### Abstract

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 $\aleph_1$. But if CH is true then such an $F$,
of cardinality $\aleph_1$, 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

BSD License*Proofs from THE BOOK*by Aigner and Ziegler.