# Order Isomorphism between Wosets is Unique

## Theorem

Let $\struct {S_1, \preceq_1}$ and $\struct {S_2, \preceq_2}$ be wosets.

Let $\struct {S_1, \preceq_1} \cong \struct {S_2, \preceq_2}$, that is, let $\struct {S_1, \preceq_1}$ and $\struct {S_2, \preceq_2}$ be order isomorphic.

Then there is exactly one mapping $f: S_1 \to S_2$ such that $f$ is an order isomorphism.

## Proof

Let $f: S_1 \to S_2$ and $g: S_1 \to S_2$ both be order isomorphisms.

By Inverse of Order Isomorphism is Order Isomorphism, the inverse $f^{-1}$ is also an order isomorphism.

Let $h = f^{-1} \circ g$ be the composition of $f^{-1}$ and $g$, which, by Composite of Order Isomorphisms is Order Isomorphism, is itself an order isomorphism.

So, by Order Isomorphism from Woset onto Subset:

- $\forall x \in S_1: x \preceq_1 \map h x$

Now we apply $f$ to $S_1$ and see that:

- $\forall x \in S_1: \map f x \preceq_2 \map f {\map h x} = \map g x$

In a similar way we can show that:

- $\forall x \in S_1: \map g x \preceq_2 \map f x$

Hence $f = g$ and the proof is complete.

$\blacksquare$

## 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: Theorem $1.7.3$