proved by transfinite induction on β: β = 0: V0 ⊆ V0. For β+1: By inductive hypothesis, Vα ⊆ Vβ. Hence, Vα ⊆ Vβ ⊆ Vβ ∪ P(Vβ) = Vβ+1. For limit β: If α < β, then Vα ⊆ ∪ξ < β Vξ = Vβ. If α = β, then Vα ⊆ Vβ. Sets enter the cumulative hierarchy through the power set P(Vβ) at step β+1. The following definitions will be needed: If x is a set, rank(x) is the least ordinal β such that x ∈ Vβ+1. The supremum of a set of ordinals A, denoted by sup A, is the least ordinal β such that α ≤ β for all α ∈ A. Zermelo's smallest model is Vω. Mathematical induction proves that Vn is finite for all n < ω: |V0| = 0. |Vn+1| = |Vn ∪ P(Vn)| ≤ |Vn| + 2 |Vn|, which is finite since Vn is finite by inductive hypothesis. Proof of Theorem 1: A set X enters Vω through P(Vn) for some n < ω, so X ⊆ Vn. Since Vn is finite, X is finite. Conversely: If a class X is finite, let N = sup {rank(x): x ∈ X}. Since rank(x) ≤ N for all x ∈ X, we have X ⊆ VN+1, so X ∈ VN+2 ⊆ Vω. Therefore, X ∈ Vω. Proof of Theorem 2: Vω is the union of countably infinitely many finite sets of increasing size. Hence, it has cardinality , which equals ω by von Neumann cardinal assignment. The sets and classes of Vω satisfy all the axioms of NBG except the axiom of infinity. The models Vκ where κ is a strongly inaccessible cardinal Two properties of finiteness were used to prove Theorems 1 and 2 for Vω: If λ is a finite cardinal, then 2λ is finite. If A is a set of ordinals such that |A| is finite, and α is finite for all α ∈ A, then sup A is finite. To find models satisfying the axiom of infinity, replace "finite" by "< κ" to produce the properties that define strongly inaccessible cardinals. A cardinal κ is strongly inaccessible if κ > ω and: If λ is a cardinal such that λ < κ, then 2λ < κ. If A is a set of ordinals such that |A| < κ, and α < κ for all α ∈ A, then sup A