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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.

The axiom of infinity fails. First prove $L_n=V_n$ for every finite ordinal $n$. The base stage is empty in both hierarchies. If $L_n=V_n$, that stage is a finite set. Every one of its subsets is definable over it with finitely many parameters, by a disjunction of equalities to the selected elements; the empty subset is defined by a false formula. HenceInduction and the limit-stage definitions givethe set of all hereditarily finite sets. Every member of this set is finite externally.

If this transitive structure satisfied the axiom of infinity, it would contain an inductive set $I$: zero belongs to $I$, and whenever $x\in I$, its successor $x\cup \{x\}$ belongs to $I$. These operations are the actual empty set and actual set successor, since they are absolute for this transitive structure and all relevant finite-rank sets lie in it. External induction would then place every finite ordinal in $I$, contradicting the finiteness of $I$.

Every individual finite ordinal does belong to $L_{\omega }$, but the set $\omega$ of all finite ordinals does not. The distinction between containing each finite ordinal and containing one inductive set is precisely the issue.

Every finite ordinal belongs to $V_{\omega }$, so $\omega \subseteq V_{\omega }$ and $\omega \in \mathcal{P}(V_{\omega })=V_{\omega +1}$. However, $\omega \notin V_{\omega }$, whose elements are the hereditarily finite sets.

Suppose $y\in V_{\omega +1}$ were an internal power set of $\omega$. Internally $\omega \subseteq \omega$, and $\omega$ is an available element of the structure. Thus the Axiom of power set would require $\omega \in y$. But $y\in \mathcal{P}(V_{\omega })$ means $y\subseteq V_{\omega }$, which would put $\omega \in V_{\omega }$, a contradiction. ThereforeThis proves power-set failure at the first rank above V omega by an internal obstruction, including the subset $\omega$ itself.

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.

Every finite ordinal belongs to $V_{\omega }$, so $\omega \in V_{\omega +1}$ but $\omega \notin V_{\omega }$. An internal power set $y$ of $\omega$ would contain $\omega$ itself. Yet $y\in V_{\omega +1}$ implies $y\subseteq V_{\omega }$, making that impossible. This example illustrates why one must check which subsets the structure sees when proving an axiom fails.