= 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 $\aleph_\alpha\leq\aleph_\beta$,
$$
\aleph_\alpha+\aleph_\beta
=\aleph_\alpha\aleph_\beta=\aleph_\beta.
$$
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.
Back to article page
