# Inverse of Infimum in Ordered Group is Supremum of Inverses

## Theorem

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

Let $x, y \in G$.

Then:

$\set {x, y}$ admits an infimum in $G$
$\set {x^{-1}, y^{-1} }$ 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:

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

$\blacksquare$