Duals of Isomorphic Ordered Sets are Isomorphic
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Theorem
Let $\struct {S, \preccurlyeq_1}$ and $\struct {T, \preccurlyeq_2}$ be ordered sets.
Let $\struct {S, \succcurlyeq_1}$ and $\struct {T, \succcurlyeq_2}$ be the dual ordered sets of $\struct {S, \preccurlyeq_1}$ and $\struct {T, \preccurlyeq_2}$ respectively.
Let $f: \struct {S, \preccurlyeq_1} \to \struct {T, \preccurlyeq_2}$ be an order isomorphism.
Then $f: \struct {S, \succcurlyeq_1} \to \struct {T, \succcurlyeq_2} $ is also an order isomorphism.
Proof
\(\ds \forall x, y \in S: \, \) | \(\ds x\) | \(\succcurlyeq_1\) | \(\ds y\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds y\) | \(\preccurlyeq_1\) | \(\ds x\) | Definition of Dual Ordering | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \map f y\) | \(\preccurlyeq_2\) | \(\ds \map f x\) | Definition of Order Isomorphism | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \map f x\) | \(\succcurlyeq_2\) | \(\ds \map f y\) | Definition of Dual Ordering |
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Theorem $14.2$
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.8$