= Solution
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 $u\geq1$ and $v\geq1$, since the two coordinate projections of the open <unit disc> are $(-1,1)$. Therefore $(1,1)\leq_P(u,v)$, proving \b[$\sup_{\leq_P}S=(1,1)$].
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 \b[there is no <least upper bound> under the <lexicographic order>]. This illustrates the <supremum of an open disc under product and lexicographic orders>.
\Image[/past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2018/discrete-mathematics-open-disc-orders.png]
{title=The same open <unit disc> has a <supremum> under the <product order> and none under the <lexicographic order>}
{description=The green corner consists of product-order <upper bounds>. Every point on the red vertical line is a lexicographic <upper bound>; moving down that line always finds a smaller one.}
Back to article page
