# Meet Preserves Directed Suprema

## Theorem

Let $\mathscr S = \struct {S, \preceq}$ be an up-complete meet semilattice such that

$\forall x \in S$, a directed subset $D$ of $S$: $x \preceq \sup D \implies x \preceq \sup \set {x \wedge y: y \in D}$

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

$\forall s, t \in S: \map f {s, t} = s \wedge t$

Then:

$f$ preserves directed suprema as a mapping from simple order product $\struct{S \times S, \precsim}$ of $\mathscr S$ and $\mathscr S$ into $\mathscr S$.

## Proof

### Lemma 2

Let $x$ be an element of $S$, $D$ be a directed subset of $S$.

Then:

$\paren {\sup D} \wedge x = \sup \set {d \wedge x: d \in D}$

$\Box$

Let $X$ be a directed subset of $S \times S$ such that

$X$ admits a supremum.
the simple order product of $\mathscr S$ and $\mathscr S$ is up-complete.
$X_1 := \map {\pr_1^\to} X$ is directed

and

$X_2 := \map {\pr_2^\to} X$ is directed

where

$\pr_1$ denotes the first projection on $S \times S$
$\pr_2$ denotes the second projection on $S \times S$
$\map {\pr_1^\to} X$ denotes the image of $X$

We will prove that

$(1): \quad \set {x \wedge \sup X_2: x \in X_1} = \set {\sup \set {x \wedge y: y \in X_2}: x \in X_1}$

First inclusion:

Let $a \in \set {x \wedge \sup X_2: x \in X_1}$.

Then

$\exists x \in X: a = x \wedge \sup X_2$
$x \wedge \sup X_2 \preceq \sup X_2$

By assumption:

$x \wedge \sup X_2 \preceq \sup \set {x \wedge \sup X_2 \wedge y: y \in X_2}$

By definition of supremum:

$\forall y \in X_2: y \preceq \sup X_2$
$\forall y \in X_2: \sup X_2 \wedge y = y$
$x \wedge \sup X_2 \preceq \sup \set {x \wedge y: y \in X_2}$
$\forall y \in X_2: x \wedge y \preceq x \wedge \sup X_2$

By definition:

$x \wedge \sup X_2$ is upper bound for $\set {x \wedge y: y \in X_2}$

By definition of supremum:

$\sup \set {x \wedge y: y \in X_2} \preceq x \wedge \sup X_2$

By definition of antisymmetry:

$a = \sup \set {x \wedge y: y \in X_2}$

Thus

$a \in \set {\sup \set {x \wedge y: y \in X_2}: x \in X_1}$

$\Box$

Second inclusion

Let $a \in \set {\sup \set {x \wedge y: y \in X_2}: x \in X_1}$.

Then

$\exists x \in X_1: a = \sup \set {x \wedge y: y \in X_2}$

Analogically to first inclusion

$a = x \wedge \sup X_2$

Thus

$a \in \set {x \wedge \sup X_2: x \in X_1}$

$\Box$

We will prove as lemma that:

$\paren {\map {f^\to} X}$ is directed.

Let $x, y \in \paren {\map {f^\to} X}$.

By image of set:

$\exists \tuple {a, b} \in X: x = \map f {a, b}$

and

$\exists \tuple {c, d} \in X: y = \map f {c, d}$

By definition of $f$:

$x = a \wedge b$ and $y = c \wedge d$

By definition of directed subset:

$\exists \tuple {g, h} \in X: \tuple {a, b} \precsim \tuple {g, h} \land \tuple {c, d} \precsim \tuple {g, h}$
$a \preceq g$, $b \preceq h$, $c \preceq g$, and $d \preceq h$
$x \preceq g \wedge h$ and $y \preceq g \wedge h$

By definition of image of set:

$g \wedge h = \map f {g, h} \in \map {f^\to} X$

Thus:

$\exists z \in \map {f^\to} X: x \preceq z \land y \preceq z$

This ends the proof of lemma.

Thus by definition of up-complete:

$\paren {\map {f^\to} X}$ admits a supremum.

Thus

 $\ds \map \sup {\map {f^\to} X}$ $=$ $\ds \sup \set {x \wedge y: x \in X_1, y \in X_2}$ Supremum of Meet Image of Directed Set $\ds$ $=$ $\ds \sup \bigcup \set {\set {x \wedge y: y \in X_2}: x \in X_1}$ $\ds$ $=$ $\ds \sup \set {\sup \set {x \wedge y: y \in X_2}: x \in X_1}$ Supremum of Suprema $\ds$ $=$ $\ds \sup \set {x \wedge \sup X_2: x \in X_1}$ the equality $(1)$ $\ds$ $=$ $\ds \paren {\sup X_1} \wedge \paren {\sup X_2}$ Lemma 2 $\ds$ $=$ $\ds \map f {\sup X_1, \sup X_2}$ Definition of $f$ $\ds$ $=$ $\ds \map f {\sup X}$ Supremum by Suprema of Directed Set in Simple Order Product

$\blacksquare$