 Solution

Induct on $\alpha$ using the transitivity and inclusion already established. The zero stage is immediate. Suppose $L_{\alpha }\cap \mathrm{Ord}=\alpha$. Being an ordinal is expressible by a bounded formula in set theory, for example transitivity, transitivity of all members, and membership comparability. It is therefore absolute for the transitive set $L_{\alpha }$. The subset of its elements that are ordinals is internally definable, soBy inclusion, all smaller ordinals are present as well. Conversely, if an ordinal $\gamma$ belongs to $L_{\alpha +1}$, then $\gamma \subseteq L_{\alpha }$, since every member of a definable power set is a subset of the preceding domain. All members of $\gamma$ are ordinals, so $\gamma \subseteq L_{\alpha }\cap \mathrm{Ord}=\alpha$ and $\gamma \le \alpha$. Hence the ordinal part of the successor level is exactly $\alpha +1$.

At a nonzero limit ordinal $\lambda$,Thus