$\kappa\otimes\kappa=\kappa\implies\lambda\otimes\kappa=max\{\lambda,\kappa\}$

Question

I was curious about cardinal and ordinal numbers so I read a bit about them in Kunens set theory book. At some point he proves that if $\kappa$ is an infinite cardinal number, then $\kappa\otimes\kappa=\kappa$ (cardinal multiplication) and he states as a corollary that if $\lambda,\kappa$ are infinite cardinals, then $\lambda\oplus\kappa =\lambda\otimes\kappa=max\{\lambda,\kappa\}$. But I don't understand if this follows.

Answer 1

If $\lambda\le\kappa$, $\lambda\otimes\kappa\color{blue}{\le}\kappa\otimes\kappa=\kappa=\max\{\lambda,\,\kappa\}$, where the $\color{blue}{\le}$ follows from the fact that, if $f$ bijects a size-$\lambda$ set $L$ into a size-$\kappa$ set $K$, $(x,\,y)\mapsto(f(x),\,y)$ similarly injects $L\times K$ into $K^2$. Similarly, if $\kappa<\lambda$ then $\lambda\otimes\kappa=\kappa\otimes\lambda=\max\{\kappa,\,\lambda\}=\max\{\lambda,\,\kappa\}$. You can handle addition in a similar way if you've proven $\kappa\oplus\kappa=\kappa$ for all transfinite $\kappa$.