Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 4 / 16I / b / Solution

Define ${\aleph }_0=|\omega |$, let ${\aleph }_{\alpha +1}$ be the least cardinal greater than ${\aleph }_{\alpha }$, and at a limit take the least cardinal above all earlier ${\aleph }_{\gamma }$. In ZFC every set is well-orderable, so every infinite cardinal is an initial ordinal and equals a unique ${\aleph }_{\alpha }$.

By transfinite induction on infinite well-ordered cardinals $\kappa$, order $\kappa \times \kappa$ first by $\max (\xi ,\eta )$ and then lexicographically. Every proper initial segment has cardinal below $\kappa$, using the inductive hypothesis for smaller infinite cardinals. Hence this well-order has cardinal at most $\kappa$, while the diagonal gives the reverse inequality, so ${\kappa }^2=\kappa$. Therefore, for ${\aleph }_{\alpha }\le {\aleph }_{\beta }$,

Finally suppose nonempty $x$ had a set $y$ containing every set equinumerous with $x$. For every ordinal $\gamma$, replacing each $z\in x$ by a tagged ordered pair $(\gamma ,z)$ gives a set $t_{\gamma }$ equinumerous with $x$ whose rank is at least $\gamma$. Then every $t_{\gamma }$ would belong to $y$, so the ranks of members of $y$ would be unbounded in the ordinals, contradicting the ordinal rank of $y$. This proves the displayed sentence.

Solved by gpt-5.6-sol high.