Relation Isomorphism preserves Total Ordering

Theorem

Let $\struct {A, \RR}$ and $\struct {B, \SS}$ be relational structures which are relationally isomorphic.

Let $\struct {A, \RR}$ be a totally ordered set.

Then $\struct {B, \SS}$ is also a totally ordered set.

Let $\struct {A, \RR}$ be a totally ordered set.

Recall the definition:

$\RR$ is a total ordering on $S$ if and only if:

$(1): \quad \RR$ is an ordering on $S$

$(2): \quad \RR$ is connected

That is, $\RR$ is an ordering with no non-comparable pairs:

$\forall x, y \in S: x \mathop \RR y \lor y \mathop \RR x$

$\SS$ is an ordering.

Let $\phi: \struct {A, \RR} \to \struct {B, \SS}$ be a relation isomorphism.

Let $x, y \in A$.

Then either $x \mathrel \RR y$ or $y \mathrel \RR x$.

From the definition of relation isomorphism, either:

$\map \phi x \mathrel \SS \map \phi y$

or:

$\map \phi y \mathrel \SS \map \phi x$

and so by definition $\SS$ is also a total ordering.

So by definition $\struct {B, \SS}$ is also a totally ordered set.

$\blacksquare$

1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.10$