Given the choice axiom and a nonempty set $X$ , we need to construct a well-ordering on $X$ . The towers simply describe the construction of a well-ordering on $X$ "step-by-step". Take any tower $T$ , let $x_{1}$ be its smallest element, then the section of the tower by $x_{1}$ , i.e. $S_{x_{1}}(T)$ , is empty, therefore, the smallest element of any tower must be $x_{1}=c(X)$ . Now take the second to least element in $T$ , $x_{2}$ , the section $S_{x_{2}}(T)$ contains $x_{1}$ only, therefore, $x_{2}=c(X-\{x_{1}\})$ . And etc. So the proof is actually to show that the construction of the tower this way can be continued up to the whole set $X$ (using transfinite induction or recursion along the way).

(a) There is an order preserving injection at least one way between the towers such that the image set of the injection is the whole range or a subsection of the range (Exercise 4(a)). So, as the hint suggests, “switching indices if necessary, there is an order preserving $h:T_{1}\rightarrow T_{2}$ such that $h(T_{1})$ equals $T_{2}$ or its section. Let $Y$ be the set of elements $y\in T_{1}$ such that $h(y)=y$ . If $S_{x}(T_{1})\subset Y$ , $x\in T_{1}$ , then $S_{x}(T_{1})=h(S_{x}(T_{1}))=S_{h(x)}(T_{2})$ (the fact proved in Exercise 4), and, therefore, $x=c(X-S_{x}(T_{1}))$ $=c(X-S_{h(x)}(T_{2}))$ $=h(x)$ . By transfinite induction, $x=h(x)$ for all points in $T_{1}$ .

(b) Take $x=c(X-T)$ and extend the relation $<$ on $T$ to $T'=T\cup\{x\}$ by taking $<'=<$ on $T$ and $z<x$ for all $z\in T$ . Then, $T'$ is well-ordered by $<'$ , $T=S_{x}(T')$ and $z=c(X-S_{z}(T'))$ holds for all $z\in T'$ including $z=x$ .

(c) In Exercise 5(b) we proved that the union of strictly ordered collection of well-ordered subsets is a well-ordered subset. Now, take any $x\in T$ , there is a tower $(T_{k},<_{k})$ containing it. For any other tower $(T_{l,}<_{l})$ containing $x$ , either $(T_{l,}<_{l})$ is a section of $(T_{k},<_{k})$ , or $(T_{k},<_{k})$ is a section of $(T_{l,}<_{l})$ (see (a)). So, $S_{x}(T)$ equals $S_{x}(T_{k})$ , and, therefore, $x=c(X-S_{x}(T_{k}))$ $=c(X-S_{x}(T))$ . Therefore, $(T,<)$ is a tower in $X$ . If $X\neq T$ , then, by (b), there is another tower in $X$ of which $T$ is a section, which is not possible, as $T$ is the union of all towers in $X$ . Hence, $T$ is a well-ordered tower such that $T=X$ .