06A12-BoundedComplete.tex

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\begin{document}
Let $P$ be a poset.  Recall that a subset $S$ of $P$ is called \emph{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 \emph{bounded complete} if every subset which is bounded from above has a supremum.

\textbf{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.

\textbf{Remark}.  A weaker concept is that of \emph{Dedekind completeness}: A poset $P$ is \emph{Dedekind complete} if every \emph{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.
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