Definition of ordinal multiplication

Question

The ordinal multiplication $\cdot$ can be defined recursively via ordinal addition $+$ for any ordinal $\alpha$ as follows:

• $\alpha\cdot 0=0$.
• $\alpha\cdot (\beta+1)=\alpha\cdot \beta+\alpha$ for any ordinal $\beta$.
• $\alpha\cdot \lambda=\bigcup \{\alpha\cdot\beta: \beta\in \lambda\}$ for any limit ordinal $\lambda$.

According to this widely accepted definition, $\alpha\cdot \beta$ can be informally interpreted as the "length" of the concatenation of $\beta$ copies of $\alpha$. For instance,

• $\omega\cdot 2$ is the length of the concatenation of 2 copies of $\omega$, which is strictly larger than $\omega$, and
• $2\cdot \omega$ is the length of the concatenation of $\omega$ copies of $2=\{0,1\}$, which is still equal to $\omega$.

Since $\cdot$ is not commutative as can be seen from the above example, $\alpha\cdot \beta$ cannot be interpreted as the "length" of the concatenation of $\alpha$ copies of $\beta$. However, I personally think that "the length of $\alpha$ copies of $\beta$" would have been a more natural way to informally interpret $\alpha\cdot\beta$. Therefore, I wonder if there are any historical reasons for this widely accepted definition instead of defining $\cdot$ the other way around, i.e., for any ordinal $\beta$

• $0\cdot \beta=0$.
• $(\alpha+1)\cdot \beta=\alpha\cdot \beta+\beta$ for any ordinal $\alpha$.
• $\lambda\cdot \beta=\bigcup \{\alpha\cdot\beta: \alpha\in \lambda\}$ for any limit ordinal $\lambda$.

Answer 1

If you look at Cantor's writing you will find the same idea. He first defined multiplication the way you suggest. But later he switched to the definition we use today. Reference:
Joseph W. Dauben. Georg Cantor, his mathematics and philosophy of the infinite. Harvard University Press, 1979

Why? My guess: When defining $\alpha + \beta$ and $\alpha^\beta$ we do it by induction in the second variable. So to be consistent, do it that way also for the definition of $\alpha\beta$.