Closed Set iff Lower and Closed under Directed Suprema in Scott Topological Ordered Set
Theorem
Let $T = \struct {S, \preceq, \tau}$ be a relational structure with Scott topology
where $\struct {S, \preceq}$ is an up-complete ordered set.
Let $A$ be a subset of $S$.
Then $A$ is closed if and only if $A$ is lower and closed under directed suprema.
Proof
Sufficient Condition
Assume that
- $A$ is closed.
By definition of closed set:
- $\relcomp S A \in \tau$
By definition of Scott topology:
- $\relcomp S A$ is upper and inaccessible by directed suprema.
- $\relcomp S {\relcomp S A} = A$
Thus by Complement of Upper Section is Lower Section:
- $A$ is a lower section.
Thus by Complement of Inaccessible by Directed Suprema Subset is Closed under Directed Suprema:
- $A$ is closed under directed suprema.
$\Box$
Necessary Condition
Assume that $A$ is lower and closed under directed suprema.
By Complement of Lower Section is Upper Section and Complement of Closed under Directed Suprema Subset is Inaccessible by Directed Suprema:
- $\relcomp S A$ is upper and inaccessible by directed suprema.
By definition of Scott topology:
- $\relcomp S A \in \tau$
Thus by definition:
- $A$ is closed set.
$\blacksquare$
Sources
- 1980: G. Gierz, K.H. Hofmann, K. Keimel, J.D. Lawson, M.W. Mislove and D.S. Scott: A Compendium of Continuous Lattices
- Mizar article WAYBEL11:7