Suprema in Ordered Group

Theorem
Let $\struct {G, \circ, \preccurlyeq}$ be an ordered group.

Let $x, y, z \in G$ be arbitrary.

Let any one of the sets $\set {x, y}$, $\set {x \circ z, y \circ z}$ or $\set {z \circ x, z \circ y}$ admit a supremum.

Then all three sets admit a supremum, and:

Proof
First we recall that by definition of ordered group, $\preccurlyeq$ is compatible with $\circ$:

Let $\set {x, y}$ admit a supremum $c$.

Then by definition of supremum:


 * $(1): \quad c$ is an upper bound of $\set {x, y}$ in $G$
 * $(2): \quad c \preccurlyeq d$ for all upper bounds $d$ of $\set {x, y}$ in $S$.

Thus we have:

Hence $\sup \set {x, y} \circ z$ is an upper bound of $\set {x \circ z, y \circ z}$.

Let $d$ be an upper bound of $\set {x \circ z, y \circ z}$.

Then as $G$ is a group we have that:
 * $d = d' \circ z$

for some $d' \in G$.

Then:

Hence $\sup \set {x, y} \circ z$ is an upper bound of $\set {x \circ z, y \circ z}$ which is smaller than an arbitrary upper bound $d$ of $\set {x \circ z, y \circ z}$.

That is, $\sup \set {x, y} \circ z$ is a supremum of $\set {x \circ z, y \circ z}$.

Let $\set {x \circ z, y \circ z}$ admit a supremum $c$.

Then by definition of supremum:


 * $(1): \quad c$ is an upper bound of $\set {x \circ z, y \circ z}$ in $G$
 * $(2): \quad c \preccurlyeq d$ for all upper bounds $d$ of $\set {x \circ z, y \circ z}$ in $S$.

As $G$ is a group, there exists $c' \in G$ such that $c' \circ z = c$.

Thus we have:

Hence $z \circ c'$ is an upper bound of $\set {z \circ x, z \circ y}$.

Let $d$ be an upper bound of $\set {z \circ x, z \circ y}$.

As $G$ is a group, there exists $d' \in G$ such that $z \circ d' = d$.

Then:

Hence $z \circ c'$ is an upper bound of $\set {z \circ x, z \circ y}$ which is smaller than an arbitrary upper bound $d$ of $\set {z \circ x, z \circ y}$.

That is, $z \circ c'$ is a supremum of $\set {z \circ x, z \circ y}$.

Let $\set {z \circ x, z \circ y}$ admit a supremum $c$.

Then by definition of supremum:


 * $(1): \quad c$ is an upper bound of $\set {z \circ x, z \circ y}$ in $G$
 * $(2): \quad c \preccurlyeq d$ for all upper bounds $d$ of $\set {z \circ x, z \circ y}$ in $S$.

As $G$ is a group, there exists $c' \in G$ such that $z \circ c' = c$.

Thus we have:

Hence $c'$ is an upper bound of $\set {x, y}$.

Let $d$ be an upper bound of $\set {x, y}$.

Then:

Hence $c'$ is an upper bound of $\set {x, y}$ which is smaller than an arbitrary upper bound $d$ of $\set {x, y}$.

That is, $c'$ is a supremum of $\set {x, y}$.

Hence by definition:
 * $z \circ \sup \set {x, y} = \sup \set {z \circ x, z \circ y}$

Thus we have shown that if any of the three sets $\set {x, y}$, $\set {x \circ z, y \circ z}$ or $\set {z \circ x, z \circ y}$ admit a supremum, they all do, and: