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.

Warning
It does not necessarily follow that if:
 * $\phi: S \to T$ is a bijection which is order-preserving

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

Example
Let $S = \mathcal P \left({\left\{{a, b}\right\}}\right), T = \left\{{1, 2, 3, 4}\right\}$.

From Subset Relation on Power Set is Partial Ordering, we have that $\left({S, \subseteq}\right)$ is a poset.

Clearly so is $\left({T, \le}\right)$ (although its ordering is in fact total, it is still technically a poset).

Let $\phi: S \to T$ be defined as:
 * $\phi \left({\varnothing}\right) = 1$
 * $\phi \left({\left\{{a}\right\}}\right) = 2$
 * $\phi \left({\left\{{b}\right\}}\right) = 3$
 * $\phi \left({\left\{{a, b}\right\}}\right) = 4$

It is easily verified that $\phi: \left({S, \subseteq}\right) \to \left({T, \le}\right)$ is a bijection which is order-preserving.

But note that while $2 \le 3$, it is not the case that $\left\{{a}\right\} \subseteq \left\{{b}\right\}$ and so $\phi^{-1}$ is not order-preserving.

Also see

 * Relation Isomorphism, from which it can be seen that order isomorphism is a special case.