Suprema in Ordered Group

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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:

\(\ds \sup \set {x \circ z, y \circ z}\) \(=\) \(\ds \sup \set {x, y} \circ z\)
\(\ds \sup \set {z \circ x, z \circ y}\) \(=\) \(\ds z \circ \sup \set {x, y}\)


Proof

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

\(\ds \forall x, y, z \in G: \, \) \(\ds x \preccurlyeq y\) \(\implies\) \(\ds x \circ z \preccurlyeq y \circ z\)
\(\ds \land \ \ \) \(\ds x \preccurlyeq y\) \(\implies\) \(\ds z \circ x \preccurlyeq z \circ y\)


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:

\(\ds x\) \(\preccurlyeq\) \(\ds c\) Definition of Upper Bound of Set
\(\, \ds \land \, \) \(\ds y\) \(\preccurlyeq\) \(\ds c\)
\(\ds \leadsto \ \ \) \(\ds x \circ z\) \(\preccurlyeq\) \(\ds c \circ z\) Definition of Relation Compatible with Operation
\(\, \ds \land \, \) \(\ds y \circ z\) \(\preccurlyeq\) \(\ds c \circ z\)

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:

\(\ds x \circ z\) \(\preccurlyeq\) \(\ds d' \circ z\) Definition of Upper Bound of Set
\(\, \ds \land \, \) \(\ds y \circ z\) \(\preccurlyeq\) \(\ds d' \circ z\)
\(\ds \leadsto \ \ \) \(\ds x \circ z \circ z^{-1}\) \(\preccurlyeq\) \(\ds d' \circ z \circ z^{-1}\) Definition of Relation Compatible with Operation
\(\, \ds \land \, \) \(\ds y \circ z \circ z^{-1}\) \(\preccurlyeq\) \(\ds d' \circ z \circ z^{-1}\)
\(\ds \leadsto \ \ \) \(\ds x\) \(\preccurlyeq\) \(\ds d'\) Group Axiom $\text G 3$: Existence of Inverse Element
\(\, \ds \land \, \) \(\ds y\) \(\preccurlyeq\) \(\ds d'\)
\(\ds \leadsto \ \ \) \(\ds c\) \(\preccurlyeq\) \(\ds d'\) Definition of Supremum of Set: $d'$ is an upper bound of $\set {x, y}$
\(\ds \leadsto \ \ \) \(\ds c \circ z\) \(\preccurlyeq\) \(\ds d' \circ z\) Definition of Relation Compatible with Operation

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 $\set {x \circ z, y \circ z}$.

$\Box$


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:

\(\ds x \circ z\) \(\preccurlyeq\) \(\ds c' \circ z\) Definition of Upper Bound of Set
\(\, \ds \land \, \) \(\ds y \circ z\) \(\preccurlyeq\) \(\ds c' \circ z\)
\(\ds \leadsto \ \ \) \(\ds x \circ z \circ z^{-1}\) \(\preccurlyeq\) \(\ds c' \circ z \circ z^{-1}\) Definition of Relation Compatible with Operation
\(\, \ds \land \, \) \(\ds y \circ z \circ z^{-1}\) \(\preccurlyeq\) \(\ds c' \circ z \circ z^{-1}\)
\(\ds \leadsto \ \ \) \(\ds x\) \(\preccurlyeq\) \(\ds c'\) Group Axiom $\text G 3$: Existence of Inverse Element
\(\, \ds \land \, \) \(\ds y\) \(\preccurlyeq\) \(\ds c'\)
\(\ds \leadsto \ \ \) \(\ds z \circ x\) \(\preccurlyeq\) \(\ds z \circ c'\) Definition of Relation Compatible with Operation
\(\, \ds \land \, \) \(\ds z \circ y\) \(\preccurlyeq\) \(\ds z \circ c'\)

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:

\(\ds z \circ x\) \(\preccurlyeq\) \(\ds d\) Definition of Upper Bound of Set
\(\, \ds \land \, \) \(\ds z \circ y\) \(\preccurlyeq\) \(\ds d\)
\(\ds z \circ x\) \(\preccurlyeq\) \(\ds z \circ d'\) Definition of $d'$
\(\, \ds \land \, \) \(\ds z \circ y\) \(\preccurlyeq\) \(\ds z \circ d'\)
\(\ds \leadsto \ \ \) \(\ds z \circ c'\) \(\preccurlyeq\) \(\ds z \circ d'\) Definition of Supremum of Set: $z \circ d'$ is an upper bound of $\set {z \circ x, z \circ y}$
\(\ds \leadsto \ \ \) \(\ds z \circ c'\) \(\preccurlyeq\) \(\ds d\) Definition of $d'$

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 $\set {z \circ x, z \circ y}$.

$\Box$


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:

\(\ds z \circ x\) \(\preccurlyeq\) \(\ds z \circ c'\) Definition of Upper Bound of Set
\(\, \ds \land \, \) \(\ds z \circ y\) \(\preccurlyeq\) \(\ds z \circ c'\)
\(\ds \leadsto \ \ \) \(\ds z^{-1} \circ z \circ x\) \(\preccurlyeq\) \(\ds z^{-1} \circ z \circ c'\) Definition of Relation Compatible with Operation
\(\, \ds \land \, \) \(\ds z^{-1} \circ z \circ y\) \(\preccurlyeq\) \(\ds z^{-1} \circ z \circ c'\)
\(\ds \leadsto \ \ \) \(\ds x\) \(\preccurlyeq\) \(\ds c'\) Group Axiom $\text G 3$: Existence of Inverse Element
\(\, \ds \land \, \) \(\ds y\) \(\preccurlyeq\) \(\ds c'\)

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


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

Then:

\(\ds x\) \(\preccurlyeq\) \(\ds d\) Definition of Upper Bound of Set
\(\, \ds \land \, \) \(\ds y\) \(\preccurlyeq\) \(\ds d\)
\(\ds z \circ x\) \(\preccurlyeq\) \(\ds z \circ d\) Definition of $d'$
\(\, \ds \land \, \) \(\ds z \circ y\) \(\preccurlyeq\) \(\ds z \circ d\)
\(\ds \leadsto \ \ \) \(\ds z \circ c'\) \(\preccurlyeq\) \(\ds z \circ d\) Definition of Supremum of Set: $z \circ d$ is an upper bound of $\set {z \circ x, z \circ y}$
\(\ds \leadsto \ \ \) \(\ds z^{-1} \circ z \circ c'\) \(\preccurlyeq\) \(\ds z^{-1} \circ z \circ d\) Definition of Relation Compatible with Operation
\(\ds \leadsto \ \ \) \(\ds c'\) \(\preccurlyeq\) \(\ds d\) Definition of $d'$

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:

$z \circ \sup \set {x, y} = \sup \set {z \circ x, z \circ y}$

$\Box$


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:

\(\ds \sup \set {x \circ z, y \circ z}\) \(=\) \(\ds \sup \set {x, y} \circ z\)
\(\ds \sup \set {z \circ x, z \circ y}\) \(=\) \(\ds z \circ \sup \set {x, y}\)

$\blacksquare$


Sources

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