Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 19 / 2 / iii / b / Solution

Use the transitive-model convention for “standard” in this assertion: the domain is a transitive set and membership is actual membership. A standard membership model of set theory without transitivity is a weaker notion and would not justify the asserted absoluteness. Ordinalhood is bounded-formula absolute, so the internal ordinals are precisely $A\cap \mathrm{Ord}$. Satisfaction for a set structure is absolute between transitive ZF models containing that structure: the domain, finite parameter tuples, formula codes and the recursive truth clauses are the same. Equivalently one may use the ${\Delta }_1$ absoluteness of the predicate defining $L_{\alpha }$.

Transfinite induction now gives $L_{\alpha }^{\mathbb{A}}=L_{\alpha }$ for each ordinal $\alpha \in A$. At successors both computations use the same definable subsets of the same set structure; at limits they take the same union over all smaller ordinals, which belong to $A$ by transitivity. ThusFor a set-sized model of ordinal height $\theta$, this is $L_{\theta }$. The restriction to actual ordinals makes the printed union precise. Mere well-foundedness of a nontransitive membership substructure would not justify these absoluteness steps; transitivity is the intended standard-model convention.