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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.

We prove the result algebraic partially ordered sets are continuous directly from the definitions. First, $a\ll b$ implies $a\le b$, by applying the way-below relation to the singleton directed set $\{b\}$. Also,if $c\le \sup D$, then $b\le \sup D$, so the witness required for $a\ll b$ also works for $a\ll c$.

Fix $x$ and putBy algebraicity, $K_x$ is a directed set with least upper bound $x$. Every $k\in K_x$ satisfies $k\ll k\le x$, hence $k\ll x$. Thus $K_x\subseteq A_x\subseteq \{a:a\le x\}$.

To prove that $A_x$ is a directed set, take $a,b\in A_x$. Applying $a\ll x$ and $b\ll x$ to $x=\sup K_x$ gives $k_a,k_b\in K_x$ such that $a\le k_a$ and $b\le k_b$. Directedness of $K_x$ gives $k\in K_x$ above both. Since $k\in A_x$, it is the required common upper bound inside $A_x$. With the usual nonempty convention, $A_x$ is nonempty since it contains $K_x$.

By directed completeness $A_x$ has a least upper bound. The inclusions above givesoThe same proof handles the paper's literal empty-set convention. If $K_x=\varnothing$, then $x=\sup \varnothing =\perp$, and the defining implication for $a\ll x$, applied to $D=K_x$, shows that $A_x$ must also be empty. Its least upper bound is again $x$. If $K_x$ is nonempty, the argument already given applies without change.

Take two increasing countable chains sharing a least element $\perp$ and a greatest element $\top$, with interiors $a_1<a_2<\cdots$ and $b_1<b_2<\cdots$ and no comparisons between branches. Each entire interior chain has $\top$ as its least upper bound. A directed set meeting both interiors must contain $\top$. Otherwise it lies within one branch, where it has either a greatest member or least upper bound $\top$. This proves that the partially ordered set is a directed-complete partial order.

The only element way-below $a_n$ is $\perp$. Indeed, an element way-below $a_n$ must be below $a_n$, but every positive $a_m\le a_n$ fails the way-below relation when tested against $\{b_1,b_2,\dots \}$: its least upper bound is $\top \ge a_n$, yet it has no member above $a_m$. Nonempty directed sets always contain an element above $\perp$, so $\perp \ll a_n$. Thus the elements way-below $a_n$ have least upper bound $\perp \ne a_n$, and this is not a continuous poset.