I am reading through my Set Theory notes and couldn't wrap my head around one particular part of our proof that $\aleph_\alpha\aleph_\alpha = \aleph_\alpha$.
In particular, I am not sure about both the validity and necessity of a line in our proof that $\omega_\alpha \times \omega_\alpha \leftrightarrow \omega_\alpha$ that I will explain now:
In brief terms, we show that there is a bijection $\omega_\alpha \leftrightarrow \omega_\alpha$ inductively. Without getting too bogged down in specifics, we do this by defining a well ordering on $\omega_\alpha \times \omega_\alpha$ by going up in squares:
Imagine $\omega_\alpha \times \omega_\alpha$ as a big square. We consider the countable sequence of little squares starting in the bottom right. These squares correspond to $\beta \times \beta$ where $\beta \leq \omega_\alpha$. We can then define a well ordering on $\omega_\alpha \times \omega_\alpha$ by comparing which two squares elements are in, and if they are in all of the same squares, we compare the components.
Given this well ordering, we consider an element $r \in \omega_\alpha \times \omega_\alpha$ and argue that the initial segment $I_r \subset \omega_\alpha \times \omega_\alpha$ is contained in $\beta \times \beta$.
By the inductive hypothesis, $\beta \times \beta \leftrightarrow \beta \leq \omega_\alpha$ and so we see that $I_r$ has order type at most $\omega_\alpha$ in $\omega_\alpha \times \omega_\alpha$
This is the line I don't understand. We don't know that the bijection between $\beta \times \beta $ and $\beta$ is order preserving, so how can we make this claim about order types?
Furthermore, is this even necessary? Can't we just say that since every initial segment of $\omega_\alpha \times \omega_\alpha$ can be injected into a proper initial segment of $\omega_\alpha$ then $\omega_\alpha \times \omega_\alpha $ can be injected into $\omega_\alpha$?
Why has my lecturer bothered to make this comment about order types here?
It would make more sense if it's supposed to say $<$ instead of $\le$ in a few places.
Since $\omega_\alpha$ is a limit ordinal, you know that actually $r$ is in $\beta\times\beta$ for some $\beta<\omega_\alpha$, not just $\beta\le\omega_\alpha$.
(Namely, $\beta$ can be taken as the successor of the larger of the two elements of the pair $r$).
Now the well-ordering of $\beta\times\beta$ is (by the induction hypothesis) of the same cardinality as $\beta$ itself -- and since $\omega_\alpha$ is an initial ordinal, that means that $|\beta\times\beta|<|\omega_\alpha|$. So it cannot be order isomorphic to anything $\ge\omega_\alpha$.
The order type of $I_r$ is what you need to look at in order to see that every initial segment maps to an initial segment of $\omega_\alpha$ in a way such that all those maps for different initial segments agree with each other.