deduce from (a) that there exists no set to which every ordinal belongs (use Exercise 14 (d)).

(c) Let $\alpha$, $\beta$, $\mu$ be three ordinals. Show that each of the relations $\mu + \alpha < \mu + \beta$, $\alpha + \mu < \beta + \mu$ implies $\alpha < \beta$; and that each of the relations $\mu\alpha < \mu\beta$, $\alpha\mu < \beta\mu$ implies $\alpha < \beta$ provided that $\mu > 0$.

(d) Show that the relation $\mu + \alpha = \mu + \beta$ implies $\alpha = \beta$, and that $\mu\alpha = \mu\beta$ implies $\alpha = \beta$ provided that $\mu > 0$.

(e) Two ordinals $\alpha$ and $\beta$ are such that $\alpha \leqslant \beta$ if and only if there exists an ordinal $\xi$ such that $\beta = \alpha + \xi$. This ordinal $\xi$ is then unique and is such that $\xi \leqslant \beta$; it is written $(-\alpha) + \beta$.

(f) Let $\alpha$, $\beta$, $\zeta$ be three ordinals such that $\zeta < \alpha\beta$. Show that there exist two ordinals $\xi$, $\eta$ such that $\zeta = \alpha\eta + \xi$ and $\xi < \alpha$, $\eta < \beta$ (cf. no. 5, Theorem 3, Corollary 3). Moreover, $\xi$ and $\eta$ are uniquely determined by these conditions.