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FALSE: the ordinals form a set

Statement

FALSE. The ordinals (Ordinal (von Neumann)) form a set: there is a set $Ω$ whose members are exactly the ordinals.

The claim is plausible because the ordinals look locally set sized. Every ordinal $α$ is itself precisely the set of all ordinals below it, so every downward closed collection of ordinals that stops somewhere is a set, and it is tempting to conclude that the collection of all of them is a set as well. In ZF that inference is unavailable: Separation produces a set only as a subset of a set already in hand, and here there is no such ambient set to start from.

Facts & Assumptions

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

[A1]

Separation carves a subset out of a set already given; it never produces a set from a defining property alone.

[L1]

No set has every ordinal as a member (Burali-Forti: there is no set of all ordinals).

[L2]

Every element of an ordinal is an ordinal, no ordinal is a member of itself, and $γ^{+} = γ \cup {γ}$ is an ordinal whose members are exactly the ordinals at most $γ$ (Basic closure properties of ordinals).

[L3]

An ordinal is a transitive set on which $\in$ is a strict well-order, and $α < β$ means $α \in β$ (Ordinal (von Neumann)).

Refutation

technique · contradiction

1.1assume-contra

Suppose the claim: there is a set $Ω$ whose members are exactly the ordinals.

1.2A1L2L3

The source of the intuition is genuine but limited: by [L3] and [L2] each ordinal $α$ equals ${β : β < α}$ , so every definable collection of ordinals bounded above by an ordinal $γ$ is a set, by Separation inside its successor $γ^{+}$ , which includes the possible endpoint $γ$ ; by [A1] this says nothing about the unbounded collection of all ordinals.

2.1step 1.1

Under the supposition, $Ω$ is a set having every ordinal as a member.

3.1step 2.1L1L2L3

No such set exists: it would be transitive by [L2], since every element of an ordinal is an ordinal and so a member of it, and $\in$ would strictly well-order it, so it would itself be an ordinal by [L3] and hence a member of itself, which [L2] forbids; this is exactly [L1].

4.1step 2.1step 3.1L1discharge-contradiction∎

Steps 2.1 and 3.1 are contradictory, so no set has the ordinals as its members: the ordinals form a proper class and the claim is false.

Remarks

Bounded is not unbounded. The honest version of the intuition is: for every ordinal $α$ , the ordinals below $α$ form a set, namely $α$ itself. Nothing in ZF upgrades a family of sets indexed by a proper class into one set, and the attempt to do so here is exactly what Burali-Forti: there is no set of all ordinals refutes.

The same trap, one level up. "The sets form a set" fails because such a set would contain every ordinal, contrary to Burali-Forti: there is no set of all ordinals. Cardinals (Cardinal (initial ordinal) and cardinality) are also unbounded among the ordinals, without choice: given an ordinal $α$ , let $H = ℵ (α)$ from Hartogs: an ordinal that does not inject into a given set. If $H \leq α$ , inclusion would inject $H$ into $α$ , so $α < H$ by ordinal comparability (Basic closure properties of ordinals). Every $β < H$ injects into $α$ by the leastness of $H$ ; a bijection $H \to β$ would compose with that injection to inject $H$ into $α$ , a contradiction. Thus $H$ is a cardinal strictly above $α$ . If the cardinals formed a set $K$ , its union would be an ordinal bounding all its members (Basic closure properties of ordinals, claims (e) and (f)), contradicting this unboundedness. In each case the correct statement replaces "set" by "proper class", which in ZF means a formula rather than an object.

What is still available. Nothing about the theory of ordinals needs them to form a set. Every construction on this page indexes by a set of ordinals, or by a single ordinal, or runs along an arbitrary well-order; Transfinite recursion and Hartogs: an ordinal that does not inject into a given set are both stated so that only sets are ever formed.

FALSE: the ordinals form a set

Statement

Facts & Assumptions

Refutation

Remarks

Depends on

Dependency tree · 2 levels

Sources