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TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)verified 2026-07-26 (claude-opus-5)
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Every well-order has a unique order type

Statement

Every well-order (W,<) (Well-order and well-ordered set) is order isomorphic (Order embedding and order isomorphism) to exactly one ordinal (Ordinal (von Neumann)), called its order type and written ot(W).

The isomorphism is the collapsing map F(a)={F(b):b<a}, and it is unique as well, by rigidity (Rigidity of well-orders).

This uses Replacement, and no form of the Axiom of Choice.

Facts & Assumptions

Given: A well-order (W,<) and the axioms of ZF, in particular Replacement. No choice principle is assumed.

[A1]

The axioms of ZF, in particular Replacement and Union, are available.

[L1]

Transfinite recursion: for a class function G there is a unique F on W with F(a)=G(F↾W<a) for all a (Transfinite recursion).

[L2]

Transfinite induction on (W,<) (Transfinite induction).

[L3]

An ordinal is a transitive set on which ∈ is a strict well-order (Ordinal (von Neumann)), and W<a={x∈W:x<a} (Initial segment of a well-order).

[L4]

Every element of an ordinal is an ordinal, α∉α, and α⊆β if and only if α∈β or α=β (Basic closure properties of ordinals).

[L5]

Any two ordinals satisfy exactly one of α∈β, α=β, β∈α, and every set of ordinals is well ordered by ∈ (Trichotomy and well-ordering of the ordinals).

[L6]

No well-order is order isomorphic to a proper initial segment of itself (Rigidity of well-orders).

Proof

technique · direct
1.1

Apply [L1] with the class function G(h)=ran(h) to obtain the unique function F on W with F(a)={F(b):b<a} for every a∈W.

L1L3construct
2.1

Every value F(a) is an ordinal, by transfinite induction: assume F(b) is an ordinal for every b<a; then F(a) is a set of ordinals by Replacement, it is transitive because F(b)={F(c):c<b}⊆F(a) whenever b<a, and ∈ well-orders it by [L5]; so F(a) is an ordinal.

step 1.1L2L5L3A1
2.2

If b<a then F(b)∈F(a), immediately from the defining equation.

step 1.1
3.1

Conversely F(b)∈F(a) forces b<a: otherwise a=b, giving F(a)∈F(a), or a<b, giving F(a)∈F(b) and hence both F(a)∈F(b) and F(b)∈F(a); each alternative contradicts [L4] or [L5].

step 2.2step 2.1L4L5
3.2

The set α={F(a):a∈W} exists by Replacement and is an ordinal: it is a set of ordinals by step 2.1, it is transitive because x∈F(a) means x=F(b) for some b<a and hence x∈α, and ∈ well-orders it by [L5].

step 2.1step 1.1L5L3A1
4.1

F is therefore a bijection from W onto α with b<a  ⟺  F(b)∈F(a), that is an order isomorphism of (W,<) onto the ordinal α ordered by membership; injectivity holds because b≠a gives b<a or a<b by trichotomy in W, hence F(b)∈F(a) or F(a)∈F(b), and F(a)∉F(a) rules out equality.

step 2.2step 3.1step 3.2L4L5
5.1

Uniqueness: suppose W≅α and W≅β with α≠β ordinals; then α≅β, and by [L5] one is a member of the other, say α∈β, so α⊆β by [L4] and α is the initial segment of β determined by α, making β order isomorphic to a proper initial segment of itself, which [L6] forbids.

step 4.1L4L5L6
6.1

Every well-order is order isomorphic to exactly one ordinal, its order type.

step 4.1step 5.1∎

Remarks

Replacement is the whole cost. The values F(a) are not subsets of any set given in advance, so Separation cannot collect them; steps 2.1 and 3.2 both invoke Replacement, exactly as Transfinite recursion does. In Zermelo set theory, which has Separation but not Replacement, the theorem fails, and the standard witness is explicit: Vω+ω satisfies Zermelo set theory, its ordinals are exactly the ordinals below ω+ω, and it contains relations on ω of order type ω⋅2, which no ordinal of the model is isomorphic to. That model is built inside ZF, so this failure carries no consistency hypothesis; it is quoted from the references below and is not proved here, since Zermelo set theory is nowhere developed in this library.

No choice, and the reason is again uniqueness. Nothing is ever selected: F is produced by recursion from a formula, and the ordinal it lands on is determined. This is what makes order type a choice-free notion, in contrast with cardinality, which needs The well-ordering theorem and hence the Axiom of Choice to be defined for an arbitrary set.

Comparability, restated. With order types available, Comparability of well-orders says exactly that the order types of two well-orders are comparable as ordinals, which is Trichotomy and well-ordering of the ordinals transported back along the collapse. Either lemma can be derived from the other, and both are proved here without choice.

The name. The general Mostowski collapse takes any well-founded extensional relation to a transitive set. The case proved here, a well-order collapsing to an ordinal, is the only one this library needs, and it is stated in that form to avoid introducing well-founded relations before they are used.

Depends on

Used by

Dependency tree · two levels

18 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources