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.

By Complement of Complement:

$\relcomp S {\relcomp S A} = A$

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