Closure of Singleton is Lower Closure of Element in Scott Topological Lattice

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Theorem

Let $T = \struct {S, \preceq, \tau}$ be a up-complete topological lattice with the Scott topology.

Let $x \in S$.


Then:

$\set x^- = x^\preceq$

where

$\set x^-$ denotes the topological closure of $\set x$
$x^\preceq$ denotes the lower closure of $x$.


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.

By Closed Set iff Lower and Closed under Directed Suprema in Scott Topological Ordered Set:

$x^\preceq$ is closed.

By definitions of lower closure of element and reflexivity:

$x \in x^\preceq$

By Singleton of Element is Subset:

$\set x \subseteq x^\preceq$

We will prove that:

for every a closed subset $C$ of $S$: $\set x \subseteq C \implies x^\preceq \subseteq C$

Let $C$ be a closed subset of $S$ such that:

$\set x \subseteq C$

By Closed Set iff Lower and Closed under Directed Suprema in Scott Topological Ordered Set:

$C$ is a lower section.

Let $y \in x^\preceq$.

By definition of lower closure of element:

$y \preceq x$

By definitions of subset and singleton:

$x \in C$

Thus by definition of lower section:

$y \in C$

$\Box$

Thus by definition of topological closure:

$\set x^- = x^\preceq$

$\blacksquare$


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