(a) At least one of the three conditions holds. Consider the well-ordered union $C$ of $A$ and $B$ as in Exercise 8 of §10, where every element of $A$ is less than every element of $B$ . Then, the identity function is an order preserving mapping from $B$ into $C$ . Using Exercise 3, there is an order preserving bijection $h$ from $B$ onto a section $S_{x}$ of $C$ or onto $C$ itself. If $x$ is in $A$ , then $S_{x}$ is a subset of $A$ , and $B$ has the order type of a section of $A$ . If $x$ equals the smallest element $z$ of $B$ , then $S_{x}=S_{z}$ equals $A$ , and $B$ has the order type of $A$ . Otherwise, there is $b\in B$ such that $h(b)=z$ , and, according to Exercise 2(a), $h(S_{b})=S_{h(b)}=S_{z}=A$ , implying there is an order preserving bijection of $A$ with the section $S_{b}$ of $B$ .

No more than one of the three conditions hold. If $A$ has the order type of a section of $B$ (there is an order preserving bijection $h:A\rightarrow S_{b}\subset B$ ), then $A$ and $B$ cannot have the same order type (there is an order preserving bijection $g:B\rightarrow A$ ), and $B$ cannot have the order type of a section of $A$ (there is an order preserving bijection $g:B\rightarrow S_{a}\subset A$ ), as in both cases this would imply that there is an order preserving bijection $g\circ h:A\rightarrow S_{g(b)}$ of $A$ with the section $S_{g(b)}$ of $A$ , which is not possible according to Exercise 2(b).

(b) This follows from (a) and the fact that there is no bijection of a countable set with an uncountable set.