(a) No $(A,<)$ equals its own section, and if $(A,<)$ equals a section of $(B,<')$ , which equals a section of $(C,<")$ , then $(A,<)$ equals a section of $(C,<")$ . So, non-reflexivity and transitivity hold.

(b) Let $x_{1},x_{2}\in(B',<')$ . Then, for $i\in\{1,2\}$ there exists $x_{i}\in(A_{i},<_{i})$ . Without loss of generality, suppose $(A_{2},<_{2})$ equals a section of $(A_{1},<_{1})$ . Then, both points $x_{1}$ and $x_{2}$ are in $(A_{1},<_{1})$ , and at least one of $x_{1}<_{1}x_{2}$ , $x_{1}=x_{2}$ and $x_{2}<_{1}x_{1}$ holds. Hence, we have comparability. Further, $x<x$ does not hold in any $(A,<)\in\mathcal{B}$ , hence, $x<'x$ does not hold, and $<'$ is irreflexive. For any three elements $x_{1},x_{2},x_{3}\in(B',<')$ such that $x_{1}<'x_{2}<'x_{3}$ , consider $(A_{i},<_{i}){}_{i=1,2}$ such that $x_{1}<_{1}x_{2}$ and $x_{2}<_{2}x_{3}$ , and suppose that $(A_{3-i},<_{3-i})$ is a section of $(A_{i},<_{i})$ ($i$ equals $1$ or $2$ ), then $x_{1}<_{i}x_{2}<_{i}x_{3}$ , implying $x_{1}<_{i}x_{3}$ and $x_{1}<'x_{3}$ , hence, $<'$ is transitive. To sum up, $<'$ is an order on $B'$ . For any nonempty subset $A$ of $B'$ , let $x\in A$ , and $(B,<)\in\mathcal{B}$ be such that $x\in(B,<)$ . Then, for $y\in A$ , $y<'x$ iff $y<x$ and $y\in B$ , and the $<$ -smallest element of $(B\cap A,<)$ is the $<'$ -smallest element of $(A,<')$ .