Let $P$ be a poset. Recall that a subset $S$ of $P$ is called bounded from above if there is an element $a\in P$ such that, for every $s\in S$ , $s\le a$ .

A poset $P$ is said to be bounded complete if every subset which is bounded from above has a supremum.

Remark. Since it is not required that the subset be non-empty, we see that $P$ has a bottom. This is because the empty set is vacuously bounded from above, and therefore has a supremum. However, this supremum is less than or equal to every member of $P$ , and hence it is the least element of $P$ .

Clearly, any complete lattice is bounded complete. An example of a non-complete bounded complete poset is any closed subset of $\mathbb{R}$ of the form $[a,\infty)$ , where $a\in \mathbb{R}$ . In addition, arbitrary products of bounded complete posets is also bounded complete.

It can be shown that a poset is a bounded complete dcpo iff it is a complete semilattice.

Remark. A weaker concept is that of Dedekind completeness: A poset $P$ is Dedekind complete if every non-empty subset bounded from above has a supremum. An obvious example is $\mathbb{R}$ , which is Dedekind complete but not bounded complete (as it has no bottom). Dedekind completeness is more commonly known as the least upper bound property.