= Solution
{c}
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. Thus
$$
\boxed{\alpha\leq\beta\quad\Longrightarrow\quad L_\alpha\subseteq L_\beta.}
$$
The <constructible hierarchy> is a <nondecreasing family of sets>.
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