Order Isomorphism is Preserved by Antilexicographic Order
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Theorem
Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be ordered sets.
Let $\struct {T_1, \preccurlyeq_{1'} }$ and $\struct {T_2, \preccurlyeq_{2'} }$ be ordered sets such that:
- $\struct {S_1, \preccurlyeq_1}$ is isomorphic to $\struct {T_1, \preccurlyeq_{1'} }$
- $\struct {S_2, \preccurlyeq_2}$ is isomorphic to $\struct {T_2, \preccurlyeq_{2'} }$
Then the antilexicographic order $\struct {S_1, \preccurlyeq_1} \otimes^a \struct {S_2, \preccurlyeq_2}$ is isomorphic to $\struct {T_2, \preccurlyeq_{2'} } \otimes^a \struct {T_2, \preccurlyeq_{2'} }$.
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 $33$