Order Automorphism on Well-Ordered Class is Identity Mapping
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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 $2.2$