Equivalence of Definitions of Order Embedding/Definition 3 implies Definition 1
Theorem
Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be ordered sets.
Let $\phi: S \to T$ be a mapping.
Let $\phi: S \to T$ be an order embedding by Definition 3:
$\phi$ is an order embedding of $S$ into $T$ if and only if both of the following conditions hold:
- $(1): \quad \phi$ is an injection
- $(2): \quad \forall x, y \in S: x \prec_1 y \iff \map \phi x \prec_2 \map \phi y$
Then $\phi: S \to T$ is an order embedding by Definition 1:
$\phi$ is an order embedding of $S$ into $T$ if and only if:
- $\forall x, y \in S: x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$
Proof
Let $\phi$ be an order embedding by definition 3.
Then by definition:
- $(1): \quad \phi$ is injective
- $(2): \quad \forall x, y, \in S: x \prec_1 y \iff \map \phi x \prec_2 \map \phi y$
Let $x \preceq_1 y$.
Then $x \prec_1 y$ or $x = y$.
If $x \prec_1 y$, then by hypothesis:
- $\map \phi x \prec_2 \map \phi y$
Thus:
- $\map \phi x \preceq_2 \map \phi y$
If $x = y$, then:
- $\map \phi x = \map \phi y$
Thus:
- $\map \phi x \preceq_2 \map \phi y$
Thus it has been shown that:
- $x \preceq_1 y \implies \map \phi x \preceq_2 \map \phi y$
$\Box$
Let $\map \phi x \preceq_2 \map \phi y$.
Then:
- $\map \phi x \prec_2 \map \phi y$
or:
- $\map \phi x = \map \phi y$
Suppose $\map \phi x \prec_2 \map \phi y$.
Then by hypothesis:
- $x \prec_1 y$
and so:
- $x \preceq_1 y$
Suppose $\map \phi x = \map \phi y$.
Then since $\phi$ is injective:
- $x = y$
and so:
- $x \preceq_1 y$
Thus in both cases:
- $x \preceq_1 y$
and so:
- $\map \phi x \preceq_2 \map \phi y \implies x \preceq_1 y$
$\Box$
Hence the result:
- $x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$
and so $\phi$ is an order embedding by definition 1.
$\blacksquare$