Exercise 17, § 2

Given an ordinal $\alpha_0$, a term $f(\xi)$ is said to be an ordinal functional symbol (with respect to $\xi$) defined for $\xi \geqslant \alpha_0$ if the relation "$\xi$ is an ordinal and $\xi \geqslant \alpha_0$" implies the relation "$f(\xi)$ is an ordinal"; $f(\xi)$ is said to be normal if the relation $\alpha_0 \leqslant \xi < \eta$ implies $f(\xi) < f(\eta)$ and if, for each family $(\xi_\iota)_{\iota \in I}$ of ordinals $\geqslant \alpha_0$, we have $\sup_{\iota \in I} f(\xi_\iota) = f(\sup_{\iota \in I} \xi_\iota)$ (cf. Exercise 14 (d)).

(a) Show that, for each ordinal $\alpha > 0$, $\alpha + \xi$ and $\alpha\xi$ are normal functional symbols defined for $\xi \geqslant 0$ (use Exercise 15 (f)).

(b) Let $w(\xi)$ be an ordinal functional symbol defined for $\xi \geqslant \alpha_0$ such that $w(\xi) \geqslant \xi$ and such that $\alpha_0 \leqslant \xi < \eta$ implies $w(\xi) < w(\eta)$. Also let $g(\xi, \eta)$ be a term such that the relation "$\xi$ and $\eta$ are ordinals $\geqslant \alpha_0$" implies the relation "$g(\xi, \eta)$ is an ordinal such that $g(\xi, \eta) > \xi$".

Define a term $f(\xi, \eta)$ with the following properties : (1) for each ordinal $\xi \geqslant \alpha_0$, $f(\xi, 1) = w(\xi)$; (2) for each ordinal $\xi \geqslant \alpha_0$ and each ordinal $\eta > 1$, $f(\xi, \eta) = \sup_{0 < \zeta < \eta} g(f(\xi, \zeta), \xi)$ (use criterion C60 of no. 2). Show that if $f_1(\xi, \eta)$ is another term with these two properties, then $f(\xi, \eta) = f_1(\xi, \eta)$ for all $\xi \geqslant \alpha_0$ and all $\eta \geqslant 1$. Prove that, for each ordinal $\xi \geqslant \alpha_0$, $f(\xi, \eta)$ is a normal functional symbol with respect to $\eta$ (defined for all $\eta \geqslant 1$). Show that $f(\xi, \eta) \geqslant \xi$ for all $\eta \geqslant 1$ and $\xi \geqslant \alpha_0$, and that $f(\xi, \eta) \geqslant \eta$ for all $\xi \geqslant \sup(\alpha_0, 1)$ and $\eta \geqslant 1$. Furthermore, for each pair $(\alpha, \beta)$ of ordinals such that $\alpha > 0$, $\alpha \geqslant \alpha_0$, and $\beta \geqslant w(\alpha)$ there exists a unique ordinal $\xi$ such that

and we have $\xi \leqslant \beta$.

(c) If we take $\alpha_0 = 0$, $w(\xi) = \xi + 1$, $g(\xi, \eta) = \xi + 1$, then $f(\xi, \eta) = \xi + \eta$. If we take $\alpha_0 = 1$, $w(\xi) = \xi$, $g(\xi, \eta) = \xi + \eta$, then $f(\xi, \eta) = \xi\eta$.

(d) Show that if the relations $\alpha_0 \leqslant \xi \leqslant \xi'$, $\alpha_0 \leqslant \eta \leqslant \eta'$ imply $g(\xi, \eta) \leqslant g(\xi', \eta')$, then the relations $\alpha_0 \leqslant \xi \leqslant \xi'$, $1 \leqslant \eta \leqslant \eta'$ imply $f(\xi, \eta) \leqslant f(\xi', \eta')$. If the relations $\alpha_0 \leqslant \xi \leqslant \xi'$, $\alpha_0 \leqslant \eta < \eta'$ imply $g(\xi, \eta) < g(\xi, \eta')$ and $g(\xi, \eta) \leqslant g(\xi', \eta)$, then the relations $\alpha_0 \leqslant \xi < \xi'$ and $\eta \geqslant 0$ imply $f(\xi, \eta + 1) < f(\xi', \eta + 1)$.

(e) Suppose that $w(\xi) = \xi$ and that the relations $\alpha_0 \leqslant \xi \leqslant \xi'$, $\alpha_0 \leqslant \eta < \eta'$ imply $g(\xi, \eta) < g(\xi, \eta')$ and $g(\xi, \eta) \leqslant g(\xi', \eta)$. Suppose, moreover, that for each $\xi \geqslant \alpha_0$, $g(\xi, \eta)$ is a normal functional symbol with respect to $\eta$ (defined for $\eta \geqslant \alpha_0$), and that, whenever $\xi \geqslant \alpha_0$, $\eta \geqslant \alpha_0$, and $\zeta \geqslant \alpha_0$, we have the associativity relation

Show that, if $\xi \geqslant \alpha_0$, $\eta \geqslant 1$, and $\zeta \geqslant 1$, then

("distributivity" of $g$ with respect to $f$) and

("associativity" of $f$).