CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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Ordinal addition exists and is unique: the clauses at $0$ , at a successor and at a limit determine one operation, and its values are ordinals

Statement

Fix an ordinal $α$ (Ordinal (von Neumann)). There is exactly one class function $β \mapsto s_{α} (β)$ , defined at every ordinal $β$ , satisfying the three clauses

$s_{α} (0) = α, s_{α} (β^{+}) = (s_{α} (β))^{+}, s_{α} (λ) = ⋃ {s_{α} (β) : β \in λ} (λ a limit ordinal),$

and every value $s_{α} (β)$ is an ordinal.

The three clauses are exhaustive and mutually exclusive, because every ordinal is exactly one of $0$ , a successor, or a limit (Successor and limit ordinals). The union in the third clause is the least upper bound of the earlier values (Basic closure properties of ordinals, claim (e)), so the limit clause reads "take the supremum of what has been built so far".

This is the well-definedness obligation discharged before ordinal addition is written down; the operation itself is named in the definition that follows. The proof is a theorem of ZF and uses no choice principle.

Facts & Assumptions

Given: A fixed ordinal $α$ and the axioms of ZF. No choice principle is assumed. For a function $h$ , $ran (h)$ is its range.

[L1]

Recursion along the ordinals: for a class function $G$ assigning a set to every function whose domain is an ordinal there is exactly one class function $F$ , defined at every ordinal, with $F (β) = G (F ↾ β)$ for all $β$ (Transfinite recursion along the ordinals: a class rule determines exactly one operation defined at every ordinal).

[L2]

Every ordinal is exactly one of: $0$ , a successor ordinal $δ^{+}$ with $δ$ uniquely determined, or a limit ordinal (Successor and limit ordinals).

[L3]

$μ^{+}$ is an ordinal whenever $μ$ is, and $⋃ A$ is an ordinal for every set $A$ of ordinals, and is their least upper bound (claims (c) and (e) of Basic closure properties of ordinals).

[L4]

Every nonempty set of ordinals has an $\in$ -least element (Trichotomy and well-ordering of the ordinals), and ${ξ \in β_{0}^{+} : P (ξ)}$ is such a set whenever $P (β_{0})$ holds and $P$ is a property of ordinals.

Proof

technique · direct

1.1

Define a class function $G$ on functions $h$ whose domain is an ordinal $β$ by: $G (h) = α$ if $β = 0$ ; $G (h) = (h (δ))^{+}$ if $β = δ^{+}$ ; and $G (h) = ⋃ ran (h)$ if $β$ is a limit ordinal.

construct

1.2

The three cases are exhaustive and mutually exclusive by [L2], and $δ$ is determined by $β = δ^{+}$ , so $G (h)$ is a well-determined set for every such $h$ and the rule is a formula.

2.1

By [L1] there is exactly one class function $s_{α}$ , defined at every ordinal, with $s_{α} (β) = G (s_{α} ↾ β)$ for every ordinal $β$ .

step 1.1step 1.2L1

3.1

Unwinding the three cases of $G$ : $s_{α} (0) = G (\emptyset) = α$ ; $s_{α} (δ^{+}) = ((s_{α} ↾ δ^{+}) (δ))^{+} = (s_{α} (δ))^{+}$ , since $δ \in δ^{+}$ ; and for a limit $λ$ , $s_{α} (λ) = ⋃ ran (s_{α} ↾ λ) = ⋃ {s_{α} (β) : β \in λ}$ .

step 2.1step 1.1

4.1

Every value is an ordinal: were $s_{α} (β_{0})$ not an ordinal for some $β_{0}$ , [L4] would give a least $μ \in β_{0}^{+}$ with $s_{α} (μ)$ not an ordinal, and each of the three cases refutes that, since $s_{α} (0) = α$ is an ordinal, $s_{α} (δ^{+}) = (s_{α} (δ))^{+}$ is an ordinal by [L3] because $δ \in μ$ makes $s_{α} (δ)$ an ordinal, and $s_{α} (λ) = ⋃ {s_{α} (β) : β \in λ}$ is a union of a set of ordinals, hence an ordinal by [L3].

step 3.1L2L3L4

4.2

Uniqueness: a class function $t$ defined at every ordinal and satisfying the three displayed clauses satisfies $t (β) = G (t ↾ β)$ for every $β$ , one case at a time, so $t = s_{α}$ by the uniqueness half of [L1].

step 3.1step 2.1L1L2

5.1

Hence exactly one class function on the ordinals satisfies the three clauses, and all its values are ordinals.

step 4.1step 4.2step 3.1∎

Remarks

Why a bridge lemma is used and not Transfinite recursion directly. The published recursion theorem is stated for a well-order, that is for a set, and delivers a function whose domain is that set. What is needed here is a rule defined at every ordinal, and the ordinals are not a set. Transfinite recursion along the ordinals: a class rule determines exactly one operation defined at every ordinal is exactly that bridge, and it is proved from the published theorem's uniqueness clause.

Three cases, and the one that is usually forgotten. Writing the recursion with a successor clause and a limit clause only leaves the operation undefined at $0$ , since $0$ is neither. The published Successor and limit ordinals states the three-way split in the form used at step 1.2, and it is cited rather than assumed.

The same argument serves multiplication and exponentiation. Only the three clauses of $G$ change. The two corollaries later on this page repeat this proof with different clauses, and the exponentiation case additionally restricts the limit clause to $0 < β < λ$ ; see Ordinal exponentiation exists and is unique, with the limit clause taken over $0 < β < λ$ so that $0^{λ} = 0$ for why that restriction is not optional.

Depends on

Used by

Ordinal multiplication exists and is unique, and its values are ordinals Corollary
Ordinal addition α + β Definition

Dependency tree · two levels

14 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

Ordinal arithmetic (Wikipedia) (standard reference, not scraped)
T. Jech, Set Theory, 3rd millennium ed., Ch. 2 (Ordinal numbers) (standard reference, not scraped)
R. Moosa, Set Theory course notes (standard reference, not scraped)

Ordinal addition exists and is unique: the clauses at 00, at a successor and at a limit determine one operation, and its values are ordinals