Order Isomorphism between Tosets is not necessarily Unique

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Theorem

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

Let $\struct {S_1, \preccurlyeq_1} \cong \struct {S_2, \preccurlyeq_2}$, that is, let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be order isomorphic.


Then it is not necessarily the case that there is exactly one mapping $f: S_1 \to S_2$ such that $f$ is an order isomorphism.


Proof

Proof by Counterexample:

Let $\Z$ denote the set of integers.

We have that Integers under Usual Ordering form Totally Ordered Set.

Let $m \in \Z$ be an arbitrary integer.


Let $f_m: \Z \to \Z$ be the order isomorphism defined as:

$\forall a \in \Z: \map {f_m} a = a + m$

It follows that there are at least as many order isomorphisms of from $\Z$ to $\Z$ as there are integers.

Hence the result

$\blacksquare$


Sources