Wetzel's Problem and the Continuum Hypothesis

Let $F$ be a set of analytic functions on the complex plane such that, for each $z\in \mathbb{C}$, the set $\{f(z)\mid f\in F\}$ is countable; must then $F$ itself be countable? The answer is yes if the Continuum Hypothesis is false, i.e., if the cardinality of $\mathbb{R}$ exceeds ${\mathrm{\aleph }}_1$. But if CH is true then such an $F$, of cardinality ${\mathrm{\aleph }}_1$, can be constructed by transfinite recursion. The formal proof illustrates reasoning about complex analysis (analytic and homomorphic functions) and set theory (transfinite cardinalities) in a single setting. The mathematical text comes from Proofs from THE BOOK by Aigner and Ziegler.