# Definition:Order Isomorphism

## Definition

### Definition 1

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

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

$\phi: S \to T$ is order-preserving
$\phi^{-1}: T \to S$ is order-preserving.

Then $\phi$ is an order isomorphism.

### Definition 2

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

Let $\phi: S \to T$ be a surjective order embedding.

Then $\phi$ is an order isomorphism.

That is, $\phi$ is an order isomorphism if and only if:

$(1): \quad \phi$ is surjective
$(2): \quad \forall x, y \in S: x \preceq_1 y \iff \phi \left({x}\right) \preceq_2 \phi \left({y}\right)$

Two ordered sets $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ are (order) isomorphic if there exists such an order isomorphism between them.

$\struct {S, \preceq_1}$ is described as (order) isomorphic to (or with) $\struct {T, \preceq_2}$, and vice versa.

This may be written $\struct {S, \preceq_1} \cong \struct {T, \preceq_2}$.

Where no confusion is possible, it may be abbreviated to $S \cong T$.

### Well-Ordered Sets

When $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ are well-ordered sets, the condition on the order preservation can be relaxed:

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

Let $\phi: S \to T$ be a bijection such that $\phi: S \to T$ is order-preserving:

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

Then $\phi$ is an order isomorphism.

## Also see

• Results about order isomorphisms can be found here.

## Linguistic Note

The word isomorphism derives from the Greek morphe (μορφή) meaning form or structure, with the prefix iso- meaning equal.

Thus isomorphism means equal structure.