Cardinality of product of ordinals without AC

Question

I want to prove, without the use of AC, that $|\alpha \cdot \beta|=|\alpha| \cdot |\beta|$, where $\alpha$ and $\beta$ are ordinals. Any help?

Answer 1

There are two points here:

1. Every proof that refers only to well-ordered sets does not use the axiom of choice. The axiom of choice is used to have these well-orders in the first place, but when the sets given are well-ordered, the use of choice is redundant. (Do note that I mean every set in the proof, not that the theorem mentions well-ordered sets, but rather that the proof uses sets with well-orders on them.)

2. Depending on how you define ordinal arithmetic, or rather what you already know about this, your question is quite literally trivial. One of the definitions of $\alpha\cdot\beta$ is the lexicographic order on $\beta\times\alpha$, it is equivalent to the recursive definition as "iterated addition".

   Since the definition of $|\alpha|\cdot|\beta|$ is, literally, $|\alpha\times\beta|$, and it is easy to see that cardinal multiplication is commutative, the result follows.

But we also have an extra point.

3. Even if you work with the definitions based on recursion, it is not hard to prove that the cardinal arithmetic behaves as expected. Again, provided that you have proved certain properties of well-ordered sets. We do this by induction on $\beta$.

   For $\beta=0$ this is trivial, in fact for the finite case it is trivial. If $\beta$ is not a cardinal, then the result is also immediate by applying the induction hypothesis to some $\gamma<\beta$ such that $|\beta|=|\gamma|$.

   If $\beta$ is an infinite cardinal, that is $\beta\geq\omega$ and it is an initial ordinal, then one needs to check there is an obvious injection from $\sup\{\alpha\cdot\gamma\mid\gamma<\beta\}$ into $\alpha\times\beta$, and vice versa.