# Order Isomorphism is Equivalence Relation/Proof 1

## Theorem

Order isomorphism between ordered sets is an equivalence relation.

So any given family of ordered sets can be partitioned into disjoint classes of isomorphic sets.

## Proof

Let $\struct {S_1, \preccurlyeq_1} \cong \struct {S_2, \preccurlyeq_2}$ denote that $\struct {S_1, \preccurlyeq_1}$ is isomorphic to $\struct {S_2, \preccurlyeq_2}$.

Checking in turn each of the criteria for equivalence:

### Reflexivity

Let $\struct {S, \preccurlyeq}$ be an ordered set.

Then $\struct {S, \preccurlyeq}$ is isomorphic to itself.

Thus $\cong$ is seen to be reflexive.

$\Box$

### Symmetric

Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be ordered sets.

Let $\struct {S_1, \preccurlyeq_1}$ be isomorphic to $\struct {S_2, \preccurlyeq_2}$.

Then $\struct {S_2, \preccurlyeq_2}$ is isomorphic to $\struct {S_1, \preccurlyeq_1}$.

Thus $\cong$ is seen to be symmetric.

$\Box$

### Transitive

Let $\struct {S_1, \preccurlyeq_1}$, $\struct {S_2, \preccurlyeq_2}$ and $\struct {S_3, \preccurlyeq_3}$ be ordered sets.

Let $\struct {S_1, \preccurlyeq_1}$ be isomorphic to $\struct {S_2, \preccurlyeq_2}$.

Let $\struct {S_2, \preccurlyeq_2}$ be isomorphic to $\struct {S_3, \preccurlyeq_3}$.

Then $\struct {S_1, \preccurlyeq_1}$ is isomorphic to $\struct {S_3, \preccurlyeq_3}$.

Thus $\cong$ is seen to be transitive.

$\Box$

$\cong$ has been shown to be reflexive, symmetric and transitive.

Hence the result.

$\blacksquare$