First, observe that is monotone. If , then , so that . As a result, if is directed, so is .
Before proving this, let’s make one additional observation:
If is continuous (under Scott topologies), then is monotone.
Suppose . We wish to show that , or . Assume the contrary. Consider . Then and is Scott open, hence is Scott open also. Since and is upper, , which implies , a contradiction. Therefore, . ∎
Now the proof of the proposition.
Suppose first that is Scott continuous. Take an open set . We want to show that is open in . In other words, is upper and that has non-empty intersection with any directed set whenever its supremum lies in . If , then some with , which implies . Since , , so , is upper. Now, suppose . So . Since is directed, there is , which means there is such that and . This shows that is Scott open.
Conversely, suppose is continuous (inverse of a Scott open set is Scott open). Let be a directed subset of and let . We want to show that . First, for any , we have that so that since is monotone. This shows . Now suppose is any upper bound of . We want to show that , or . Assume not. Then lies in , a Scott open set. So , also Scott open, which implies some with , or . This means , a contradiction. Thus , and the proof is complete. ∎
- 1 G. Gierz, K. H. Hofmann, K. Keimel, J. D. Lawson, M. W. Mislove, D. S. Scott, Continuous Lattices and Domains, Cambridge University Press, Cambridge (2003).