# Join Semilattice is Ordered Structure

## Contents

## Theorem

Let $\left({S, \vee, \preceq}\right)$ be a join semilattice.

Then $\left({S, \vee, \preceq}\right)$ is an ordered structure.

That is, $\preceq$ is compatible with $\vee$.

## Proof 1

For $\struct {S, \vee, \preceq}$ to be an ordered structure is equivalent to, for all $a, b, c \in S$:

- $a \preceq b \implies a \vee c \preceq b \vee c$
- $a \preceq b \implies c \vee a \preceq c \vee b$

Since Join is Commutative, it suffices to prove the first of these implications.

By definition of join:

- $a \vee c = \sup \set {a, c}$

where $\sup$ denotes supremum.

- $b \preceq b \vee c$
- $c \preceq b \vee c$

Now also $a \preceq b$, and by transitivity of $\preceq$ we find that:

- $a \preceq b \vee c$

Thus $b \vee c$ is an upper bound for $\set {a, c}$.

Hence:

- $a \vee c \preceq b \vee c$

by definition of supremum.

$\blacksquare$

## Proof 2

Let $a, b, c \in S$.

Let $a \preceq b$.

By the definition of join semilattice:

- $a \vee b = b$

Thus:

- $\left({a \vee b}\right) \vee c = b \vee c$

Since $\vee$ is associative, commutative, and idempotent:

- $\left({a \vee c}\right) \vee \left({b \vee c}\right) = b \vee c$

Therefore, $a \vee c \preceq b \vee c$.

From Join is Commutative, we conclude that:

- $c \vee a \preceq c \vee b$

$\blacksquare$