I am working on Exercise 3.6 from Jech's Set Theory.
Ex. 3.6: There is a well ordering of the class of all finite sequences of ordinals such that for each $\alpha$, the set of all finite sequences in $\omega_{\alpha}$ is an initial segment and its order-type is $\omega_{\alpha}$.
With the help of this post, I have the well-ordering that gives $[\omega_{\alpha}]^{<\omega}$ as an initial sequence. Now I must show that the order type is $\omega_{\alpha}$.
The post linked above and also this one say to use transfinite induction, but I'm not so sure that it is even true (given this well-ordering.) For example, if $\alpha=0$, then in $[\omega]^{<\omega}$, the single-term sequence $\langle2\rangle$ has infinitely many predecessors, (all finite sequences of just $0$ and $1$). This means that the order type of $[\omega]^{<\omega}$ is strictly greater than $\omega$.
I haven't thought too much about $\alpha>0$, having been disheartened when the base case failed. If transfinite induction is the way to go, then $\alpha=1$ needs to be proven on its own, (as a new base case.) I don't have a whole lot of intuition for that, but my counterexample above doesn't seem to generalize, so it may be doable.
Any advice would be greatly appreciated. Specifically, I want to know how to resolve the issue of $\alpha=0$. (Is it possible that $\alpha=0$ is an edge case Jech forgot to mention?) Clearing that up will probably clear up the rest.
Thanks!
EDIT: I think I can actually do it for $\alpha>0$, using transfinite induction with $\alpha=1$ as a base case. Perhaps the order can be modified to make $\alpha=0$ work too?
EDIT 2: Okay, so I did it for $\alpha>0$ without induction by looking at how many sequences less than a sequence of order $\omega_\alpha$ one would have remaining. Since there are only countably many finite sequences of finite ordinals, one can simply redefine $<$ to work for $\alpha=0$ without disturbing the rest. I am not confident that this is the best solution - I am happy to hear what anyone else has to say.