Erdős Problem #1119

Let \(\mathfrak{m}\) be an infinite cardinal with \(\aleph_0 < \mathfrak{m} < \mathfrak{c} = 2^{\aleph_0}\). Let \(\{f_\alpha\}\) be a family of entire functions such that, for every \(z_0 \in \mathbb{C}\), there are at most \(\mathfrak{m}\) distinct values of \(f_\alpha(z_0)\). Must \(\{f_\alpha\}\) have cardinality at most \(\mathfrak{m}\)? This is Problem 2.46 in [Ha74], where it is attributed to Erdős. The question is **independent of ZFC**: the headline statement is neither provable nor refutable from the usual axioms of set theory, so its `sorry` cannot be filled either way. The answer is yes if \(\mathfrak{m}^+ < \mathfrak{c}\) (see `erdos_1119.variants.easy_case`), so the question reduces to the case \(\mathfrak{m}^+ = \mathfrak{c}\), where it is undecidable: Kumar and Shelah [KuSh17] produced a model of \(\mathfrak{c} = \aleph_2\) in which the answer is yes (with \(\mathfrak{m} = \aleph_1\)), while Schilhan and Weinert [ScWe24] produced a different model of \(\mathfrak{c} = \aleph_2\) in which the answer is no.