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# bounded complete

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 $s\in S$, $s\leq a$.

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.

## Mathematics Subject Classification

06A12*no label found*06B23

*no label found*03G10

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