Complement of Lower Closure of Element is Open 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 $x \in S$.

Then $\relcomp S {x^\preceq}$ is topologically open,

where

$x^\preceq$ denotes the lower closure of $x$,

$\relcomp S {x^\preceq}$ denotes the relative complement of $x^\preceq$.

Proof

By Lower Closure of Element is Topologically Closed in Scott Topological Ordered Set:

$x^\preceq$ is closed.

By definition of closed set:

$\relcomp S {x^\preceq} \in \tau$

Thus by definition:

$\relcomp S {x^\preceq}$ is a open 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 WAYBEL11:12