= Solution
\b[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. Hence
$$
\operatorname{Def}(L_n)=\mathcal P(V_n)=V_{n+1}.
$$
Induction and the limit-stage definitions give
$$
\boxed{L_\omega=V_\omega,}
$$
the 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.
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