Order Isomorphism is Preserved by Antilexicographic Order

From ProofWiki
Jump to navigation Jump to search

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




Sources