Operator Generated by Closure System Preserves Directed Suprema iff Closure System Inherits Directed Suprema

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Theorem

Let $L = \struct {X, \vee, \wedge, \preceq}$ be a complete lattice.

Let $S = \struct {Y, \precsim}$ be a closure system on $L$.


Then $\map {\operatorname {operator} } S$ preserves directed suprema if and only if $S$ inherits directed suprema.

where $\map {\operatorname {operator} } S$ denotes the operator generated by $S$.


Proof

Sufficient Condition

Assume that $\map {\operatorname {operator} } S$ preserves directed suprema.

Let $Z$ be directed subset of $Y$ such that

$Z$ admits a supremum in $L$.

By Image of Operator Generated by Closure System is Set of Closure System:

$\map {\operatorname {operator} } S \sqbrk X = Y$

By Operator Generated by Closure System is Closure Operator:

$\map {\operatorname {operator} } S$ is closure operator.

By definition of closure operator/idempotent:

$\map {\operatorname {operator} } S \sqbrk Z = Z$

By definitions of Definition:Mapping Preserves Supremum/Directed and Definition:Mapping Preserves Supremum/Subset:

$\sup_L Z = \map {\map {\operatorname {operator} } S} {\sup_L Z}$

Thus by definition of image of mapping:

$\sup_L Z \in Y$

$\Box$


Necessary Condition

Assume that

$S$ inherits directed suprema.

Define $f := \map {\operatorname {operator} } S$.

By Image of Operator Generated by Closure System is Set of Closure System:

$f \sqbrk X = Y$

Let $A$ be a directed subset of $X$ such that

$A$ admits a supremum in $L$.

Thus by definition of complete lattice:

$f \sqbrk A$ admits a supremum in $L$.

We will prove that

$\map f {\sup_L A}$ is upper bound for $f \sqbrk A$

Let $x \in f \sqbrk A$.

By definition of image of subset:

$\exists y \in A: x = \map f y$

By definitions of supremum and upper bound:

$y \preceq \sup_L A$

Thus by definition of closure operator/increasing:

$x \preceq \map f {\sup_L A}$

$\Box$


By definition of supremum:

$\map {\sup_L} {f \sqbrk A} \preceq \map f {\sup_L A}$

We will prove that

$\map {\sup_L} {f \sqbrk A}$ is upper bound for $A$

Let $x \in A$.

By definition of image of set:

$\map f x \in f \sqbrk A$

By definitions of supremum and upper bound:

$\map f x \preceq \map {\sup_L} {f \sqbrk A}$

By definition of closure operator/inflationary:

$x \preceq \map f x$

Thus by definition of transitivity:

$x \preceq \map {\sup_L} {f \sqbrk A}$

$\Box$


By definition of supremum:

$\sup_L A \preceq \map {\sup_L} {f \sqbrk A}$

By Image of Directed Subset under Increasing Mapping is Directed:

$f \sqbrk A$ is directed in $L$.

By definition of ordered subset:

$f \sqbrk A$ is directed in $S$.

By definition of directed suprema inheriting:

$\map {\sup_L} {f \sqbrk A} \in Y$

By definition of closure operator/idempotent:

$\map f {\map {\sup_L} {f \sqbrk A} } = \map {\sup_L} {f \sqbrk A}$

By definition of closure operator/increasing:

$\map f {\sup_L A} \preceq \map {\sup_L} {f \sqbrk A}$

Thus by definition of antisymmetry:

$\map f {\sup_L A} = \map {\sup_L} {f \sqbrk A}$

$\blacksquare$


Sources

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