Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing

Theorem
Let $$\left({S; \le_1}\right)$$ be a totally ordered set and let $$\left({T; \le_2}\right)$$ be a poset.

A mapping $$\phi: \left({S; \le_1}\right) \to \left({T; \le_2}\right)$$ is an order monomorphism iff $$\phi$$ is strictly increasing.

Proof

 * Let $$\phi$$ be an order monomorphism:


 * Now let $$\phi$$ be strictly increasing.

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

Then $$\phi$$ is injective, by Monotone Mapping is Injective.

By Strictly Increasing is Increasing, $$x \le_1 y \Longrightarrow \phi \left({x}\right) \le_2 \phi \left({y}\right)$$.

Now suppose $$\phi \left({x}\right) \le_2 \phi \left({y}\right)$$.

Then, if $$y <_1 x$$, by the fact that $$\phi$$ is strictly increasing, we would have $$\phi \left({y}\right) <_2 \phi \left({x}\right)$$, which contradicts $$\phi \left({x}\right) \le_2 \phi \left({y}\right)$$.

Thus $$\phi \left({x}\right) \le_2 \phi \left({y}\right) \iff x \le_1 y$$ and $$\phi$$, being injective, has been proved to be an order monomorphism from its definition.