# 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 {T, \succcurlyeq_2} $ is also an order isomorphism.

## Proof

\(\displaystyle \forall x, y \in S: \ \ \) | \(\displaystyle x\) | \(\succcurlyeq_1\) | \(\displaystyle y\) | ||||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle y\) | \(\preccurlyeq_1\) | \(\displaystyle x\) | Definition of Dual Ordering | |||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle \map f y\) | \(\preccurlyeq_2\) | \(\displaystyle \map f x\) | Definition of Order Isomorphism | |||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle \map f x\) | \(\succcurlyeq_2\) | \(\displaystyle \map f y\) | Definition of Dual Ordering |

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 14$: Theorem $14.2$