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 $T$ both symmetric and transitive; it is reflexive from the root solution. Therefore $T$ is an equivalence relation containing $R$ and $S$.

If $U$ is any other equivalence relation containing both, a witness $xRy,ySz$ gives $xUy,yUz$, and transitivity gives $xUz$. Hence $T\subseteq U$. Thus (iii) implies the unique-smallest assertion in (iv).