Image of Compact Subset under Directed Suprema Preserving Closure Operator is Subset of Compact Subset

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Theorem

Let $L = \struct {S, \preceq}$ be an algebric lattice.

Let $c: S \to S$ be a closure operator that preserves directed suprema.


Then:

$c \sqbrk {\map K L} \subseteq \map K {\struct {c \sqbrk S, \precsim} }$

where:

$\map K L$ denotes the compact subset of $L$
$c \sqbrk S$ denotes the image of $S$ under $c$
$\mathord \precsim = \mathord \preceq \cap \paren {c \sqbrk S \times \map c S}$


Proof

Let $x \in c \sqbrk {\map K L}$.

By definition of image of set:

$\exists y \in \map K L: x = \map c y$

and

$x \in c \sqbrk S$

By definition of compact subset:

$y$ is compact in $L$.

By definition of compact element:

$y \ll y$

where $\ll$ denotes the way below relation.

Define $P = \struct {c \sqbrk S, \precsim}$ as an ordered subset of $L$.

We will prove that:

for every directed subset $D$ of $c \sqbrk S$: $x \precsim \sup_P D \implies \exists d \in D: d \precsim x$

Let $D$ be a directed subset of $c \sqbrk S$.

By definition of ordered subset:

$D$ is directed in $L$.

By definition of algebraic ordered set:

$L$ is up-complete.

By definition of up-complete:

$D$ admits a supremum in $L$.
\(\ds \map c { {\sup}_L D}\) \(=\) \(\ds \map { {\sup}_L} {c \sqbrk D}\) Definition of Mapping Preserves Supremum
\(\ds \) \(=\) \(\ds {\sup}_L D\) Definition of Closure Operator

By definition of image of set:

$\sup_L D \in c \sqbrk S$

By Supremum in Ordered Subset:

$\sup_L D = \sup_P D$

By definition of ordered subset:

$x \preceq \sup_L D$

By definition of closure operator/inflationary:

$y \preceq x$

By definition of transitivity:

$y \preceq \sup_L D$

By definition of way below relation:

$\exists d \in D: y \preceq d$

By definition of subset:

$d \in c \sqbrk S$

By definition of closure operator/idempotent:

$d = \map c d$

By definition of closure operator/increasing:

$x = \map c y \preceq \map c d = d$

Thus by definition of ordered subset:

$\exists d \in D: x \precsim d$

$\Box$


By definition of way below relation:

$x \ll_P x$

By definition:

$x$ is a compact element in $P$.

Thus by definition of compact subset:

$x \in \map K P$

$\blacksquare$


Sources