I am currently trying to practice the technique of transfinite induction with the following problem:
Suppose that $X$ is a non-empty subset of an ordinal $\alpha$, so that $X$ is well-ordered by $\in$. Show that $\text{type}(X; \in) \leq \alpha$.
My approach thus far:
Let $\beta = \text{type}(X; \in)$ and $f: X \rightarrow \beta$ be an order-preserving isomorphism. Now we show that $f(\xi) \leq \xi$ for all $\xi \in X$ by transfinite induction.
Base Case: Let $\xi_{0} \in X$ be minimal with respect to $\in$ in $X$. As $f$ preserves order, it must be the case that $f(\xi_{0}) = \emptyset$ and so $f(\xi_{0}) \leq \xi_{0}$.
Inductive Step: Suppose that $f(\xi) \leq \xi$ for all $\xi < \gamma$ for some $\gamma$. Now we deduce that $f(\gamma) \leq \gamma$.
My question is how to prove this crucial step $f(\gamma) \leq \gamma$.
After proving this, then by transfinite induction we have that $f(\xi) \leq \xi$ for all $\xi \in X$ and so $\beta = f(X) \subseteq \alpha$ and so $\text{type}(X; \in) = \beta \leq \alpha$ as desired.