Definition:Order Isomorphism

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

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.

That is, $\phi$ is an order isomorphism iff:


 * $\forall x, y \in S: x \mathop {\preceq_1} y \iff \phi \left({x}\right) \mathop {\preceq_2} \phi \left({y}\right)$

So an order isomorphism can be described as a bijection that preserves ordering in both directions.

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

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

This may be written $\left({S, \preceq_1}\right) \cong \left({T, \preceq_2}\right)$.

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

Well-Ordered Sets
When $\left({S, \preceq_1}\right)$ and $\left({T, \preceq_2}\right)$ are well-ordered sets, the condition on the order preservation can be relaxed:

Also see

 * Relation Isomorphism, from which it can be seen that order isomorphism is a special case.
 * Inverse of Increasing Bijection need not be Increasing