Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing

Theorem
Let $\left({S, \preceq_1}\right)$ be a totally ordered set and let $\left({T, \preceq_2}\right)$ be a poset.

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

Then $\phi$ is an order embedding iff $\phi$ is strictly increasing.

That is, iff:


 * $\forall x, y \in S: x \prec_1 y \implies \phi \left({x}\right) \prec_2 \phi \left({y}\right)$

Forward Implication
Let $\phi$ be an order embedding.

Let $x, y \in S$ with $x \prec y$.

Then:

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

Necessary Condition
Now let $\phi$ be strictly increasing.

Then $\phi$ is a strictly monotone mapping by definition.

Thus $\phi$ is injective, by Strictly Monotone Mapping is Injective.

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

As $\left({S, \prec_1}\right)$ is a Linearly ordered set:
 * Either $y \prec_1 x$, $y = x$, or $x \prec_1 y$.

Suppose for the sake of contradiction that $y \prec_1 x$.

By the definition of a strictly increasing mapping:


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

which contradicts the fact that $\phi \left({x}\right) \preceq_2 \phi \left({y}\right)$.

Therefore $y \not \prec_1 x$.

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

Hence:


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

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