Equivalence of Definitions of Order Isomorphism

Theorem
Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be ordered sets.

Definition 1 implies Definition 2
Let $\phi: S \to T$ be an order isomorphism by Definition 1.

Then $\phi$ is bijective, and thus trivially surjective.

Let $x, y \in S$.

Then by Definition 1:


 * $x \preceq_1 y \implies \map \phi x \preceq_2 \map \phi y$

Suppose that $\map \phi x \preceq_2 \map \phi y$.

Then by Definition 1:


 * $\map {\phi^{-1} } {\map \phi x} \preceq_1 \map {\phi^{-1} } {\map \phi y}$

By the definition of inverse:


 * $x \preceq_1 y$

Thus by Rule of Implication:


 * $\map \phi x \preceq_2 \map \phi y \implies x \preceq_1 y$.

So:


 * $x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$

As this holds for all $x, y \in S$ and $\phi$ is surjective, $\phi$ is an order isomorphism by Definition 2.

Definition 2 implies Definition 1
Let $\phi: S \to T$ be an order isomorphism by Definition 2.

Then $\phi$ is a surjective order embedding.

By Order Embedding is Injection, $\phi$ is injective.

As it is also surjective, $\phi$ is bijective, and thus satisfies the first condition of Definition 1.

Since $\phi$ is an order embedding:
 * $\forall x, y \in S: \paren {x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y}$

Thus
 * $\forall x, y \in S: \paren {x \preceq_1 y \implies \map \phi x \preceq_2 \map \phi y}$

This satisfies the second condition of Definition 1.

Furthermore:


 * $\forall x, y \in S: \paren {\map \phi x \preceq_2 \map \phi y \implies x \preceq_1 y}$

Let $p, q \in T$ and let $p \preceq_2 q$.

Then since $\phi$ is bijective, it has an inverse $\phi^{-1}$ such that


 * $\map \phi {\map {\phi^{-1} } p} = p$
 * $\map \phi {\map {\phi^{-1} } q} = q$

Thus:
 * $\map \phi {\map {\phi^{-1} } p} \preceq_2 \map \phi {\map {\phi^{-1} } q}$

Therefore:


 * $\map {\phi^{-1} } p \preceq_1 \map {\phi^{-1} } q$

Thus we see that:


 * $\forall p, q \in T: \paren {p \preceq_2 q \implies \map {\phi^{-1} } p \preceq_2 \map {\phi^{-1} } q}$

Thus we have satisfied the final condition of Definition 1.

Definition 1 iff Definition 3
Let $\phi: S \to T$ be an order isomorphism by Definition 1.

Then by Definition 1:

That is:
 * $\forall x, y \in S: \map \phi x \preceq_2 \map \phi y \implies x \preceq_1 y$

which together with:
 * $\forall x, y \in S: x \preceq_1 y \implies \map \phi x \preceq_2 \map \phi y$

gives:
 * $\forall x, y \in S: x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$

Hence $\phi$ is an order isomorphism by Definition 3.

The above argument reverses.