Arithmetic iff Way Below Relation is Multiplicative in Algebraic Lattice

Theorem

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

Then:

$L$ is arithmetic

if and only if:

$\ll$ is a multiplicative relation

where $\ll$ denotes the way below relation of $L$.

Proof

Sufficient Condition

Let $L$ be arithmetic.

Let $a, x, y \in S$ such that

$a \ll x$ and $a \ll y$

By Algebraic iff Continuous and For Every Way Below Exists Compact Between:

$\exists c \in \map K L: a \preceq c \preceq x$

and

$\exists k \in \map K L: a \preceq k \preceq y$

where $\map K L$ denotes the compact subset of $L$.

By Meet Semilattice is Ordered Structure:

$c \wedge k \preceq x \wedge y$

By definition of arithmetic:

$\map K L$ is meet closed.

By definition of meet closed:

$c \wedge k \in \map K L$

By definition of compact subset:

$c \wedge k$ is compact.

By definition of compact:

$c \wedge k \ll c \wedge k$

By Meet is Idempotent:

$a \wedge a = a$

By Meet Semilattice is Ordered Structure:

$a \preceq c \wedge k$

Thus by Preceding and Way Below implies Way Below

$a \ll x \wedge y$

$\Box$

Necessary Condition

Let $\ll$ be a multiplicative relation.

Thus $L$ is algebraic.

It remains to prove that

$\map K L$ is meet closed.

Let $x, y \in \map K L$.

By definition of compact subset:

$x$ and $y$ are compact.

By definition of compact:

$x \ll x$ and $y \ll y$

By Way Below Relation is Auxiliary Relation:

$\ll$ is auxiliary relation.

By Multiplicative Auxiliary Relation iff Congruent:

$x \wedge y \ll x \wedge y$

By definition:

$x \wedge y$ is compact.

Thus by definition of compact subset:

$x \wedge y \in \map K L$

$\blacksquare$

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_8:20