# Order Isomorphic Sets are Equivalent

## Theorem

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

Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be order isomorphic.

Then $S$ and $T$ are equivalent.

## Proof

By definition, an order isomorphism is a bijection $\phi$ such that:

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

So, by definition, there exists a bijection between $S$ and $T$.

The result follows by definition of set equivalence.

$\blacksquare$