# Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing/Reverse Implication/Proof 1

## 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

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:

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

$\blacksquare$