Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/ia/paper-4/5e/iii/solution

The inverse argument in part (i) proves that the one-sided inclusion in (iii) already implies equality:
Parts (i) and (ii) then make both symmetric and transitive; it is reflexive from the root solution. Therefore is an equivalence relation containing and .
If is any other equivalence relation containing both, a witness gives , and transitivity gives . Hence . Thus (iii) implies the unique-smallest assertion in (iv).

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