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A partially ordered set in which the compact elements below each element form a directed set with that element as their least upper bound.

Let $K_x$ be the directed set of compact elements below $x$, with least upper bound $x$, and put $A_x=\{y:y\ll x\}$. A compact element $k\le x$ belongs to $A_x$: a directed set whose least upper bound is at least $x$ also has least upper bound at least $k$, so the definition of $k\ll k$ supplies an element above $k$. Hence $K_x\subseteq A_x$.

If $y_1,y_2\in A_x$, apply the way-below relation to $K_x$ to find $k_1,k_2\in K_x$ above them. Since $K_x$ is directed, it contains a common upper bound $k$ for $k_1,k_2$. This $k$ belongs to $A_x$, proving that $A_x$ is directed. Every member of $A_x$ is below $x$, while $K_x\subseteq A_x$ has least upper bound $x$. Therefore $x$ is the least upper bound of $A_x$, which is the defining property of a continuous poset.

Give the real interval $[0,1]$ its usual total order. Every nonempty directed set has a least upper bound in the real interval, so it is a directed-complete partial order. If $x<y\le \sup D$, some $d\in D$ exceeds $x$; otherwise $x$ would be an upper bound for $D$. Also $0\ll 0$, since every member of a nonempty $D$ is at least zero. For $x>0$, the increasing sequence $x(1-1/n)$, $n\ge 1$, has least upper bound $x$ but never reaches it, so $x/\ll x$. Together with $x\ll y\Rightarrow x\le y$, this proves the displayed way-below relation.

For each $y>0$, the elements way-below $y$ form $[0,y)$, a directed set with least upper bound $y$; for $y=0$ they form $\{0\}$. Thus this is a continuous poset. Its only compact element is zero, so it is not an algebraic partially ordered set.

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: