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