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 injection such that for all $p, q \in S$:


 * $p \prec_1 q \iff \phi \left({p}\right) \prec_2 \phi \left({q}\right)$

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 2 then it is also an order embedding by Definition 1.

Proof
Let $\phi: S \to T$ be a injection such that for all $p, q \in S$:


 * $p \prec_1 q \iff \phi \left({p}\right) \prec_2 \phi \left({q}\right)$

Forward implication
Suppose that $p \preceq_1 q$.

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

If $p \prec_1 q$, then by the premise, $\phi \left({p}\right) \prec_2 \phi \left({q}\right)$.

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

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

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

We have shown that:


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

Reverse implication
Suppose that $\phi \left({p}\right) \preceq_2 \phi \left({q}\right)$.

Then $\phi \left({p}\right) \prec_2 \phi \left({q}\right)$ or $\phi \left({p}\right) = \phi \left({q}\right)$

If $\phi \left({p}\right) \prec_2 \phi \left({q}\right)$, then by the premise:


 * $p \prec_1 q$, so $p \preceq_1 q$.

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


 * $p = q$, so $p \preceq_1 q$.