Question about proof of Hessenberg: $\kappa \cdot \lambda = \lambda$

Question

The following is a theorem:

(Hessenberg) Let $1 \le \kappa \le \lambda$ where $\lambda$ is an infinite cardinal. Then $\kappa \cdot \lambda = \lambda$.

The proof in the book proceeds by transfinite induction showing $\lambda \cdot \lambda \le \lambda$. I have a question about the induction step. In the proof they define a well-order on $\lambda \times \lambda$ and then show that every proper initial segment of $\lambda \times \lambda$ has cardinality less than $\lambda$. It seems long-ish.

Why can't one argue like this: By the inductive assumption we have $\kappa \cdot \kappa \le \kappa$ for all $\kappa < \lambda$. Hence $\sup (\kappa \cdot \kappa) \le \sup \kappa $. But $\sup \kappa = \lambda$ and $\sup \kappa \cdot \kappa = \lambda \cdot \lambda$ which proves the claim.

Thanks for your help.

Answer 1

It doesn't work for successor cardinals. If $\lambda=\mu^+$, then $\sup_{\kappa<\lambda}\kappa=\mu<\lambda$.

For limit cardinals it doesn't work quite so easily either. You'd only have that $\sup_{\kappa<\lambda} \kappa\cdot \kappa=\lambda$, but you don't know if the former is actually $\lambda\cdot\lambda$.

Answer 2

Note that you're not exactly arguing about the cardinality as much as you are arguing about the order type of the Hessenberg sum.$\newcommand{\otp}{\operatorname{otp}}$

Furthermore if you have $\otp(\kappa\boxplus\kappa)=\kappa$ (where $\otp$ denotes order-type and $\boxplus$ is the Hessenberg sum) then you get stuck in the successor step:

If $\lambda=\kappa^+$ then your suggestion fails to even go beyond any limit ordinal above $\kappa$.

For limit cardinals you still don't know that $\otp(\kappa\boxplus\lambda)$ is less or more than $\lambda$, and you cannot use that for $\otp(\lambda\boxplus\lambda)=\lambda$.