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A set of ordinals that is not an ordinal: ${0, 1, ω}$

Statement refuted

Refuted claim: every set of ordinals is itself an ordinal, since membership well-orders any such set.

The witness is $A = {0, 1, ω}$ . Its members are ordinals (Ordinal (von Neumann)), and membership restricted to $A$ is a strict well-order: it is irreflexive, transitive as a relation, trichotomous, and every nonempty subset of $A$ has an $\in$ -least element (Trichotomy and well-ordering of the ordinals). Yet $A$ is not transitive, and therefore not an ordinal: $2 \in ω \in A$ , but $2 \notin A$ , because $2$ differs from each of $0$ , $1$ and $ω$ . So the well-ordering clause of the definition of an ordinal does not imply the transitivity clause, and the two clauses are genuinely independent: conversely $B = {0, 1, {1}}$ is transitive but membership does not compare its distinct members $0$ and ${1}$ . No choice principle is used.

Facts & Assumptions

Given: The von Neumann naturals $N = ω$ with $0 = \emptyset$ , $σ (n) = n \cup {n}$ , $1 = {0}$ , $2 = {0, 1}$ , and the ordinals of Ordinal (von Neumann); the set $A = {0, 1, ω}$ .

[F1]

A set $α$ is an ordinal when it is transitive and $\in$ restricted to $α$ is a strict well-order; $0 = \emptyset$ is an ordinal, $α^{+} = α \cup {α}$ , and for ordinals $α < β$ means $α \in β$ (Ordinal (von Neumann)).

[F2]

For ordinals $α, β$ : every element of an ordinal is an ordinal, and $α \notin α$ (Basic closure properties of ordinals).

[F3]

Any two ordinals satisfy exactly one of $α \in β$ , $α = β$ , $β \in α$ ; every nonempty set of ordinals has an $\in$ -least element; and a set of ordinals ordered by inclusion is well ordered with strict part $\in$ (Trichotomy and well-ordering of the ordinals).

[F4]

Every natural number is an ordinal; $ω$ is an ordinal and a limit ordinal; and every ordinal in $ω$ is $0$ or a successor ordinal ( $ω$ is the least limit ordinal).

[F5]

$N = ω$ is inductive: $0 = \emptyset \in ω$ and $n \in ω$ implies $σ (n) = n \cup {n} \in ω$ ; hence $1 = {0}$ and $2 = {0, 1}$ are members of $ω$ (The natural numbers $N$ (von Neumann)).

[F6]

An ordinal is a successor ordinal when it equals $β^{+}$ for some ordinal $β$ , and a limit ordinal when it is nonzero and not a successor (Successor and limit ordinals).

[F7]

A well-order is a total order in which every nonempty subset has a least element (Well-order and well-ordered set).

Counterexample

technique · direct

1.1F1F4F5F6

The members of $A = {0, 1, ω}$ are ordinals, and the membership relations among them are as follows: $0$ and $1$ and $2$ are natural numbers, hence ordinals by [F4]; $ω$ is an ordinal by [F4]; $1 = {0}$ and $2 = {0, 1}$ by [F5], and $0 \in 1$ , $1 \in ω$ , $2 \in ω$ , $0 \in ω$ because $0, 1, 2$ are members of the inductive set $ω$ ; moreover $2 \neq ω$ , since $2 = σ (1) = 1^{+}$ is a successor ordinal while $ω$ is a limit ordinal by [F4] and [F6].

1.2F1F5

Conversely $B = {0, 1, {1}}$ is transitive: its members $0 = \emptyset$ , $1 = {0}$ and ${1}$ are all subsets of $B$ . But $0 \neq {1}$ , since ${1}$ is nonempty, and neither $0 \in {1}$ (because $0 \neq 1$ ) nor ${1} \in 0$ holds. Thus membership on $B$ is not trichotomous and is not a strict well-order. Thus transitivity alone does not imply the well-ordering clause of the ordinal definition.

2.1step 1.1F2

Membership restricted to $A$ is irreflexive: each of $0$ , $1$ , $ω$ is an ordinal by step 1.1, and no ordinal is a member of itself by [F2].

2.2step 1.1

Membership restricted to $A$ is transitive as a relation: the only pairs of members of $A$ linked by membership are $0 \in 1$ , $1 \in ω$ and $0 \in ω$ by step 1.1, so the only two-step chain is $0 \in 1 \in ω$ , and it is completed by $0 \in ω$ .

2.3step 1.1F3

Membership restricted to $A$ is trichotomous: any two distinct members of $A$ are ordinals by step 1.1, so exactly one of $ξ \in η$ , $ξ = η$ , $η \in ξ$ holds by [F3].

2.4step 1.1F3

Every nonempty subset of $A$ has an $\in$ -least element: subsets of $A$ are sets of ordinals by step 1.1, so the least-element clause of [F3] applies.

2.5step 1.1F2F5

The set $A$ is not transitive: $2 \in ω$ and $ω \in A$ by step 1.1, but $2 \notin A$ , because $2 \neq 0$ ( $2 = {0, 1}$ has members while $0 = \emptyset$ has none), $2 \neq 1$ ( $1 = {0}$ but $1 \in 2$ , so $1 \neq 2$ since no ordinal is a member of itself by [F2]), and $2 \neq ω$ by step 1.1.

3.1step 2.1step 2.2step 2.3step 2.4step 2.5step 1.2F1F7∎

Steps 2.1 to 2.4 exhibit membership restricted to $A$ as a strict well-order in the sense of [F1] and [F7]: irreflexive, transitive as a relation, trichotomous, with a least element in every nonempty subset. Step 2.5 shows that the set $A$ is not transitive. By [F1] an ordinal must satisfy both clauses, so $A$ is not an ordinal even though it is a set of ordinals; the refuted claim is false, and the well-ordering clause alone does not suffice; step 1.2 proves the converse independence as well. No choice principle is used.

Depends on

Used by

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Dependency tree · two levels

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Sources

A. Marks, Set Theory (Berkeley lecture notes, 2026) (standard reference, not scraped)
J. T. Moore, MATH 6870: Set Theory (Cornell lecture notes) (standard reference, not scraped)

A set of ordinals that is not an ordinal: {0,1,ω}{0,1,ω}