31. Define kard $A$ to be the collection of all sets $B$ such that (i) $A$ is equinumerous to $B$, and (ii) nothing of rank less than rank $B$ is equinumerous to $B$.
(a) Show that kard $A$ is a set.
(b) Show that kard $A$ is nonempty.
(c) Show that for any sets $A$ and $B$,$$\text{kard }A=\text{kard }B\quad\text{iff}\quad A\approx B$$
I understand the exercise by rewriting it into the following Theorem (which I believe is essentially what's being asked to prove ignoring (c) which should be proved as a corollary)
Theorem (Scott's Trick) For each set $A$, there exists a unique collection $\mathcal{C}$ of sets whose elements are exactly those elements $B$ such that $B\sim A$ and for each set $C$, if $\mathrm{rank}(C)<\mathrm{rank}(B)$, then $C\nsim B$.
My instincts tell me we should start with the ordinal $\mathrm{rank}(A)$, and find the least ordinal $\alpha$ less than or equal such that $V_\alpha$ has the property for $\mathcal{C}$. Is this right? All this does is prove (a), but how does one show (b) and (c)?
Any insight or rigorous solution is much appreciated!
The definition of $\mathrm{rank}(A)$ is the least ordinal $\beta $ such that $A\subseteq V_\beta$ where $V_\beta$ is the $\beta$-stage in the cumulative hierarchy of sets; that is, $V_\beta=\bigcup_{\alpha\in\beta}\mathcal{P}(V_\alpha)$ where $V_0=\varnothing$.
$V_\alpha$ itself probably won’t have the desired property, but using the rank of $A$ to limit the possibilities is the right idea.
Let $\alpha=\operatorname{rank}(A)$; then $A\in V_{\alpha+1}$. Let $\mathscr{A}=\{B\in V_{\alpha+1}:B\approx A\}$; $\mathscr{A}$ is a set, and $A\in\mathscr{A}\ne\varnothing$. Let $\beta=\min\{\eta\in\alpha+2:\mathscr{A}\cap V_\eta\ne\varnothing\}$; then $\operatorname{kard}A=\mathscr{A}\cap V_\beta$. In particular, the definition of the von Neumann hierarchy ensures that $\beta=\xi+1$ for some ordinal $\xi$, so $\operatorname{rank}(B)=\xi$ for each $B\in\operatorname{kard}A$, and if $\operatorname{rank}(C)<\xi$, then $C\notin\operatorname{kard}A$.