= Solution
Take the <natural numbers> including zero with their usual <total order>:
$$
\boxed{P=(\{0,1,2,\ldots\},\le).}
$$
A nonempty subset with an <upper bound> $m\in P$ is contained in the finite <set> $\{0,1,\ldots,m\}$. It therefore has a <greatest element>, which is its <least upper bound>. The empty subset has <least upper bound> $0$. Thus $P$ is a <bounded-complete partially ordered set>.
But $D=P$ is a <directed set>, since $\max(x,y)\in P$ is above any two members. It has no <upper bound> in $P$: for any proposed bound $m$, the member $m+1$ is larger. It therefore has no <least upper bound>. Hence
$$
\boxed{P\text{ is bounded-complete but not directed-complete}.}
$$
Back to article page
