Order Isomorphic Sets are Equivalent

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

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