4. Let 1,...,N be preference relations, and dene by a: 3; if, and only if :1: , 3; for every :2 = 1, . . . ,N. o Is complete? a Is transitive? Answer: is transitive but not complete. is transitive: By assumption, each , is transitive, so if a: , 3; and 3; , 3 for all i = 1,2, . . . ,n (equivalently, if a: 3; and 3; 2;), then 3: g 2 for all i : 1,2,...,n, which is equivalent to 3: 2;. is not complete: The simplest way to see this is a counterexample with N = 2 and X : {ray}. As 1 is complete, either a: 1 y or y l a: must be true (or possibly both). Suppose 3: 1 3;. Since g is complete, either a: g 3; or y g :1: (or possibly both). There are three possible cases: (1) .1: g 3,: and 3; g 3:, (2) a: g 3; and :(y g 33), which is equivalent to 3: >2 3;, and (3) I(:}3 g 3;) and :1; g :13, which is equivalent to y >g 3:. In case (3), as dened in this problem will not be complete. The argument for y l a: is analogous