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.


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 $\lambda\cdot\lambda$.

  • $\begingroup$ 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! :-) $\endgroup$ – Asaf Karagila Dec 13 '12 at 13:42
  • 1
    $\begingroup$ @AsafKaragila: Well, I guess it happens to everyone from time to time. :) The startling, that is. $\endgroup$ – 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 $\otp(\lambda\boxplus\lambda)=\lambda$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.