Continuous iff Meet-Continuous and There Exists Smallest Auxiliary Approximating Relation

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Theorem

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


Then:

$L$ is continuous

if and only if:

$L$ is meet-continuous and there exists the smallest auxiliary approximating relation on $S$


That is:

$L$ is continuous

if and only if:

$L$ is meet-continuous and there exists an auxiliary approximating relation $\RR$ on $S$
for every auxiliary approximating relation $\QQ$ on $S$: $\RR \subseteq \QQ$


Proof

Sufficient Condition

Let $L$ be continuous.

Thus by Continuous Lattice is Meet-Continuous:

$L$ is meet-continuous.

Thus by Way Below is Approximating Relation and Way Below Relation is Auxiliary Relation:

$\ll$ is auxiliary approximating relation on $S$.

Thus by Continuous Lattice iff Auxiliary Approximating Relation is Superset of Way Below Relation:

for every auxiliary approximating relation $\QQ$ on $S$: $\ll \subseteq \QQ$

$\Box$


Necessary Condition

Let $L$ be meet-continuous.

Assume that:

there exists an auxiliary approximating relation $\RR$ on $S$
for every auxiliary approximating relation $\QQ$ on $S$: $\RR \subseteq \QQ$

Let $x \in S$.

By Intersection of Relation Segments of Approximating Relations equals Way Below Closure:

$\ds \bigcap \set {x^\QQ: \QQ \in \map {\operatorname{App} } L} = x^\ll$

where:

$x^\QQ$ denotes the $\QQ$-segment of $x$,
$\map {\operatorname{App} } L$ denotes the set of all auxiliary approximating relations on $S$.

By Intersection is Subset/General Result:

$x^\ll \subseteq x^\RR$

By definition of approximating relation:

$x = \map \sup {x^\RR}$

We will prove that:

$\forall a \in \set {x^\QQ: \QQ \in \map {\operatorname{App} } L}: x^\RR \subseteq a$

Let $a \in \set {x^\QQ: \QQ \in \map {\operatorname{App} } L}$

Then:

$\exists \QQ \in \map {\operatorname{App} } L: a = x^\QQ$

By assumption:

$\RR \subseteq \QQ$

Thus by Relation Segment is Increasing:

$x^\RR \subseteq a$

$\Box$

By Intersection is Largest Subset/General Result:

$x^\RR \subseteq x^\ll$

By definition of set equality:

$x^\RR = x^\ll$

Thus

$x = \map \sup {x^\ll}$

By definition:

$L$ satisfies the axiom of approximation.

By Way Below Closure is Directed in Bounded Below Join Semilattice:

$\forall x \in S: x^\ll$ is directed

By definition of complete lattice:

$L$ is up-complete.

Hence $L$ is continuous.

$\blacksquare$


Sources

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