# Inverse of Supremum in Ordered Group is Infimum of Inverses

## Theorem

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

Let $x, y \in G$.

Then:

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

in which case:

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

## Proof

Let:

 $\ds a$ $=$ $\ds x^{-1}$ $\ds b$ $=$ $\ds y^{-1}$

Recall from Inverse of Group Inverse:

 $\ds a^{-1}$ $=$ $\ds x$ $\ds b^{-1}$ $=$ $\ds y$

Then:

$\set {a, b}$ admits an infimum in $G$
$\set {a^{-1}, b^{-1} }$ admits a supremum in $G$

in which case:

$\paren {\inf \set {a, b} }^{-1} = \sup \set {a^{-1}, b^{-1} }$

Substituting back for $a$ and $b$:

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

in which case:

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

Hence from Group Axiom $\text G 3$: Existence of Inverse Element:

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

$\blacksquare$