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:


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

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

Two posets are (order) isomorphic if there exists such an order isomorphism between them.

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.