Strictly Order-Preserving and Order-Reversing Mapping on Strictly Totally Ordered Set is Injection

Theorem

Let $\struct {S, \prec_1}$ and $\struct {T, \prec_2}$ be strictly totally ordered sets.

Let $\phi: S \to T$ be a mapping.

Let $\pi: S \to T$ be a mapping with the property that:

$\forall x, y \in S: x \prec_1 y \iff \map \pi x \prec_2 \map \pi y$

Then $\pi$ is an injection.

Proof

Aiming for a contradiction, suppose $\pi$ is not an injection.

Hence:

$\exists x, y \in S: \map \pi x = \map \pi y$

As $S$ is strictly totally ordered:

$x \prec_1 y$ or $y \prec_1 x$

Without loss of generality, let $x \prec_1 y$.

Then we have:

$\map \pi x = \map \pi y$

But by hypothesis:

$\map \pi x \prec_2 \map \pi y$

Because $\prec_2$ is a strict ordering, it follows that:

$\map \pi x \ne \map \pi y$

It follows by Proof by Contradiction that it cannot be the case that $\pi$ is not an injection.

Hence the result.

$\blacksquare$

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 $20 \ \text {(c)}$