Semilattice has Unique Ordering such that Operation is Supremum

Theorem
Let $\struct {S, \vee}$ be a semilattice.

Then there exists a unique ordering $\preccurlyeq$ on $S$ such that:
 * $x \vee y = \sup \set {x, y}$

where $\sup \set {x, y}$ is the supremum of $\set {x, y}$ $\preccurlyeq$.

Proof
Let us define a relation $\RR$ on $S$ as:


 * $\forall x, y \in S: x \mathrel \RR y \iff x \vee y = y$

From Semilattice Induces Ordering, we have that $\RR$ is an ordering.

Let us denote this ordering by $\preccurlyeq$, and recall its definition:


 * $x \vee y = y \iff x \preccurlyeq y$

Supremum
Let $x, y \in S$ be arbitrary.

Let $x \vee y = z$.

We have:

Hence we see that $x \vee y$ is an upper bound of $\set {x, y}$ $\preccurlyeq$.

Now let $w$ be an arbitrary upper bound of $\set {x, y}$ $\preccurlyeq$.

That is:
 * $x \preccurlyeq w$ and $y \preccurlyeq w$

Then:

So:
 * $x \vee y$ is an upper bound of $\set {x, y}$ $\preccurlyeq$
 * for every upper bound $w$ of $\set {x, y}$ $\preccurlyeq$, we have that $x \vee y \preccurlyeq w$

Hence, by definition, $x \vee y$ is the supremum of $\set {x, y}$ $\preccurlyeq$.

Uniqueness
It remains to be proved that the ordering $\preccurlyeq$, with the property that:
 * $x \vee y = \sup \set {x, y}_\preccurlyeq$

is unique.

Indeed, suppose there exists another ordering $\preccurlyeq'$ such that:
 * $x \vee y = \sup \set {x, y}_{\preccurlyeq'}$

We have:

The proof is complete.

Also see

 * Definition:Join (Order Theory)