Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 121 / 2 / ii / b / Solution

The argument in part (a) established $L_{\alpha }\subseteq L_{\alpha +1}$ for every ordinal $\alpha$. A transfinite induction on $\beta$ now proves that every earlier level is contained in $L_{\beta }$. If $\beta =\gamma +1$, use the inductive inclusions into $L_{\gamma }$ followed by $L_{\gamma }\subseteq L_{\gamma +1}$. At a limit stage, every earlier level is one of the sets in the defining union. Equality of indices is trivial. ThusThe constructible hierarchy is a nondecreasing family of sets.