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.

$\blacksquare$

Sources

• 1968: A.N. Kolmogorov and S.V. Fomin: Introductory Real Analysis ... (previous) ... (next): $\S 3.3$: Ordered sets. Order types