Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 16H / b / Solution

Under the axiom of choice, identify every infinite cardinal $\kappa$ with its initial ordinal. We prove ${\kappa }^2=\kappa$ by transfinite induction on infinite cardinals. The countable case follows from the usual diagonal enumeration of $\mathbb{N}\times \mathbb{N}$.

For the induction step, well-order the Cartesian product $\kappa \times \kappa$ by first comparingand then using lexicographic order among pairs with the same maximum. The predecessors of $(\alpha ,\beta )$ lie in $(\eta +1)\times (\eta +1)$, where $\eta =\max \{\alpha ,\beta \}<\kappa$. Put $\mu =|\eta +1|<\kappa$. If $\mu$ is infinite, the induction hypothesis makes this predecessor set have cardinal at most ${\mu }^2=\mu <\kappa$; if $\mu$ is finite, the same conclusion is immediate.

Let $\theta$ be the order type of this well-order. Every proper initial segment of $\theta$ therefore has cardinal less than $\kappa$. Hence $\theta <{\kappa }^{+}$, since otherwise its initial segment of order type $\kappa$ would have cardinal $\kappa$. Thus $|\kappa \times \kappa |=|\theta |\le \kappa$. The map $\alpha \mapsto (\alpha ,0)$ gives the reverse injection, and the Cantor-Schröder-Bernstein theorem yields

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