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

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.

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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 $\lambda\cdot\lambda$.

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I wanted to give the same argument for limit cardinals, but suddenly the banner with the new answer startled me and I lost my tracks... :-) In the meantime I added it to my answer. After all you made me forget it, now you made me recall it... seems fair! :-) –  Asaf Karagila Dec 13 '12 at 13:42
@AsafKaragila: Well, I guess it happens to everyone from time to time. :) The startling, that is. –  tomasz Dec 13 '12 at 13:45
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 $\otp(\lambda\boxplus\lambda)=\lambda$.