Solution
= Solution
\b[True.] For each $y\in x$, the <singleton set> $\{y\}$ is a finite subset of $x$, so the hypothesis gives
$$
\operatorname{rank}(y)+1
=\operatorname{rank}(\{y\})\leq c.
$$
Taking the <supremum> over all $y\in x$ yields
$$
\boxed{\operatorname{rank}(x)
=\sup_{y\in x}\bigl(\operatorname{rank}(y)+1\bigr)
\leq c.}
$$