In an effort to finally learn set theory rigourously, I've decided to start plowing through Jech's Set theory, making sure to do each of the exercises.
Here is Jech's problem 3.13 (pg. 34 in the third edition):
($\mathbf{ZF}$) Show that $\omega_2$ is not a countable union of countable sets.
He then gives a hint:
Assume $\omega_2 = \bigcup_{n<\omega} S_n$ with $S_n$ countable and let $\alpha_n$ denote the order-type of $S_n$. Then $\alpha = \sup_n \alpha_n \leq \omega_1$ and there is a mapping $\omega_1\times \alpha$ onto $\omega_2$.
The point is to do this without choice (where it is consistent, I believe, that $\omega_1$ is a countable union of countable sets). In fact, choice won't even be introduced for another 2 chapters.
I almost completely understand the hint and I see how to use the hint to solve the exercise - I'm just missing one small detail:
What is meant by the order type of $S_n$?
I understand what the order type of a well-ordered set is, and I understand that, since $S_n$ is countable, there is a bijection between $S_n$ and $\aleph_0$, so $S_n$ is well-orderable. But without choosing a bijection for each $S_n$ (infinitely many choices - not allowed), I don't see how $S_n$ has an order type.
I'm sure I'm missing something stupid, I just don't see it.
Thanks!
Every set of ordinals is well-ordered by $\in$, and therefore is isomorphic to a unique ordinal, and in fact the isomorphism is unique and does not require the axiom of choice. In fact this is the Mostowski collapse of $S_n$. The minimum goes to $0$, and by recursion we continue and collapse more and more ordinals until we finish collapsing $S_n$. Being countable the result is a countable ordinal.
So there is a unique way to choose these bijections between $S_n$ and $\alpha_n$. Of course, if we wanted to argue by cardinality, then you would have to choose a bijection for each $S_n$, but luckily this is not the case.