Ordered Set of Auxiliary Relations is Complete Lattice

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Theorem

Let $L = \left({S, \vee, \preceq}\right)$ be a bounded below join semilattice.

Let $\operatorname {Aux} \left({L}\right)$ be the set of all auxiliary relations on $S$.

Let $P = \left({\operatorname {Aux} \left({L}\right), \precsim}\right)$ be an ordered set where $\precsim \mathop = \subseteq \restriction_{\operatorname {Aux} \left({L}\right) \times \operatorname {Aux} \left({L}\right)}$


Then

$P$ is a complete lattice.


Proof

Let $X \subseteq \operatorname {Aux} \left({L}\right)$

In the case when $X \ne \O$:

By Intersection of Auxiliary Relations is Auxiliary Relation:

$\bigcap X \in \operatorname {Aux} \left({L}\right)$

By Intersection is Largest Subset, $\bigcap X$ is the infimum of $X$.

In case when $X = \O$:

By proof of Preceding is Top in Ordered Set of Auxiliary Relations:

$\O$ admits an infimum in $P$

Then:

$X$ (empty or non-empty) admits an infimum in $P$

By duality of Lattice is Complete iff it Admits All Suprema:

$P$ is a complete lattice.

$\blacksquare$


Sources