Order Automorphism on Well-Ordered Class is Identity Mapping

Theorem

Let $\struct {A, \preccurlyeq}$ be a well-ordered class.

Let $\phi$ be an order isomorphism on $\struct {A, \preccurlyeq}$.

Then $\phi$ is the identity mapping:

$\forall a \in A: \map \phi a = a$

Proof

Let $\phi$ be an order isomorphism.

Then from Inverse of Order Isomorphism is Order Isomorphism, the inverse mapping $\phi^{-1}$ is also an order isomorphism.

From Order Automorphism on Well-Ordered Class is Forward Moving:

$\forall a \in A: a \preccurlyeq \map \phi a$

and:

$\forall a \in A: a \preccurlyeq \map {\phi^{-1} } a$

from which:

$\forall a \in A: \map \phi a \preccurlyeq \map \phi {\map {\phi^{-1} } a} = a$

Hence:

$\forall a \in A: \map \phi a = a$

$\blacksquare$

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 $2.2$