Let $x \subseteq \alpha$ be a subset of a (von Neumann) ordinal $\alpha$ taken with the induced ordering. Using the recursion theorem or otherwise show that $x$ is order isomorphic to a unique ordinal $\mu(x)$. Suppose that $x \subseteq y \subseteq \alpha$. Show that $\mu(x) \leqslant \mu(y) \leqslant \alpha$.

Suppose that $x_{0} \subseteq x_{1} \subseteq x_{2} \subseteq \cdots$ is an increasing sequence of subsets of $\alpha$, with $x_{i}$ an initial segment of $x_{j}$ whenever $i<j$. Show that $\mu\left(\bigcup_{n} x_{n}\right)=\bigcup_{n} \mu\left(x_{n}\right)$. Does this result still hold if the condition on initial segments is omitted? Justify your answer.

Suppose that $x_{0} \supseteq x_{1} \supseteq x_{2} \supseteq \cdots$ is a decreasing sequence of subsets of $\alpha$. Why is the sequence $\mu\left(x_{n}\right)$ eventually constant? Is it the case that $\mu\left(\bigcap_{n} x_{n}\right)=\bigcap_{n} \mu\left(x_{n}\right)$ ? Justify your answer.