a space is connected under the ordered topology if and only if it is a linear continuum.
Suppose is not bounded-complete. Let be a nonempty subset of that is bounded above by , but has no least upper bound. Let be the set of upper bounds of . If , then is not a least upper bound of , so there is a such that and is an upper bound of . Then the set is open and contains . Furthermore, all of its elements exceed , so it is a subset of . Thus, is open. contains , so it is not empty. Let . Then is not an upper bound of , so there is an such that . The set is open, and contains no upper bounds of , so it is a subset of . Thus contains a neighborhood of each of its points, and is therefore open. Since and are open, X is not connected.
Suppose the ordering of is not dense, so there are and in with so that there is no in with (there is a gap between and ). Let and let . Because there are no elements between and , . By transitivity and trichotomy, . and are both open. and , so neither nor is empty. Thus, and separate , so is not connected.
Therefore, if is connected, then is a linear continuum.
Now suppose that is disconnected and bounded-complete, and that and are (nonempty, open and closed) sets separating . Suppose that , and suppose also that there is an element such that (if there is none, swap the names of and , or reverse the ordering). is open (it is the intersection of two open sets), and contains (so it is not empty). is bounded below by , so it has a greatest lower bound .
If , then, since is open, there is an interval in containing , which must, to exclude , be of the form or of the form , for some and perhaps . But then would be a lower bound of , contradicting the fact that is the infimum of .
If , then, since is open, there is an interval in containing , which must, to exclude , be of the form or of the form for some and perhaps . In either case, and the set is a subset of . If , then and , so , contradicting the fact that is the infimum of . Thus, there are no elements of between and , so the order on is not dense. This proves that if is a linear continuum, then is connected.
|Title||a space is connected under the ordered topology if and only if it is a linear continuum.|
|Date of creation||2013-03-22 17:38:38|
|Last modified on||2013-03-22 17:38:38|
|Last modified by||dfeuer (18434)|