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 $2.4$