Bijection on Total Ordering is Order Isomorphism

Theorem

Let $\struct {S, \preccurlyeq}$ be a totally ordered set.

Let $f: S \to T$ be a bijection to an arbitrary set $T$.

Let $\RR$ be a relation on $T$ defined such that:

$\RR: = \set {\tuple {\map f x, \map f y}: x \preccurlyeq y}$

Then $f$ is an order isomorphism between $\struct {S, \preccurlyeq}$ and $\struct {T, \RR}$.

Proof

From Bijection on Total Ordering reflects Total Ordering, we have that $\RR$ is a total ordering.

We have that $f$ is a bijection.

Hence a fortiori $f$ is also a surjection.

Let $x, y \in S$ such that $x \preccurlyeq y$.

Then by definition of $\RR$:

$\map f x \mathrel \RR \map f y$

Now let $\map f x \mathrel \RR \map f y$.

Then again by definition of $\RR$:

$x \preccurlyeq y$

That is, $f$ is by definition an order isomorphism between $\struct {S, \preccurlyeq}$ and $\struct {T, \RR}$.

$\blacksquare$

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 $41 \ \text {(d)}$