Past exam of the year 1 of the computer science course of the University of Oxford / 2018 / Discrete mathematics, continuous mathematics, and probability / 3 / d

Under the product order, every point of the open unit disc has both coordinates below one, so $(1,1)$ is an upper bound. Any upper bound $(u,v)$ must have $u\ge 1$ and $v\ge 1$, since the two coordinate projections of the open unit disc are $(-1,1)$. Therefore $(1,1){\le }_P(u,v)$, proving ${\sup }_{{\le }_P}S=(1,1)$.

Under the lexicographic order, every point $(1,v)$, with arbitrary real $v$, is an upper bound, because every point of $S$ has first coordinate strictly below one. Conversely, an upper bound cannot have first coordinate $u<1$: a point $(x,0)\in S$ with $x>u$ would exceed it. If an upper bound has $u>1$, the upper bound $(1,0)$ is strictly smaller. If it is $(1,v)$, the upper bound $(1,v-1)$ is strictly smaller. Hence there is no least upper bound under the lexicographic order. This illustrates the supremum of an open disc under product and lexicographic orders.