Well-Ordered Classes are Isomorphic at Most Uniquely

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Theorem

Let $\struct {A, \preccurlyeq_A}$ and $\struct {B, \preccurlyeq_B}$ be well-ordered classes.

Then there exists at most one order isomorphism from $\struct {A, \preccurlyeq_A}$ to $\struct {B, \preccurlyeq_B}$

Proof

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Let $\phi$ and $\psi$ be order isomorphisms from $A$ to $B$.

Then from Inverse of Order Isomorphism is Order Isomorphism:

the inverse mapping $\psi^{-1}$ is an order isomorphism from $B$ to $A$.

Hence from Composite of Order Isomorphisms is Order Isomorphism:

$\phi \circ \psi^{-1}$ is an order isomorphism from $A$ to $A$.

But from Order Automorphism on Well-Ordered Class is Identity Mapping:

$\phi \circ \psi^{-1}$ is the identity mapping.

From Composite of Mapping with Inverse of Another is Identity implies Mappings are Equal:

$\phi = \psi$

$\blacksquare$

Sources

• 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 2$ Isomorphisms of well orderings: Corollary $2.3$