bounded completeBoundedComplete2007-04-30 00:16:542007-05-13 18:46:02Definitioncomplete latticeDedekind complete\usepackage{amssymb,amscd}
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\newcommand{\pdiff}[2]{\frac{\partial #1}{\partial #2}}
\newcommand{\mpdiff}[3]{\frac{\partial^#1 #2}{\partial #3^#1}}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.