Inverse of Supremum in Ordered Group is Infimum 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 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:
\(\ds a^{-1}\) | \(=\) | \(\ds x\) | ||||||||||||
\(\ds b^{-1}\) | \(=\) | \(\ds y\) |
From Inverse of Infimum in Ordered Group is Supremum of Inverses:
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:
- $\inf \set {x^{-1}, x^{-1} } = \paren {\sup \set {x, y} }^{-1}$
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Exercise $15.10 \ \text {(d)}$