Past exam of the year 1 of the computer science course of the University of Oxford / 2005 / Calculus, linear algebra, discrete mathematics, logic and proof, and models of computation / 6 / b / Solution

The continuous nonalgebraic unit interval isEvery nonempty directed set has a least upper bound by completeness of the real numbers, and that least upper bound lies in $[0,1]$. Thus this is a directed-complete partial order.

For $0\le a<b\le 1$, we have $a\ll b$. Indeed, if $b\le \sup D$ and no $d\in D$ were at least $a$, then $a$ would be an upper bound for $D$, implying $\sup D\le a<b$, a contradiction. If $a>b$, the singleton directed set $\{b\}$ excludes $a\ll b$. If $a=b>0$, the increasing sequence $d_n=b(1-1/n)$, $n\ge 1$, has least upper bound $b$ and has no member at least $b$, so $b/\ll b$.

Under the usual nonempty convention, $0\ll 0$. Therefore the elements way-below $b>0$ form $[0,b)$, and those way-below $0$ form $\{0\}$. These are directed sets with least upper bounds $b$ and $0$, respectively, proving that $[0,1]$ is a continuous directed-complete partial order. Its only compact element is $0$. For any $b>0$, the compact elements below $b$ have least upper bound $0$, rather than $b$, so it is not an algebraic partially ordered set.

Under the literal empty-set convention of the paper, $0/\ll 0$ as well. The elements way-below zero then form the empty set, which is directed in that convention and has least upper bound zero. There are no compact elements, and their empty set still fails to approximate any positive $b$. Thus both conventions give the same requested classification: