Editing Total order (section)

===Completeness===<!-- This section is linked from [[Completely distributive lattice]]. See [[WP:MOS#Section management]] -->

A totally ordered set is said to be '''[[Completeness (order theory)|complete]]''' if every nonempty subset that has an [[upper bound]], has a [[least upper bound]]. For example, the set of [[real number]]s '''R''' is complete but the set of [[rational number]]s '''Q''' is not. In other words, the various concepts of [[Completeness (order theory)|completeness]] (not to be confused with being "total") do not carry over to [[Binary relation|restrictions]]. For example, over the [[real number]]s a property of the relation {{char|≤}} is that every [[Empty set|non-empty]] subset ''S'' of '''R''' with an [[upper bound]] in '''R''' has a [[Supremum|least upper bound]] (also called supremum) in '''R'''. However, for the rational numbers this supremum is not necessarily rational, so the same property does not hold on the restriction of the relation {{char|≤}} to the rational numbers.

There are a number of results relating properties of the order topology to the completeness of X:
* If the order topology on ''X'' is connected, ''X'' is complete.
* ''X'' is connected under the order topology if and only if it is complete and there is no ''gap'' in ''X'' (a gap is two points ''a'' and ''b'' in ''X'' with ''a'' < ''b'' such that no ''c'' satisfies ''a'' < ''c'' < ''b''.)
* ''X'' is complete if and only if every bounded set that is closed in the order topology is compact.

A totally ordered set (with its order topology) which is a [[complete lattice]] is [[Compact space|compact]]. Examples are the closed intervals of real numbers, e.g. the [[unit interval]] [0,1], and the [[affinely extended real number system]] (extended real number line). There are order-preserving [[homeomorphism]]s between these examples.