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

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

Then $\phi$ is an order embedding if and only if $\phi$ is strictly increasing.

That is:

$\forall x, y \in S: x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$
$\forall x, y \in S: x \prec_1 y \implies \map \phi x \prec_2 \map \phi y$

## Proof

### Forward Implication

Let $\phi$ be an order embedding.

Let $x, y \in S$ with $x \prec_1 y$, where $\prec_1$ is the strict predecessor relation.

Then:

 $\ds$  $\ds x \prec_1 y \land x \ne y$ Definition of Strictly Precedes $\ds$ $\leadsto$ $\ds \map \phi x \preceq_2 \map \phi y$ Definition of Order Embedding $\ds$ $\leadsto$ $\ds \map \phi x \prec_2 \map \phi y$ $\phi$ is injective

So by definition, $\phi$ is strictly increasing.

$\Box$

### Reverse Implication

#### 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.

$\blacksquare$