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:
- $\paren {\inf \set {x, y} }^{-1} = \sup \set {x^{-1}, y^{-1} }$
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Exercise $15.10 \ \text {(c)}$