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 $G$

if and only if:

$\set {x^{-1}, y^{-1} }$ 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 $G$

if and only if:

$\set {a^{-1}, b^{-1} }$ 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 $G$

if and only if:

$\set {x, y}$ 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

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