Order types of subsets of ordinals and transfinite induction 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. 
 A: Things will go a bit smoother if you strengthen the induction hypothesis to include "... and $f(X\cap\gamma)$ is an initial segment of $\beta$".
If $\gamma\notin X$ then there's nothing new to prove. So assume that $\gamma\in X$.
Now, for $f$ to be an order isomorphism, $f(\gamma)$ has to be the smallest ordinal that is not an $f(\xi)$ with $\xi<\gamma$. Due to the induction hypothesis we cannot have $f(\xi)=\gamma$ for any such $\xi$, so $f(\gamma)$ is the minimum of a class of ordinals that includes $\gamma$ itself. Therefore $f(\gamma)\le \gamma$.

By the way, you don't need any explicit base case when you're doing ordinal induction. The induction scheme itself supplies the necessary base case -- the induction hypothesis "such-and-such holds for all $\xi<\alpha$" is vacuously true (and therefore not helpful) when $\alpha=0$.
A: It’s easier if you replace $f$ by its inverse and work with an order-isomorphism $f:\beta\to X$. Then $f$ is an order-isomorphism of $\beta$ into $\alpha$, and you want to show that $\xi\le f(\xi)$ for each $\xi\in\beta$. If not, let $\gamma\in\beta$ be minimal such that $f(\gamma)<\gamma$. Then by the minimality of $\gamma$ and the fact that $f$ is order-preserving you have
$$f(\gamma)\le f\big(f(\gamma)\big)<f(\gamma)\;,$$
which is absurd. Hence $\xi\le f(\xi)$ for all $\xi\in\beta$.
