Well-Ordered Class is not Isomorphic to Initial Segment
Theorem
Let $\struct {A, \preccurlyeq}$ be a well-ordered class.
There exists no order isomorphism from $\struct {A, \preccurlyeq}$ to an initial segment of $A$.
Proof
Let $a \in A$.
Let $A_a$ be the initial segment of $A$ determined by $a$.
Aiming for a contradiction, suppose $\phi: A \to A_a$ is an order isomorphism from $A$ to $A_a$.
From Order Automorphism on Well-Ordered Class is Forward Moving:
- $a \preccurlyeq \map \phi a$
But by definition of initial segment:
- $\map \phi a \notin A_a$
Hence $\phi$ cannot be an order isomorphism.
Hence by Proof by Contradiction there exists no such order isomorphism.
$\blacksquare$
Also presented as
Some sources present this result in the context of well-ordered sets, and do not take into account the concept of classes.
The result is the same whether the domain of any hypothetical order isomorphism is a set or a propr class.
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 $1.7.4$
- 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 $2.4$