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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (openai/gpt-5.4)verified 2026-07-26 (claude-opus-5)
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Burali-Forti: there is no set of all ordinals

Statement

There is no set whose members are exactly the ordinals (Ordinal (von Neumann)). More strongly, no set has every ordinal as a member: the ordinals form a proper class.

Facts & Assumptions

Given: The axioms of ZF, in particular the Separation schema. No choice principle is used.

[A1]

Separation: for every set XX and every formula φ\varphi, the collection {x∈X:φ(x)}\{x \in X : \varphi(x)\} is a set.

[L1]

An ordinal is a transitive set on which ∈\in is a strict well-order (Ordinal (von Neumann)).

[L2]

Every element of an ordinal is an ordinal, and α∉α\alpha \notin \alpha for every ordinal α\alpha (Basic closure properties of ordinals).

[L3]

Any two ordinals satisfy exactly one of α∈β\alpha \in \beta, α=β\alpha = \beta, β∈α\beta \in \alpha, and every nonempty set of ordinals has an ∈\in-least element (Trichotomy and well-ordering of the ordinals).

Proof

technique · contradiction
1.1

Suppose, for contradiction, that some set XX has every ordinal as a member.

assume-contra
2.1

By Separation, Ω={x∈X:x is an ordinal}\Omega = \{x \in X : x \text{ is an ordinal}\} is a set, and by the supposition its members are exactly the ordinals.

step 1.1A1
3.1

Ω\Omega is a transitive set: if α∈Ω\alpha \in \Omega and x∈αx \in \alpha then xx is an ordinal, hence x∈Ωx \in \Omega; so α⊆Ω\alpha \subseteq \Omega.

step 2.1L2
3.2

The relation ∈\in is a strict well-order of Ω\Omega: it is irreflexive there because α∉α\alpha \notin \alpha for ordinals, transitive there because x∈y∈zx \in y \in z with zz an ordinal gives y⊆zy \subseteq z and so x∈zx \in z, trichotomous there by [L3], and every nonempty subset of Ω\Omega is a nonempty set of ordinals and so has an ∈\in-least element.

step 2.1L1L2L3
4.1

Hence Ω\Omega is an ordinal, so Ω∈Ω\Omega \in \Omega by step 2.1, contradicting Ω∉Ω\Omega \notin \Omega; therefore no set has every ordinal as a member, and in particular there is no set of all ordinals.

step 3.1step 3.2step 2.1L1L2discharge-contradiction∎

Remarks

Why this is a theorem and not a paradox. In naive set theory the same computation is a contradiction, because unrestricted comprehension guarantees that the ordinals form a set. In ZF, Separation only carves subsets out of sets already given, so the argument instead refutes the assumption that some set collects them all. The historical statement, Burali-Forti 1897, predates that distinction, which is why it is remembered as a paradox.

Nothing about size is being said. The obstruction is not that there are "too many" ordinals in any measurable sense; it is that the supposed set would be transitive and well ordered by membership, which are exactly the two clauses of Ordinal (von Neumann), so it would have to be one of its own members. The same shape of argument shows there is no set of all sets.

Consequences used later. Since no set contains all ordinals, for any set AA there must be ordinals lying outside every construction indexed by AA, which is the crude form of the fact sharpened by Hartogs: an ordinal that does not inject into a given set. The false statement this theorem refutes is recorded as FALSE: the ordinals form a set.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 20 results over 10 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources