Lower Closure of Element is Topologically Closed in Scott Topological Ordered Set
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Theorem
Let $\struct {S, \preceq}$ be an up-complete ordered set.
Let $T = \struct {S, \preceq, \tau}$ be a relational structure with the Scott topology.
Let $x \in S$.
Let $x^\preceq$ denote the lower closure of $x$.
Then $x^\preceq$ is topologically closed.
Proof
By Lower Closure of Element is Closed under Directed Suprema:
- $x^\preceq$ is closed under directed suprema.
By Lower Closure of Singleton:
- $\set x^\preceq = x^\preceq$
By Lower Closure is Lower Section:
- $x^\preceq$ is a lower section.
Thus by Closed Set iff Lower and Closed under Directed Suprema in Scott Topological Ordered Set:
- $x^\preceq$ is closed.
$\blacksquare$
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 WAYBEL11:11