Given the maximum principle and a nonempty set $X$ we need to construct a well-ordering on $X$ . Consider the strict partial order on the collection $\mathcal{A}$ of well-ordered subsets of $X$ as in Exercise 5. The collection $\mathcal{A}$ is not empty as, for example, there are finite well-ordered subsets of $X$ . By the maximum principle, there exists a maximal ordered subcollection $\mathcal{B}$ of $\mathcal{A}$ . Take the union of all subsets of $X$ in $\mathcal{B}$ , a well-ordered subset $(B',<')$ of $X$ (Exercise 5(b)). If there is $x\notin B'$ , then $(B'\cup\{x\},<")$ , where $<"$ is the extension of $<'$ defined by $b<x$ for all $b\in B'$ , is a well-ordered subset of $X$ , and it is strictly $\prec$ -greater than any subset in $\mathcal{B}$ , as $B'=S_{x}$ . But this contradicts the maximality of $\mathcal{B}$ , as $(B'\cup\{x\},<")\notin\mathcal{B}$ . So, $X=B'$ and $<'$ is a well-ordering on $X$ .

Given the well-ordering theorem and a nonempty set $X$ together with a strict partial order $<$ , we need to prove that there is a maximal ordered subset of $X$ . Let $\prec$ be a well-ordering on $X$ , and by the general principle of recursive definition (Exercise 1), we construct $h:X\rightarrow\mathcal{P}(X)$ using $$\rho(f:S_{x}\rightarrow\mathcal{P}(X))=\begin{cases} f(S_{x})\cup\{x\} & \mbox{if }f(S_{x})\cup\{x\}\mbox{ is }<\mbox{-ordered},\\ f(S_{x}) & \mbox{otherwise}. \end{cases}$$ Then, $h(\alpha)=\rho(h|S_{\alpha})=h(S_{\alpha})\cup\{\alpha\}$ if it is $<$ -ordered, and $h(S_{\alpha})$ otherwise (sections are all with respect to $\prec$ ). The idea is exactly as in the text: take all elements one-by-one in some order, and see if you can add each one to the existing ordered subset so that the resulting subset is still ordered. Note, that $h(\alpha)$ is an $<$ -ordered subset of $X$ , and $h(S_{\alpha})$ is always a suborder of $h(\alpha)$ . Hence, the union $Z=\cup_{\alpha\in X}h(\alpha)$ is an ordered subset of $X$ (this is similar but even either than Exercise 5(b)). Further, it is a maximal ordered subset of $X$ . Indeed, if some $x\notin Z$ , then $h(S_{x})\cup\{x\}\subset Z\cup\{x\}$ is not ordered, neither is $Z\cup\{x\}$ .