Order Isomorphism on Totally Ordered Set preserves Total Ordering

Theorem
Let $\struct {S, \preccurlyeq_1}$ and $\struct {T, \preccurlyeq_2}$ be ordered sets.

Let $\phi: \struct {S, \preccurlyeq_1} \to \struct {T, \preccurlyeq_2}$ be an order isomorphism.

Then $\struct {S, \preccurlyeq_1}$ is a totally ordered set $\struct {T, \preccurlyeq_2}$ is also a totally ordered set.

Necessary Condition
Let $\struct {S, \preccurlyeq_1}$ be a totally ordered set

Then by definition $\preccurlyeq_1$ is a total ordering.

Let $x, y \in S$.

Then either $x \preccurlyeq_1 y$ or $y \preccurlyeq_1 x$.

From the definition of order isomorphism, either:
 * $\map \phi x \preccurlyeq_2 \map \phi y$

or:
 * $\map \phi y \preccurlyeq_2 \map \phi x$

and so by definition $\preccurlyeq_2$ is also a total ordering.

So by definition $\struct {T, \preccurlyeq_2}$ is also a totally ordered set.

Sufficient Condition
By Inverse of Order Isomorphism is Order Isomorphism, if $\phi$ is an order isomorphism then so is $\phi^{-1}$.

Applying the other implication (proven above) to $\phi^{-1}$, it follows that if $\struct {T, \preccurlyeq_2}$ is a totally ordered set then so is $\struct {S, \preccurlyeq_1}$.