Meet-Continuous iff Meet Preserves Directed Suprema

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Theorem

Let $\mathscr S = \struct {S, \vee, \wedge, \preceq}$ be an up-complete lattice.

Let $\struct {S \times S, \precsim}$ be the simple order product of $\mathscr S$ and $\mathscr S$.

Let $f: S \times S \to S$ be a mapping such that

$\forall x, y \in S: \map f {x, y} = x \wedge y$


Then:

$\mathscr S$ is meet-continuous

if and only if:

$f$ preserves directed suprema


Proof

Sufficient Condition

Assume that:

$\mathscr S$ is meet-continuous.

We will prove that:

for every element $x$ of $S$, a directed subset $D$ of $S$ if $x \preceq \sup D$, then $x = \sup \set {x \wedge d: d \in D}$

Let $x \in S$, $D$ be a directed subset of $S$ such that:

$x \preceq \sup D$

Thus

\(\ds x\) \(=\) \(\ds x \wedge \sup D\) Preceding iff Meet equals Less Operand
\(\ds \) \(=\) \(\ds \sup \set {x \wedge d: d \in D}\) Definition of Meet-Continuous Lattice

$\Box$

By definition of reflexivity:

for every an element $x$ of $S$, a directed subset $D$ of $S$ if $x \preceq \sup D$, then $x \preceq \sup \set {x \wedge d: d \in D}$

By Meet is Directed Suprema Preserving:

$f$ preserves directed suprema.

$\Box$


Necessary Condition

Assume that:

$f$ preserves directed suprema.

By Meet is Directed Suprema Preserving implies Meet of Suprema equals Supremum of Meet of Directed Subsets:

$\paren {\sup D_1} \wedge \paren {\sup D_2} = \sup \set {d_1 \wedge d_2: d_1 \in D_1, d_2 \in D_2}$

Thus by Meet-Continuous iff Meet of Suprema equals Supremum of Meet of Directed Subsets:

$\mathscr S$ is meet-continuous.

$\blacksquare$


Sources

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