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The finite von Neumann ordinals, and ω as the least limit ordinal

Example

With 0=∅ and σ(n)=n∪{n} (The natural numbers N (von Neumann)) and the ordinals of Ordinal (von Neumann):

(i) 0,1,2,3 are ordinals, 1={0}, 2={0,1}, 3={0,1,2}=2∪{2}=2+, and for each k≤3 the set k is exactly the set of ordinals below k, so k={ξ:ξ<k};

(ii) ω=N is an ordinal, ω is a limit ordinal, the union of the members of ω is ω itself, and every element of ω is either 0 or a successor, so no element of ω is a limit ordinal (Successor and limit ordinals, ω is the least limit ordinal);

(iii) ω is the least limit ordinal: ω≤γ for every limit ordinal γ.

No choice principle is used.

Facts & Assumptions

Given: The von Neumann naturals N=ω with 0=∅ and σ(n)=n∪{n}, and the ordinal notation of Ordinal (von Neumann): a set α is an ordinal when it is transitive and ∈ strictly well-orders it, with α<β meaning α∈β and α+=α∪{α}.

[F1]

An ordinal is a transitive set on which ∈ is a strict well-order, 0:=∅ is an ordinal, α+=α∪{α}, and for ordinals α<β means α∈β while α≤β means α∈β or α=β (Ordinal (von Neumann)).

[F2]

For every ordinal α: every element of α is an ordinal, and α+ is an ordinal (Basic closure properties of ordinals).

[F3]

N=ω is the smallest inductive set, with 0=∅, σ(n)=n∪{n}, 1={0}, 2={0,1}, 3={0,1,2} and in general n={0,1,…,n−1}; ω is inductive, so 0∈ω and n∈ω implies σ(n)∈ω (The natural numbers N (von Neumann)).

[F4]

Every natural number is an ordinal, and ω is an ordinal; ω is a limit ordinal; every ordinal α∈ω is 0 or a successor ordinal; and ω⊆γ for every limit ordinal γ (ω is the least limit ordinal).

[F5]

An ordinal α is a successor ordinal when α=β+ for some ordinal β, and a limit ordinal when α≠0 and α is not a successor (Successor and limit ordinals).

[F6]

Every natural number is a transitive set and satisfies n∉n (Every natural number is a transitive set and is not a member of itself).

Verification

technique · direct
1.1F1F2F3F4

Claim (i): 0=∅ is an ordinal by [F1]; 1=0+={0}, 2=1+={0,1} and 3=2+={0,1,2} by [F3], and each is an ordinal by the successor clause of [F2], so 3=2∪{2}=2+ as displayed. For k≤3: every element ξ of k is a natural number, hence an ordinal by [F4], and conversely every ordinal ξ with ξ<k satisfies ξ∈k by [F1]; the two sides of k={ξ:ξ<k} therefore have the same members, since the elements of k are among the ordinals below k and every ordinal below k is an element of k.

1.2F1F3F4F5F6

Claim (ii): ω is an ordinal by [F4] and is a limit ordinal by [F4], and since ω≠0 by [F3] we may compute the union of its members. First ⋃ω⊆ω: a member of ⋃ω is a member of some n∈ω, and n⊆ω because ω is transitive as an ordinal while n∈ω, so that member lies in ω as well. Conversely ω⊆⋃ω: for n∈ω the successor σ(n)=n∪{n} again lies in ω by inductivity, and n∈σ(n), so every member of ω lies in a member of ω. Hence ⋃ω=ω by extensionality. Finally every element of ω is 0 or a successor by [F4], and a limit ordinal is nonzero and not a successor by [F5], so no element of ω is a limit ordinal.

1.3F1F4

Claim (iii): [F4] states that ω⊆γ, that is ω≤γ in the ordinal order of [F1], for every limit ordinal γ; with ω itself a limit ordinal by [F4], this says ω is the least limit ordinal.

2.1step 1.1step 1.2step 1.3∎

Claims (i) to (iii) are established: the finite levels 0,1,2,3 are ordinals with k={ξ:ξ<k} by step 1.1, the ordinal ω is a limit ordinal, equals the union of its members, and has no limit ordinal among its elements by step 1.2, and it is least among limit ordinals by step 1.3. Every ingredient is a theorem of ZF applied to the given sets, and nothing is selected from a family of sets, so no choice principle is used.

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