Absorption: for cardinals $\kappa ,\lambda$ with $\kappa$ infinite and $\lambda \le \kappa$, $\kappa \oplus \lambda =\kappa$, and $\kappa \otimes \lambda =\kappa$ when $\lambda \mathrm{\ne }0$

Let $\kappa$ be an infinite cardinal and $\lambda$ a cardinal with $\lambda \le \kappa$ (Cardinal (initial ordinal) and cardinality). Then

$\kappa \oplus \lambda =\kappa ,\text{and}\kappa \otimes \lambda =\kappa \text{ whenever }\lambda \mathrm{\ne }0$

(Cardinal sum $\kappa \oplus \lambda$, product $\kappa \otimes \lambda$ and exponentiation ${\kappa }^{\lambda }$, and why they are written apart from the ordinal operations). The exception at $\lambda =0$ is not an artefact: $\kappa \otimes 0=0$.

This is a theorem of ZF, inherited from Hessenberg: $\kappa \otimes \kappa =\kappa$ for every infinite cardinal $\kappa$, proved in ZF from the canonical well-order of $\kappa \times \kappa$, which is choice free. In particular the ordinary arithmetic of infinite cardinals collapses completely for $\oplus$ and $\otimes$: below the level of exponentiation, the larger argument simply swallows the smaller one.

What absorption costs. Nothing beyond Hessenberg: $\kappa \otimes \kappa =\kappa$ for every infinite cardinal $\kappa$, proved in ZF from the canonical well-order of $\kappa \times \kappa$: the only extra input is the injection of step 1.2, which folds two copies of $\kappa$ into a rectangle of width $2$. So absorption is choice free wherever Hessenberg's theorem is, that is, for cardinals.

Why the hypothesis is $\lambda \le \kappa$ and not $\lambda <\kappa$. The case $\lambda =\kappa$ is the interesting one and is used constantly: $\kappa \oplus \kappa =\kappa$ and $\kappa \otimes \kappa =\kappa$. Stating the corollary with $\le$ avoids a separate appeal to Hessenberg's theorem at every later use.

Absorption destroys cancellation. From $\kappa \oplus \lambda =\kappa$ for every $\lambda \le \kappa$ it follows at once that $\oplus$ cannot be cancellative on infinite cardinals, and the companion false statement FALSE: $\kappa \oplus \mu =\lambda \oplus \mu$ implies $\kappa =\lambda$ records exactly that. The same collapse does not reach exponentiation: assuming the Axiom of Choice, Assuming the Axiom of Choice, $2^{\kappa }=\mid \mathcal{P}(\kappa )\mid$, and Cantor's theorem in cardinal form: $\kappa <2^{\kappa }$ gives a strict increase at every cardinal.