Past exam of the analytic topology and axiomatic set theory course of the University of Oxford / 2010 / B / 4 / b / Solution

Use transfinite induction. The assertion is immediate at stage zero, and at a limit stage it follows by taking the increasing union of the earlier stages.

Assume $L_{\alpha }\cap \mathrm{On}=\alpha$. Any ordinal $\gamma \in L_{\alpha +1}$ is a subset of $L_{\alpha }$, since members of the definable power set are such subsets. Every element of $\gamma$ is an ordinal, sowhich implies $\gamma \le \alpha$. This gives the upper bound on the ordinals appearing at the successor stage.

For the reverse inclusion, the smaller ordinals are already in $L_{\alpha }$ and hence in $L_{\alpha +1}$. The remaining ordinal $\alpha$ itself is the definable subset $L_{\alpha }\cap \mathrm{On}$. The fact used here is absoluteness of ordinalhood between a transitive set and the universe: ordinalhood has a bounded definition, and bounded formulas are absolute for transitive sets. Thus the formula asserting that a member is an ordinal, interpreted over $L_{\alpha }$, defines exactly $\alpha$. Therefore $\alpha \in \mathrm{Def}(L_{\alpha })=L_{\alpha +1}$.

This provesIn particular each ordinal $\alpha$ is a member of $L_{\alpha +1}$, so $\boxed{\mathrm{On}\subseteq L}$.