Join Semilattice is Ordered Structure

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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.


By Join Succeeds Operands:

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

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:

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

$\blacksquare$


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