Total Semilattice has Unique Total Ordering

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

Let $\struct {S, \vee}$ be such that for all subsets $T$ of $S$, $\struct {T, \vee}$ is closed.

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

where:
 * $\max \set {x, y} := \begin {cases} b & : a \preccurlyeq b \\ a & : b \preccurlyeq a \end {cases}$

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


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

It is to be shown that $\RR$ is a total ordering.

Checking in turn each of the criteria:

Reflexivity
From :
 * $\forall a \in S: a \vee a = a$

Hence by definition of $\RR$:


 * $\forall a \in S: a \mathrel \RR a = a$

So $\RR$ has been shown to be reflexive.

Transitivity
Let $a, b, c \in S$ such that $a \mathrel \RR b$ and $b \mathrel \RR c$.

Thus:

So $\RR$ has been shown to be transitive.

Antisymmetry
Let $a, b \in S$ such that $a \mathrel \RR b$ and $b \mathrel \RR a$.

So $\RR$ has been shown to be antisymmetric.

$\RR$ has been shown to be reflexive, transitive and antisymmetric.

Hence by definition it is an ordering.

Connectedness
Let $a, b \in S$ be arbitrary such that $a \ne b$.

We have that:
 * $\forall x, y \in \set {a, b}: x \vee y \in \set {a, b}$

That is, either:
 * $a \vee b = a$

or:
 * $a \vee b = b$

Hence, by definition of $\RR$, either:
 * $a \mathrel \RR b$

or:
 * $b \mathrel \RR a$

That is, $\RR$ is a connected relation.

Summarising:
 * $\RR$ is an ordering

and
 * $\RR$ is connected.

Hence by definition $\RR$ is a total ordering.

Let us denote this total ordering by $\preccurlyeq$.

We have:

and:

Hence by definition:
 * $x \vee y = \max \set {x, y}$

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

is unique.

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

We have:

The proof is complete.