 Solution

Take the five-element partially ordered set $P=\{\perp ,a,b,u,v\}$. Its strict comparisons are $\perp <a$, $\perp <b$, and $a<u$, $a<v$, $b<u$, $b<v$, together with their transitive consequences. The elements $a,b$ are incomparable, as are $u,v$.

A finite directed set has a greatest element whenever it is nonempty: list its elements and repeatedly use directedness to find an internal upper bound of the previously chosen element and the next one. The final element is above every member of the directed set, and is its least upper bound. Thus every nonempty directed set in $P$ has a least upper bound. The empty subset also has a least upper bound, namely $\perp$. Therefore $P$ satisfies the paper's completeness condition under either convention.

However, $\{a,b\}$ is bounded above by both $u$ and $v$. A least upper bound would have to be an upper bound lying below both $u$ and $v$. Neither $u$ nor $v$ has that property, and none of $\perp ,a,b$ is an upper bound of $\{a,b\}$. Hence