Wetzel's Problem and the Continuum Hypothesis

Lawrence C. Paulson

February 18, 2022

This is a development version of this entry. It might change over time and is not stable. Please refer to release versions for citations.

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 Proofs from THE BOOK by Aigner and Ziegler.
BSD License

Topics

Theories of Wetzels_Problem