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A Von Neumann ordinal $x$ is finite exactly when $x$ and every member of $x$ are either zero or successor ordinals. Every finite ordinal has that property. Conversely, an infinite ordinal is either $\omega$, which is a nonzero limit, or is greater than $\omega$ and contains that nonzero limit ordinal. The test can be expressed by a bounded first-order formula: ordinality uses transitivity and membership comparability, and a nonempty ordinal is a successor exactly when it has a greatest member. This defines the class of natural numbers without quantifying over arbitrary infinite inductive sets.

A finite ordinal is the order type of a finite well-order; in the von Neumann representation it is one of $0,1,2,\dots$. They are exactly the ordinals $x$ such that $x$ and each member of $x$ are either zero or successor ordinals. If an ordinal is infinite, it either is the nonzero limit $\omega$ or contains $\omega$ as a member, so it fails this test. Thus the finite ordinals give the usual representation of the natural numbers.

Use the bounded successor characterization of finite ordinals. Let $\mathrm{Ord}(x)$ assert that $x$ is transitive, each of its members is transitive, and any two members are equal or comparable by membership. This is a bounded first-order formula; foundation makes its linear membership order well-founded, so it is precisely the Von Neumann ordinal predicate. Defineand putFor an ordinal, having a greatest member $y$ says exactly $x=y\cup \{y\}$. The displayed definition makes no reference to an infinite inductive set or to $\omega$; all quantifiers are bounded within the candidate and its members. Occurrences of $x=\varnothing$ can also be expressed as $(\forall u\in x)u\ne u$, so no unbounded quantifier is hidden in empty-set notation.

Every usual natural number satisfies this predicate, by induction: it is a finite ordinal, and every nonzero initial segment is a successor. Conversely, suppose an ordinal $x$ is not a usual natural number. Ordinal comparability gives $x\ge \omega$. If $x=\omega$, it has no greatest member, contradicting ${\mathrm{Succ}}_0(x)$. If $x>\omega$, then $\omega \in x$ has no greatest member, contradicting the corresponding bounded requirement on members of $x$. Hence $\mathrm{Nat}(x)$ holds exactly for the usual natural numbers. The axiom of infinity and separation therefore collect this class as the usual set $\omega$. This is a bounded definition of membership in the natural-number class; it does not claim that the infinite set $\omega$ itself can be produced without the infinity axiom.