# Strictly Monotone Mapping with Totally Ordered Domain is Injective

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## Theorem

Let $\struct {S, \preceq_1}$ be a totally ordered set.

Let $\struct {T, \preceq_2}$ be an ordered set.

Let $\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$ be a strictly monotone mapping.

Then $\phi$ is injective.

## Proof

\(\displaystyle x, y\) | \(\in\) | \(\displaystyle S\) | |||||||||||

\(\, \displaystyle \land \, \) | \(\displaystyle x\) | \(\ne\) | \(\displaystyle y\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x\) | \(\prec_1\) | \(\displaystyle y\) | ||||||||||

\(\, \displaystyle \lor \, \) | \(\displaystyle y\) | \(\prec_1\) | \(\displaystyle x\) | Trichotomy Law | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \map \phi x\) | \(\prec_2\) | \(\displaystyle \map \phi y\) | ||||||||||

\(\, \displaystyle \lor \, \) | \(\displaystyle \map \phi y\) | \(\prec_2\) | \(\displaystyle \map \phi x\) | $\phi$ is strictly monotone | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \map \phi x\) | \(\ne\) | \(\displaystyle \map \phi y\) | Definition of $\prec_2$ |

Hence the result.

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 14$: Theorem $14.9: \ 1^\circ$