Extension Theorem for Total Orderings

Theorem

Let the following conditions be fulfilled:

$(1):\quad$ Let $\struct {S, \circ, \preceq}$ be a totally ordered commutative semigroup
$(2):\quad$ Let all the elements of $\struct {S, \circ, \preceq}$ be cancellable
$(3):\quad$ Let $\struct {T, \circ}$ be an inverse completion of $\struct {S, \circ}$.

Then:

$(1):\quad$ The relation $\preceq'$ on $T$ satisfying $\forall x_1, x_2, y_1, y_2 \in S: x_1 \circ {y_1}^{-1} \preceq' x_2 \circ {y_2}^{-1} \iff x_1 \circ y_2 \preceq x_2 \circ y_1$ is a well-defined relation
$(2):\quad$ $\preceq'$ is the only total ordering on $T$ compatible with $\circ$
$(3):\quad$ $\preceq'$ is the only total ordering on $T$ that induces the given ordering $\preceq$ on $S$.

Proof

every element of $T$ is of the form $x \circ y^{-1}$ where $x, y \in S$.

Proof that Relation is Well-Defined

First we need to show that $\preceq'$ is well-defined.

So we need to show that if $x_1, x_2, y_1, y_2, z_1, z_2, w_1, w_2 \in S$ satisfy:

 $\ds x_1 \circ {y_1}^{-1}$ $=$ $\ds x_2 \circ {y_2}^{-1}$ $\ds z_1 \circ {w_1}^{-1}$ $=$ $\ds z_2 \circ {w_2}^{-1}$ $\ds x_1 \circ w_1$ $\preceq$ $\ds y_1 \circ z_1$

then:

$x_2 \circ {w_2}^{-1} = y_2 \circ {z_2}^{-1}$

We have:

 $\ds x_1 \circ {y_1}^{-1}$ $=$ $\ds x_2 \circ {y_2}^{-1}$ $\ds \leadstoandfrom \ \$ $\ds x_1 \circ y_2$ $=$ $\ds x_2 \circ y_1$

and:

 $\ds z_1 \circ {w_1}^{-1}$ $=$ $\ds z_2 \circ {w_2}^{-1}$ $\ds \leadstoandfrom \ \$ $\ds z_1 \circ w_2$ $=$ $\ds z_2 \circ w_1$

So:

 $\ds x_2 \circ w_2 \circ y_1 \circ z_1$ $=$ $\ds x_1 \circ w_1 \circ y_2 \circ z_2$ $\ds$ $\preceq$ $\ds y_1 \circ z_1 \circ y_2 \circ z_2$ $\ds \leadsto \ \$ $\ds x_2 \circ w_2$ $\preceq$ $\ds y_2 \circ z_2$ Strict Ordering Preserved under Product with Cancellable Element

Thus $\preceq'$ is a well-defined relation on $T$.

$\Box$

Proof that Relation is Transitive

To show that $\preceq'$ is transitive:

 $\ds \forall x, y, z, w, u, v \in S: \,$ $\ds x \circ y^{-1}$ $\preceq'$ $\ds z \circ w^{-1}$ $\ds z \circ w^{-1}$ $\preceq'$ $\ds u \circ v^{-1}$ $\ds \leadsto \ \$ $\ds x \circ w \circ v$ $\preceq$ $\ds y \circ z \circ v$ $\ds$ $\preceq$ $\ds y \circ w \circ u$ $\ds \leadsto \ \$ $\ds x \circ v$ $\preceq$ $\ds y \circ u$ Strict Ordering Preserved under Product with Cancellable Element $\ds \leadsto \ \$ $\ds x \circ y^{-1}$ $\preceq'$ $\ds u \circ v^{-1}$

Thus $\preceq'$ is transitive.

$\Box$

Proof that Relation is Total Ordering

To show that $\preceq'$ is a total ordering on $T$ compatible with $\circ$:

Let $z_1, z_2 \in T$.

Then:

$\exists x_1, x_2, y_1, y_2 \in S: z_1 = x_1 \circ y_1^{-1}, z_2 = x_2 \circ y_2^{-1}$

Then:

$z_1 \circ y_1 = x_1$
$z_2 \circ y_2 = x_2$

As $\preceq$ is a total ordering on $S$, either:

$z_1 \circ y_1 \circ y_2 \preceq z_2 \circ y_2 \circ y_1$

or:

$z_2 \circ y_2 \circ y_1 \preceq z_1 \circ y_1 \circ y_2$

Without loss of generality, suppose that:

$z_1 \circ y_1 \circ y_2 \preceq z_2 \circ y_2 \circ y_1$

So:

 $\ds z_1 \circ y_1 \circ y_2$ $\preceq$ $\ds z_2 \circ y_2 \circ y_1$ $\ds \leadsto \ \$ $\ds x_1 \circ y_2$ $\preceq$ $\ds x_2 \circ y_1$ $\ds \leadsto \ \$ $\ds x_1 \circ y_1^{-1}$ $\preceq'$ $\ds x_2 \circ y_2^{-1}$ $\ds \leadsto \ \$ $\ds z_1$ $\preceq'$ $\ds z_2$

and it is seen that $\preceq'$ is a total ordering on $T$.

$\Box$

Proof that Relation is Compatible with Operation

Let $x_1, x_2, y_1, y_2 \in T$ such that $x_1 \preceq' x_2, y_1 \preceq' y_2$.

We need to show that $x_1 \circ y_1 \preceq' x_2 \circ y_2$.

Let:

$x_1 = r_1 \circ s_1^{-1}, x_2 = r_2 \circ s_2^{-1}$
$y_1 = u_1 \circ v_1^{-1}, y_2 = u_2 \circ v_2^{-1}$

We have:

 $\ds x_1$ $\preceq'$ $\ds x_2$ $\ds \leadsto \ \$ $\ds r_1 \circ s_1^{-1}$ $\preceq'$ $\ds r_2 \circ s_2^{-1}$ $\ds \leadsto \ \$ $\ds r_1 \circ s_2$ $\preceq$ $\ds r_2 \circ s_1$

and

 $\ds y_1$ $\preceq'$ $\ds y_2$ $\ds \leadsto \ \$ $\ds u_1 \circ v_1^{-1}$ $\preceq'$ $\ds u_2 \circ v_2^{-1}$ $\ds \leadsto \ \$ $\ds u_1 \circ v_2$ $\preceq$ $\ds u_2 \circ v_1$

Because of the compatibility of $\preceq$ with $\circ$ on $S$:

 $\ds \paren {r_1 \circ s_2} \circ \paren {u_1 \circ v_2}$ $\preceq$ $\ds \paren {r_2 \circ s_1} \circ \paren {u_2 \circ v_1}$ $\ds \leadsto \ \$ $\ds \paren {r_1 \circ u_1} \circ \paren {s_2 \circ v_2}$ $\preceq'$ $\ds \paren {r_2 \circ u_2} \circ \paren {s_1 \circ v_1}$ $\ds \leadsto \ \$ $\ds \paren {r_1 \circ u_1} \circ \paren {s_1 \circ v_1}^{-1}$ $\preceq'$ $\ds \paren {r_2 \circ u_2} \circ \paren {s_2 \circ v_2}^{-1}$ $\ds \leadsto \ \$ $\ds \paren {r_1 \circ s_1^{-1} } \circ \paren {u_1 \circ v_1^{-1} }$ $\preceq'$ $\ds \paren {r_2 \circ s_2^{-1} } \circ \paren {u_2 \circ v_2^{-1} }$ $\ds \leadsto \ \$ $\ds x_1 \circ y_1$ $\preceq'$ $\ds x_2 \circ y_2$

Thus compatibility is proved.

$\Box$

To show that the restriction of $\preceq'$ to $S$ is $\preceq$:

 $\ds \forall x, y \in S: \,$ $\ds x$ $\preceq$ $\ds y$ $\ds \leadsto \ \$ $\ds \forall a \in S: \,$ $\ds \paren {x \circ a} \circ a$ $\preceq$ $\ds \paren {y \circ a} \circ a$ $\ds \leadsto \ \$ $\ds x$ $=$ $\ds \paren {x \circ a} \circ a^{-1}$ $\ds$ $\preceq$ $\ds \paren {y \circ a} \circ a^{-1}$ $\ds$ $=$ $\ds y$

Conversely:

 $\ds \forall x, y \in S: \,$ $\ds x$ $\preceq'$ $\ds y$ $\ds \leadsto \ \$ $\ds \exists u, v, z, w \in S: \,$ $\ds x$ $=$ $\ds u \circ v^{-1}$ $\ds y$ $=$ $\ds z \circ w^{-1}$ $\ds u \circ w$ $\preceq$ $\ds z \circ v$ $\ds \leadsto \ \$ $\ds x \circ v \circ w$ $=$ $\ds u \circ w$ $\ds$ $\preceq$ $\ds z \circ v$ $\ds$ $=$ $\ds y \circ v \circ w$ $\ds \leadsto \ \$ $\ds x$ $\preceq$ $\ds y$ Strict Ordering Preserved under Product with Cancellable Element

$\Box$

Proof that Relation is Unique

To show that $\preceq'$ is unique:

Let $\preceq^*$ be any ordering on $T$ compatible with $\circ$ that induces $\preceq$ on $S$.

Then:

 $\ds \forall x, y, z, w \in S: \,$ $\ds x \circ y^{-1}$ $\preceq^*$ $\ds z \circ w^{-1}$ $\ds \leadstoandfrom \ \$ $\ds x \circ w$ $\preceq$ $\ds y \circ z$

Then:

 $\ds x \circ y^{-1}$ $\preceq^*$ $\ds z \circ w^{-1}$ $\ds \leadsto \ \$ $\ds x \circ w$ $=$ $\ds \paren {x \circ y^{-1} } \circ \paren {y \circ w}$ $\ds$ $\preceq$ $\ds \paren {z \circ w^{-1} } \circ \paren {y \circ w}$ $\ds$ $=$ $\ds y \circ z$

 $\ds x \circ w$ $\preceq$ $\ds y \circ z$ $\ds \leadsto \ \$ $\ds x \circ y^{-1}$ $=$ $\ds \paren {x \circ w} \circ w^{-1} \circ y^{-1}$ $\ds$ $\preceq^*$ $\ds \paren {y \circ z} \circ w^{-1} \circ y^{-1}$ $\ds$ $=$ $\ds z \circ w^{-1}$

So:

 $\ds \forall x, y, z, w \in S: \,$ $\ds x \circ y^{-1}$ $\preceq^*$ $\ds z \circ w^{-1}$ $\ds \leadstoandfrom \ \$ $\ds x \circ w$ $\preceq$ $\ds y \circ z$

Hence as every element of $T$ is of the form $x \circ y^{-1}$ where $x, y \in S$, the orderings $\preceq^*$ and $\preceq'$ are identical.

$\blacksquare$