Order Types of Duals of Isomorphic Sets are Equal

Theorem

Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be ordered sets.

Let:

$\map \ot {S_1, \preccurlyeq_1} = \map \ot {S_2, \preccurlyeq_2}$

where $\ot$ denotes the order type operator.

Let $\struct {S_1, \succcurlyeq_1}$ and $\struct {S_2, \succcurlyeq_2}$ denote the dual ordered sets of $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$.

Then:

$\map \ot {S_1, \succcurlyeq_1} = \map \ot {S_2, \succcurlyeq_2}$

Proof

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Sources

• 1996: Winfried Just and Martin Weese: Discovering Modern Set Theory. I: The Basics ... (previous) ... (next): Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $27 \ \text {(d)}$