(Characterization of Ordinal Exponentiation) Let $\alpha$ and $\beta$ be ordinals. For f:$\alpha \rightarrow \beta$, let s(f)={$\epsilon < \alpha|f(\epsilon) \not= 0$}. Let S($\alpha, \beta)$={$f|f:\alpha \rightarrow \beta \land s(f)$: finite}.
Define $<_S$ on S($\alpha, \beta)$ as follows: f$<_S$g if and only if there is $\epsilon_0<\alpha$ such that f($\epsilon_0)<g(\epsilon_0$) and f($\epsilon$)=g($\epsilon$) for all $\epsilon$ > $\epsilon_0$.
Show that (S($\alpha, \beta)$, $<_S$) is isomorphic to ($\beta^\alpha$, <).
This is also a question that is listed in "Introduction to Set Theory" by Hrbacek and Jech on page 123. And here is my trial: First, I had to check if $<_S$ is well-ordered. I just checked the linearity and passed the well-ordering part, though. Then I tried to find a bijective function that preserves the order, which I couldn't.
To visualize easily, if S and $\beta$ has greatest element, then the greatest element in S would have the form (0, 0, 0, $\cdots$, $\gamma$, $\gamma$, $\gamma$) something like this(only finitely many $\gamma$s) where $\gamma^+ =\beta$. But I couldn't think of a suitable mapping of this element into element of $\beta^\alpha$.
Actually, I have no idea why the given conditions allow only finitely many elements of f to have its value other than 0. I just got stuck. Any hint is also appreciated!
HINT:
You can prove this by induction on $\alpha$. The limit case should be fairly easy, as well the case $\alpha=0$.
To do the successor case, recall that $\beta^{\alpha+1}=\beta^\alpha\cdot\beta$, try to find an isomorphism between $S(\alpha+1,\beta)$ and $\beta\times S(\alpha,\beta)$.
(The fact that only finitely many points is non-zero is essential for the proof that the relation is well-ordered. Consider what happens if we take all the functions from $\omega$ to itself instead.)