Preceding is Top in Ordered Set of Auxiliary Relations

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $L = \struct {S, \vee, \preceq}$ be a bounded below join semilattice.

Let $\map {\it Aux} L$ be the set of all auxiliary relations on $S$.

Let $P = \struct {\map {\it Aux} L, \precsim}$ be an ordered set

where $\precsim \mathop = \subseteq \restriction_{\map {\it Aux} L \times \map {\it Aux} L}$


Then

$\preceq \mathop = \top_P$

where $\top_P$ denotes the greatest element in $P$.


Proof

By Preceding is Auxiliary Relation:

$\preceq \mathop \in \map {\it Aux} L$

By definition:

$\preceq$ is lower bound for $\O$ in $P$

We will prove that:

$\forall R \in \map {\it Aux} L: R$ is lower bound for $\O \implies R \mathop \precsim \preceq$

Let $R \in \map {\it Aux} L$

By condition $(i)$ of definition of auxiliary relation:

$R \mathop \subseteq \preceq$

Thus by definition of $\precsim$:

$R \mathop \precsim \preceq$

$\Box$


By definition of infimum:

$\preceq \mathop = \inf_P \O$

Thus by Infimum of Empty Set is Greatest Element:

$\preceq \mathop = \top_P$

$\blacksquare$


Sources