Complement of Lower Closure is Prime Element in Inclusion Ordered Set of Scott Sigma

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $L = \struct {S, \preceq, \tau}$ be a complete Scott topological lattice.

Let $D = \struct {\map \sigma L, \precsim}$ be an inclusion ordered set of the Scott sigma of $L$.

Let $x \in S$.


Then:

$\relcomp S {x^\preceq}$ is a prime element in $D$

and:

$\relcomp S {x^\preceq} \ne S$


Proof

By Scott Topology equals to Scott Sigma:

$\tau = \map \sigma L$

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

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

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

By Topological Closure of Singleton is Irreducible:

$x^\preceq$ is topologically irreducible

Thus by Complement of Irreducible Topological Subset is Prime Element:

$\relcomp S {x^\preceq}$ is a prime element in $D$.

By definitions of reflexivity and lower closure of element:

$x \in x^\preceq$

By definitions of relative complement and difference:

$x \notin \relcomp S {x^\preceq}$

Hence by definition set equality:

$\relcomp S {x^\preceq} \ne S$

$\blacksquare$


Sources