In fact, a more general result is true (Exercise 1 of Supplementary Exercises). Let $C$ be any set (not necessary well-ordered), let $\mathcal{F}$ be the set of all functions mapping a section of $J$ into $C$ , and let $\rho:\mathcal{F}\rightarrow C$ be a function that assign to each function from $\mathcal{F}$ an element in $C$ . Then, there exists a unique function $h:J\rightarrow C$ such that $$\begin{array}{llcrr} (\ast) & & h(x)=\rho(h|S_{x})\end{array}$$ for all $x\in J$ .

This is a more general result, as in this particular exercise we are asked to prove it for the particular case when $\rho(f:S_{x}\rightarrow C)=smallest[C-f(S_{x})]$ . This function is well-defined, because (i) we are given that there is no surjection from $J$ onto $C$ , so $C-f(S_{x})$ is nonempty, and (ii) in our case $C$ is assumed to be well-ordered, so that for every nonempty subset of $C$ there exists its smallest element.

(a) (This is an analogue to Lemma 8.2.) If $h$ and $k$ are defined and equal on a section $S_{\alpha}$ , and $\alpha$ is in both domains, then $h(\alpha)=k(\alpha)$ , because they both satisfy $(\ast)$ . Hence, if $h$ and $k$ are defined on $A$ , which is a section $S_{\beta}$ or all of $J$ , then they must be equal on $A$ : consider the set of $x$ ’s for which $h(x)=k(x)$ , note that it is inductive (use the fact that every section of $A$ is a section of $J$ ) and apply transfinite induction on $A$ . The result follows from the fact that the common domain of $h$ and $k$ must be a section or all of $J$ .

(b) (This is an analogue to Lemma 8.1.) Extend $h$ by the formula $(\ast)$ (in the specific case of $\rho$ given in the exercise, the extension is well-defined because, basically, (i) and (ii) hold; in the more general case, it is well-defined because $\rho$ is assumed to be well-defined).

(c) (This is an analogue to the second part of the proof of Theorem 8.3.) This follows from (a). Take any element $\beta$ in the union of sections, there is a section $S_{\alpha}$ (may be not unique) that contains the element and the corresponding function $h_{\alpha}$ . Define $k(\beta)=h_{\alpha}(\beta)$ (the choice of the section does not matter by (a)), and show that $k$ satisfies $(\ast)$ (which is almost identical to the last paragraph of the proof of Theorem 8.3).

(d) (This is again an analogue to Lemma 8.1, but including transfinite induction.) Consider the set of all those $\beta$ ’s for which there does not exists the function, then take its smallest element, and use the hint to show that there does exist such function. Or, use the transfinite induction: let $J_{0}$ be the set of all those $\beta$ ’s for which the function exists, then if a section $S_{\alpha}$ is in the $J_{0}$ then $\alpha\in J_{0}$ (consider the two cases again using the hint). When using the hint, the two cases correspond to (b) and (c) proved earlier.

(e) (This is an analogue to Theorem 8.3.) Existence. (d) proves that for every section there is a function satisfying $(\ast)$ . (c) proves that, therefore, there is a function satisfying $(\ast)$ on the union of all sections of $J$ . The union contains all elements of $J$ except for the largest element $m$ in $J$ , if the one exists (every other element has a larger element in the set; thanks to Fran for the comment). But if there is a largest element $m$ in $J$ , we can extend the function from the union of all sections, which is equal to $S_{m}$ , to the whole $J=S_{m}\cup\{m\}$ using (b). The uniqueness follows from (a).