Equivalence of Definitions of Order Embedding/Definition 3 implies Definition 1

Theorem
Let $\left({S, \preceq_1}\right)$ and $\left({T, \preceq_2}\right)$ be ordered sets.

Let $\phi: S \to T$ be a mapping such that:


 * $(1): \quad \phi$ is injective


 * $(2): \quad \forall x, y, \in S: x \prec_1 y \iff \phi \left({x}\right) \prec_2 \phi \left({y}\right)$

Then for all $p, q \in S$:


 * $p \preceq_1 q \iff \phi \left({p}\right) \preceq_2 \phi \left({q}\right)$

That is, if $\phi$ is an order embedding by Definition 3 then it is also an order embedding by Definition 1.

Proof
Let $\phi$ be an order embedding by definition 3.

Then by definition:
 * $(1): \quad \phi$ is injective


 * $(2): \quad \forall x, y, \in S: x \prec_1 y \iff \phi \left({x}\right) \prec_2 \phi \left({y}\right)$

Let $p \preceq_1 q$.

Then $p \prec_1 q$ or $p = q$.

If $p \prec_1 q$, then by hypothesis:
 * $\phi \left({p}\right) \prec_2 \phi \left({q}\right)$

Thus:
 * $\phi \left({p}\right) \preceq_2 \phi \left({q}\right)$

If $p = q$, then:
 * $\phi \left({p}\right) = \phi \left({q}\right)$

Thus:
 * $\phi \left({p}\right) \preceq_2 \phi \left({q}\right)$

Thus it has been shown that:


 * $p \preceq_1 q \implies \phi \left({p}\right) \preceq_2 \phi \left({q}\right)$

Let $\phi \left({p}\right) \preceq_2 \phi \left({q}\right)$.

Then:
 * $\phi \left({p}\right) \prec_2 \phi \left({q}\right)$

or:
 * $\phi \left({p}\right) = \phi \left({q}\right)$

Suppose $\phi \left({p}\right) \prec_2 \phi \left({q}\right)$.

Then by hypothesis:
 * $p \prec_1 q$

and so:
 * $p \preceq_1 q$

Suppose $\phi \left({p}\right) = \phi \left({q}\right)$.

Then since $\phi$ is injective:
 * $p = q$

and so:
 * $p \preceq_1 q$

Thus in both cases:
 * $p \preceq_1 q$

and so:
 * $\phi \left({p}\right) \preceq_2 \phi \left({q}\right) \implies p \preceq_1 q$

Hence the result:
 * $p \preceq_1 q \iff \phi \left({p}\right) \preceq_2 \phi \left({q}\right)$

and so $\phi$ is an order embedding by definition 1.