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 monomorphism iff $\phi$ is strictly increasing.

That is, iff:


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

Sufficient Condition
Let $\phi$ be an order monomorphism:

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

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

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

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

By Strictly Increasing Mapping is Increasing:


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

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

Then, since if $\phi$ is strictly increasing:


 * $y \prec_1 x \implies \phi \left({y}\right) \prec_2 \phi \left({x}\right)$

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

Thus $y \not \prec_1 x$.

As $\left({S, \prec_1}\right)$ is totally ordered, this means $x \preceq_1 y$.

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 monomorphism.