Erdős Problem 1119

References:

• erdosproblems.com/1119
• [Ha74] Hayman, W. K., Research problems in function theory: new problems. (1974), 155--180.
• [Er64g] Erdős, P., An interpolation problem associated with the continuum hypothesis. Michigan Math. J. (1964), 9--10.
• [KuSh17] Kumar, Ashutosh and Shelah, Saharon, On a question about families of entire functions. Fund. Math. (2017), 279--288.
• [ScWe24] Schilhan, Jonathan and Weinert, Thilo, Wetzel families and the continuum. J. Lond. Math. Soc. (2) (2024), Paper No. e12918, 27.

theorem Erdos1119.erdos_1119 : sorry ↔ ∀ (m : Cardinal.{0}), Cardinal.aleph0 < m → m < Cardinal.continuum → ∀ (F : Set (ℂ → ℂ)), (∀ f ∈ F, Differentiable ℂ f) → (∀ (z₀ : ℂ), Cardinal.mk ↑{y : ℂ | ∃ f ∈ F, f z₀ = y} ≤ m) → Cardinal.mk ↑F ≤ m

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, so the headline statement carries answer(sorry): it is neither provable nor refutable from the usual axioms of set theory.

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.

theorem Erdos1119.erdos_1119.variants.easy_case (m : Cardinal.{0}) (hm : Cardinal.aleph0 < m) (hsucc : Order.succ m < Cardinal.continuum) (F : Set (ℂ → ℂ)) (hF : ∀ f ∈ F, Differentiable ℂ f) (hval : ∀ (z₀ : ℂ), Cardinal.mk ↑{y : ℂ | ∃ f ∈ F, f z₀ = y} ≤ m) : Cardinal.mk ↑F ≤ m

The 'easy' case of Erdős Problem 1119: if moreover $\mathfrak{m}^+ < \mathfrak{c}$, then any family of entire functions taking at most $\mathfrak{m}$ distinct values at each point has cardinality at most $\mathfrak{m}$. In [Ha74] it is written that this is 'easy to see'.

theorem Erdos1119.erdos_1119.variants.erdos_wetzel (h : Cardinal.aleph 1 < Cardinal.continuum) (F : Set (ℂ → ℂ)) (hF : ∀ f ∈ F, Differentiable ℂ f) (hval : ∀ (z₀ : ℂ), {y : ℂ | ∃ f ∈ F, f z₀ = y}.Countable) : F.Countable

Erdős's theorem [Er64g], answering a question of Wetzel: if $\mathfrak{c} > \aleph_1$, then every family of entire functions taking only countably many distinct values at each point $z_0 \in \mathbb{C}$ is itself countable.

theorem Erdos1119.erdos_1119.variants.erdos_wetzel_ch (h : Cardinal.continuum = Cardinal.aleph 1) : ∃ (F : Set (ℂ → ℂ)), (∀ f ∈ F, Differentiable ℂ f) ∧ (∀ (z₀ : ℂ), {y : ℂ | ∃ f ∈ F, f z₀ = y}.Countable) ∧ ¬F.Countable

Erdős [Er64g] also showed that the previous statement fails under the continuum hypothesis: if $\mathfrak{c} = \aleph_1$, then there is an uncountable family of entire functions taking only countably many distinct values at each point $z_0 \in \mathbb{C}$.

theorem Erdos1119.erdos_1119.test.empty_family (m : Cardinal.{0}) (_hm : Cardinal.aleph0 < m) : (∀ f ∈ ∅, Differentiable ℂ f) ∧ (∀ (z₀ : ℂ), Cardinal.mk ↑{y : ℂ | ∃ f ∈ ∅, f z₀ = y} ≤ m) ∧ Cardinal.mk ↑∅ ≤ m

Sanity check: the empty family of entire functions satisfies both the value-bound hypothesis and the cardinality conclusion of erdos_1119, for any infinite $\mathfrak{m}$.