Relation Isomorphism preserves Ordering
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Theorem
Let $\struct {A, \RR}$ and $\struct {B, \SS}$ be relational structures which are relationally isomorphic.
Let $\struct {A, \RR}$ be an ordered set.
Then $\struct {B, \SS}$ is also an ordered set.
Proof
Let $\struct {A, \RR}$ be an ordered set.
Recall the definition:
$\RR$ is an ordering on $S$ if and only if $\RR$ satisfies the ordering axioms:
\((1)\) | $:$ | $\RR$ is reflexive | \(\ds \forall a \in S:\) | \(\ds a \mathrel \RR a \) | |||||
\((2)\) | $:$ | $\RR$ is transitive | \(\ds \forall a, b, c \in S:\) | \(\ds a \mathrel \RR b \land b \mathrel \RR c \implies a \mathrel \RR c \) | |||||
\((3)\) | $:$ | $\RR$ is antisymmetric | \(\ds \forall a, b \in S:\) | \(\ds a \mathrel \RR b \land b \mathrel \RR a \implies a = b \) |
From Relation Isomorphism Preserves Reflexivity:
- $\SS$ is reflexive.
From Relation Isomorphism Preserves Antisymmetry:
- $\SS$ is antisymmetric.
From Relation Isomorphism Preserves Transitivity:
- $\SS$ is transitive.
So by definition $\struct {B, \SS}$ is an ordered set.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.10$