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
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
- 1993: Keith Devlin: The Joy of Sets: Fundamentals of Contemporary Set Theory (2nd ed.) ... (previous) ... (next): $\S 1$: Naive Set Theory: $\S 1.7$: Well-Orderings and Ordinals