# Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing/Reverse Implication

## Theorem

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

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

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

Then $\phi$ is an order embedding.

## Proof 1

Let $x \preceq_1 y$.

Then $x = y$ or $x \prec_1 y$.

Let $x = y$.

Then

- $\map \phi x = \map \phi y$

so:

- $\map \phi x \preceq_2 \map \phi y$

Let $x \prec_1 y$.

Then by the definition of strictly increasing mapping:

- $\map \phi x \prec_2 \map \phi y$

so by the definition of $\prec_2$:

- $\map \phi x \preceq_2 \map \phi y$

Thus:

- $x \preceq_1 y \implies \map \phi x \preceq_2 \map \phi y$

It remains to be shown that:

- $\map \phi x \preceq_2 \map \phi y \implies x \preceq_1 y$

Suppose that $x \npreceq_1 y$.

Since $\preceq_1$ is a total ordering:

- $y \prec_1 x$

Thus since $\phi$ is strictly increasing:

- $\map \phi y \prec_1 \map \phi x$

Thus:

- $\map \phi x \npreceq_1 \map \phi y$

Therefore:

- $x \npreceq_1 y \implies \map \phi x \npreceq_2 \map \phi y$

By the Rule of Transposition:

- $\map \phi x \preceq_2 \map \phi y \implies x \preceq y$

$\blacksquare$

## Proof 2

Let $\phi$ be strictly increasing.

Let $\map \phi x \preceq_2 \map \phi y$.

As $\struct {S, \prec_1}$ is a strictly totally ordered set:

- Either $y \prec_1 x$, $y = x$, or $x \prec_1 y$.

Aiming for a contradiction, suppose that $y \prec_1 x$.

By the definition of a strictly increasing mapping:

- $\map \phi y \prec_2 \map \phi x$

which contradicts the fact that $\map \phi x \preceq_2 \map \phi y$.

Therefore $y \nprec_1 x$.

Thus $y = x$, or $x \prec_1 y$, so $x \preceq_1 y$.

Hence:

- $\map \phi x \preceq_2 \map \phi y \iff x \preceq_1 y$

and $\phi$ has been proved to be an order embedding.

$\blacksquare$