Way Below in Meet-Continuous Lattice
Theorem
Let $\mathscr S = \struct {S, \vee, \wedge, \preceq}$ be a meet-continuous bounded below lattice.
Let $x, y \in S$.
Then $x \ll y$ if and only if
- $\forall I \in \map {\operatorname {Ids} } {\mathscr S}: y = \sup I \implies x \in I$
where
- $\ll$ denotes the way below relation,
- $\map {\operatorname {Ids} } {\mathscr S}$ denotes the set of all ideals in $\mathscr S$.
Proof
Sufficient Condition
Let $x \ll y$
Let $I \in \map {\operatorname {Ids} } {\mathscr S}$ such that
- $y = \sup I$
By definition of reflexivity:
- $y \preceq \sup I$
By definition of meet-continuous:
- $\mathscr S$ is up-complete.
Thus by Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal:
- $x \in I$
$\Box$
Necessary Condition
Assume:
- $\forall I \in \map {\operatorname {Ids} } {\mathscr S}: y = \sup I \implies x \in I$
We will prove that:
- $\forall I \in \map {\operatorname {Ids} } {\mathscr S}: y \preceq \sup I \implies x \in I$
Let $I \in \map {\operatorname {Ids} } {\mathscr S}$ such that
- $y \preceq \sup I$
By definition of ideal:
Define $Y := \set {y \wedge d: d \in I}$
By Set of Finite Suprema is Directed:
- $\set {\sup A: A \in \map {\operatorname {Fin} } Y \land A \ne \O}$ is directed.
where $\map {\operatorname {Fin} } Y$ denotes the set of all finite subsets of $Y$.
By definition of up-complete:
- $\set {\sup A: A \in \map {\operatorname {Fin} } Y \land A \ne \O}$ admits a supremum.
By definition of union:
- $Y = \bigcup \map {\operatorname {Fin} } Y$
\(\ds y\) | \(=\) | \(\ds y \wedge \sup I\) | Preceding iff Meet equals Less Operand | |||||||||||
\(\ds \) | \(=\) | \(\ds \sup Y\) | Definition of Meet-Continuous Lattice | |||||||||||
\(\ds \) | \(=\) | \(\ds \sup \set {\sup A: A \in \map {\operatorname {Fin} } Y \land A \ne \O}\) | Supremum of Suprema | |||||||||||
\(\ds \) | \(=\) | \(\ds \sup \set {\sup A: A \in \map {\operatorname {Fin} } Y \land A \ne \O}^\preceq\) | Supremum of Lower Closure of Set |
where $^\preceq$ denotes the lower closure of set.
By Lower Closure of Directed Subset is Ideal:
- $\set {\sup A: A \in \map {\operatorname {Fin} } Y \land A \ne \O}^\preceq$ is an ideal in $\mathscr S$
By assumption:
- $x \in \set {\sup A: A \in \map {\operatorname {Fin} } Y \land A \ne \O}^\preceq$
By definition of lower closure of set:
- $\exists z \in \set {\sup A: A \in \map {\operatorname {Fin} } Y \land A \ne \O}: x \preceq z$
Then
- $\exists A \in \map {\operatorname {Fin} } Y: z = \sup A \land A \ne \O$
We will prove as a sublemma that:
- $A \subseteq I$
Let $a \in A$.
By definition of subset:
- $a \in Y$
By definition of $Y$:
- $\exists i \in I: a = y \wedge i$
- $a \preceq i$
Thus by definition of lower section:
- $a \in I$
This ends the proof of sublemma.
By Directed iff Finite Subsets have Upper Bounds:
- $\exists h \in I: \forall a \in A: a \preceq h$
By definition:
- $h$ is upper bound for $A$
By definition of supremum:
- $z = \sup A \preceq h$
By definition of transitivity:
- $x \preceq h$
Thus by definition of lower section:
- $x \in I$
$\Box$
Thus by Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal:
- $x \ll y$
$\blacksquare$
Sources
- 1980: G. Gierz, K.H. Hofmann, K. Keimel, J.D. Lawson, M.W. Mislove and D.S. Scott: A Compendium of Continuous Lattices
- Mizar article WAYBEL_3:22