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:

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

if and only if:

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


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:

\(\displaystyle \) \(\) \(\displaystyle x \prec_1 y \land x \ne y\) Definition of Strictly Precedes
\(\displaystyle \) \(\leadsto\) \(\displaystyle \map \phi x \preceq_2 \map \phi y\) Definition of Order Embedding
\(\displaystyle \) \(\leadsto\) \(\displaystyle \map \phi x \prec_2 \map \phi y\) $\phi$ 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

$\phi \left({x}\right) = \phi \left({y}\right)$

so:

$\phi \left({x}\right) \preceq_2 \phi \left({y}\right)$


Let $x \prec_1 y$.

Then by the definition of strictly increasing mapping:

$\phi \left({x}\right) \prec_2 \phi \left({y}\right)$

so by the definition of $\prec_2$:

$\phi \left({x}\right) \preceq_2 \phi \left({y}\right)$

Thus:

$x \preceq_1 y \implies \phi \left({x}\right) \preceq_2 \phi \left({y}\right)$


It remains to be shown that:

$\phi \left({x}\right) \preceq_2 \phi \left({y}\right) \implies x \preceq_1 y$

Suppose that $x \npreceq_1 y$.

Since $\preceq_1$ is a total ordering:

$y \prec_1 x$

Thus since $\phi$ is strictly increasing:

$\phi \left({y}\right) \prec_1 \phi \left({x}\right)$

Thus:

$\phi \left({x}\right) \not\preceq_1 \phi \left({y}\right)$

Therefore:

$x \npreceq_1 y \implies \phi \left({x}\right) \npreceq_2 \phi \left({y}\right)$

By the Rule of Transposition:

$\phi \left({x}\right) \preceq_2 \phi \left({y}\right) \implies x \preceq y$

$\blacksquare$


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.

$\blacksquare$


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