Inverse of Infimum in Ordered Group is Supremum of Inverses

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Theorem

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

Let $x, y \in G$.


Then:

$\set {x, y}$ admits an infimum in $G$

if and only if:

$\set {x^{-1}, y^{-1} }$ admits a supremum in $G$

in which case:

$\paren {\inf \set {x, y} }^{-1} = \sup \set {x^{-1}, y^{-1} }$


Proof

Sufficient Condition

Let $\set {x, y}$ admits an infimum $c$ in $G$.

Then:

\(\ds c\) \(\preccurlyeq\) \(\ds x\) as $c$ is a lower bound of $\set {x, y}$
\(\, \ds \land \, \) \(\ds c\) \(\preccurlyeq\) \(\ds y\)
\(\ds \leadsto \ \ \) \(\ds x^{-1}\) \(\preccurlyeq\) \(\ds c^{-1}\) Inversion Mapping Reverses Ordering in Ordered Group
\(\, \ds \land \, \) \(\ds y^{-1}\) \(\preccurlyeq\) \(\ds c^{-1}\)

and so $c^{-1}$ is an upper bound of $\set {x^{-1}, y^{-1} }$.


Suppose $d$ is an upper bound of $\set {x^{-1}, y^{-1} }$.

Then:

\(\ds x^{-1}\) \(\preccurlyeq\) \(\ds d\) Definition of Upper Bound of Set
\(\, \ds \land \, \) \(\ds y^{-1}\) \(\preccurlyeq\) \(\ds d\)
\(\ds \leadsto \ \ \) \(\ds d^{-1}\) \(\preccurlyeq\) \(\ds x\) Inversion Mapping Reverses Ordering in Ordered Group
\(\, \ds \land \, \) \(\ds d^{-1}\) \(\preccurlyeq\) \(\ds y\)

That is, $d^{-1}$ is a lower bound of $\set {x, y}$.

But because $c$ is an infimum of $\set {x, y}$:

\(\ds c\) \(\preccurlyeq\) \(\ds d^{-1}\) Definition of Infimum of Set
\(\ds \leadsto \ \ \) \(\ds c^{-1}\) \(\preccurlyeq\) \(\ds d\) Inversion Mapping Reverses Ordering in Ordered Group

That is, for an arbitrary upper bound $d$ of $\set {x^{-1}, y^{-1} }$:

$c^{-1} \preccurlyeq d$

and so $c^{-1}$ is a supremum of $\set {x^{-1}, y^{-1} }$ by definition.


That is:

$\paren {\inf \set {x, y} }^{-1} = \sup \set {x^{-1}, y^{-1} }$

$\Box$


Necessary Condition

Let $\set {x^{-1}, y^{-1} }$ admit a supremum $c$ in $G$.

Then:

\(\ds x^{-1}\) \(\preccurlyeq\) \(\ds c\) as $c$ is an upper bound of $\set {x, y}$
\(\, \ds \land \, \) \(\ds y^{-1}\) \(\preccurlyeq\) \(\ds c\)
\(\ds \leadsto \ \ \) \(\ds c^{-1}\) \(\preccurlyeq\) \(\ds x\) Inversion Mapping Reverses Ordering in Ordered Group
\(\, \ds \land \, \) \(\ds c^{-1}\) \(\preccurlyeq\) \(\ds y\)

and so $c^{-1}$ is a lower bound of $\set {x, y}$.


Suppose $d$ is a lower bound of $\set {x, y}$.

Then:

\(\ds d\) \(\preccurlyeq\) \(\ds x\) Definition of Upper Bound of Set
\(\, \ds \land \, \) \(\ds d\) \(\preccurlyeq\) \(\ds y\)
\(\ds \leadsto \ \ \) \(\ds x^{-1}\) \(\preccurlyeq\) \(\ds d^{-1}\) Inversion Mapping Reverses Ordering in Ordered Group
\(\, \ds \land \, \) \(\ds y^{-1}\) \(\preccurlyeq\) \(\ds d^{-1}\)

That is, $d^{-1}$ is an upper bound of $\set {x, y}$.

But because $c$ is a supremum of $\set {x^{-1}, y^{-1} }$:

\(\ds d^{-1}\) \(\preccurlyeq\) \(\ds c\) Definition of Supremum of Set
\(\ds \leadsto \ \ \) \(\ds d\) \(\preccurlyeq\) \(\ds c^{-1}\) Inversion Mapping Reverses Ordering in Ordered Group

That is, for an arbitrary lower bound $d$ of $\set {x, y}$:

$d \preccurlyeq c^{-1}$

and so $c^{-1}$ is an infimum of $\set {x, y}$ by definition.


That is:

$\paren {\inf \set {x, y} } = \sup \set {x^{-1}, y^{-1} }^{-1}$

from which it follows from Group Axiom $\text G 3$: Existence of Inverse Element that:

$\paren {\inf \set {x, y} }^{-1} = \sup \set {x^{-1}, y^{-1} }$

$\blacksquare$


Sources

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