Past exam of the year 1 of the computer science course of the University of Oxford / 2021 / Discrete mathematics, continuous mathematics, and probability / 2 / c

Take $u=(1,1)$ and $v=(1,2)$. Their first coordinates are equal, so both $uRv$ and $vRu$ hold, although $u\ne v$. Thus $R$ is not an antisymmetric relation, and hence it is not a partial order. The failure arises because its comparisons ignore the second coordinate entirely.

Check each partial order axiom. For every $(a,b)$, $a\le a$ and $b\ge b$, so $S$ is a reflexive relation. If both $(a,b)S(c,d)$ and $(c,d)S(a,b)$, the first coordinates satisfy $a\le c\le a$ and the second coordinates satisfy $b\ge d\ge b$. Therefore $a=c$ and $b=d$, which proves $S$ is an antisymmetric relation.

Finally, if $(a,b)S(c,d)$ and $(c,d)S(e,f)$, then $a\le c\le e$ and $b\ge d\ge f$. Hence $(a,b)S(e,f)$, making $S$ a transitive relation. All three axioms hold, so $S$ is a partial order. It is a product order, with the usual order on the first coordinate and the reversed order on the second.