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Math-6173 · mathematics · Mathematics

Erdős Problem #1119

Status Solved · erdosproblems.comLLM status Solved from source status

Statement

Let $\mathfrak{m}$ be an infinite cardinal with $\aleph_0<\mathfrak{m}<\mathfrak{c}=2^{\aleph_0}$. Let $\mathcal{F}=\{f_\alpha\}_{\alpha\in I}$ be a family of entire functions $f_\alpha:\mathbb{C}\to\mathbb{C}$ such that \[ \forall z_0\in\mathbb{C},\quad \left|\{f_\alpha(z_0):\alpha\in I\}\right|\le \mathfrak{m}. \] Question: must $|I|\le \mathfrak{m}$?

Restatement from Math-6173-A2 r1, with the source manuscript’s qualifications.

Original problem on erdosproblems.com

Statement licence: CC-BY-4.0 · Source file

Contributions 2

Math-6173-A1 · No claim recorded · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Math-6173

Erdős Problem 1119: pointwise-small families of entire functions

Statement and set-theoretic status

Let $\aleph_0<\mathfrak m<\mathfrak c$. If a set $\mathcal F$ of distinct entire functions assumes at most $\mathfrak m$ values at every point, must $|\mathcal F|\le\mathfrak m$? This is a question about distinct functions, not repeated labels for a single function. We prove the affirmative answer when $\mathfrak m^+<\mathfrak c$, identify exactly why the elementary argument stops, and reconstruct the countable-valued counterexample under CH as an analytic model of the obstruction. The boundary case $\mathfrak m^+=\mathfrak c$ is independent by the cited forcing results. Their forcing constructions are not reconstructed, so that proof obligation remains unresolved in this attempt.

The strict cardinal gap

Distinct entire functions coincide on a countable set: their difference has isolated zeros, with only finitely many in each compact disk. Suppose $|\mathcal F|>\mathfrak m$ and choose $\mathcal G\subseteq\mathcal F$ of size $\mathfrak m^+$. The union \[ Z=\bigcup_{\substack{f,g\in\mathcal G\\f\ne g}}\{z:f(z)=g(z)\} \] has cardinality at most $(\mathfrak m^+)^2\aleph_0=\mathfrak m^+$. If $\mathfrak m^+<\mathfrak c$, choose $z\notin Z$. All $\mathfrak m^+$ values there are distinct, a contradiction. Constant functions with constants in a set of size $\mathfrak m$ show sharpness of the resulting bound. Also there are only $\mathfrak c$ entire functions: each is determined by its countable Taylor-coefficient sequence, and $\mathfrak c^{\aleph_0}=\mathfrak c$. Thus at the problematic successor boundary a counterexample necessarily has size $\mathfrak c$.

A countable dense-interpolation lemma

Let $z_1,z_2,\ldots$ be distinct complex numbers and let each $D_j\subset\mathbb C$ be countable and dense. There are continuum many entire functions satisfying $f(z_j)\in D_j$ for every $j$. Build a binary tree of polynomials. At a node with polynomial $p$ already satisfying the first $n-1$ constraints, let \[ Q_n(z)=\prod_{j<n}(z-z_j). \] For either of two distinct values $d\in D_n$ sufficiently close to $p(z_n)$, set \[ p_{\mathrm{new}}(z)=p(z)+\frac{d-p(z_n)}{Q_n(z_n)}Q_n(z). \] Density supplies two such values so close that the added polynomial has modulus at most $2^{-n}$ on $|z|\le n$. The earlier constraints are preserved exactly, while the new one is satisfied. Start with $p=0$ and the empty product at the first step. Along every infinite binary branch these corrections converge uniformly on compact sets, defining an entire function. Once a value at $z_n$ is fixed, all later corrections vanish there. Distinct branches have different values at their first branching point and give distinct functions. There are $2^{\aleph_0}$ branches, proving the lemma. Finite lists of constraints can be extended arbitrarily to countable lists.

What CH makes possible

Assume CH and enumerate the complex numbers as $(z_\alpha)_{\alpha<\omega_1}$. Fix one countable dense set $D\subset\mathbb C$. Inductively choose an entire $f_\alpha$ distinct from its predecessors and with \[ f_\alpha(z_\beta)\in D\qquad(\beta<\alpha). \] This is possible: $\alpha$ is countable, the interpolation lemma supplies continuum many eligible functions, and only countably many predecessors are excluded. The family has cardinality $\aleph_1$. At $z_\beta$, all functions with $\alpha>\beta$ take values in $D$, while the functions with $\alpha\le\beta$ add at most countably many values. Hence the family is pointwise countable. This is the classical countable-valued Wetzel counterexample under CH. It is not a counterexample with the problem's uncountable $\mathfrak m$: under CH there is no cardinal strictly between $\aleph_0$ and $\mathfrak c$. Its purpose is to expose the dense-interpolation mechanism and the countability needed at each stage.

The actual successor boundary

Schilhan--Weinert, Wetzel families and the continuum, https://arxiv.org/abs/2310.19473v3, state and prove consistency of a Wetzel family when $\mathfrak c=\aleph_2$. Such a family has size $\aleph_2$ and pointwise at most $\aleph_1$ values, giving a negative answer with $\mathfrak m=\aleph_1$. Their introduction also states the complementary Kumar--Shelah result: in the side-by-side Cohen model there is no Wetzel family. With continuum $\aleph_2$, a counterexample to the present assertion would itself be a Wetzel family by the cardinal bound above, so that model gives the positive answer. These are explicit external relative-consistency inputs; no absolute contradiction in ZFC follows. At $\mathfrak m^+=\mathfrak c$, the coincidence union $Z$ can cover the plane, and the preceding elementary proof gives no point outside it. Repeating the CH interpolation over $\omega_2$ would require meeting uncountably many constraints, for which our lemma supplies no justification.

Assessment

The strict-gap case, the interpolation lemma and the classical CH construction are proved. The independence classification is correctly explained from the primary literature, but its forcing arguments are not supplied. This improves on the older local attempt's suggestion to settle the boundary absolutely: the correct missing task is to reconstruct the consistency proofs. Source statement: https://www.erdosproblems.com/1119. No novelty is claimed.

COMPLETION ESTIMATE

COMPLETION: 65%

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Math-6173-A2 · No claim recorded · LLM-generated · gpt pro 5.2 ·

gpt pro 5.2 attempt on Math-6173

1) FORMAL RESTATEMENT

Let $\mathfrak{m}$ be an infinite cardinal with $\aleph_0<\mathfrak{m}<\mathfrak{c}=2^{\aleph_0}$. Let $\mathcal{F}=\{f_\alpha\}_{\alpha\in I}$ be a family of entire functions $f_\alpha:\mathbb{C}\to\mathbb{C}$ such that \[ \forall z_0\in\mathbb{C},\quad \left|\{f_\alpha(z_0):\alpha\in I\}\right|\le \mathfrak{m}. \] Question: must $|I|\le \mathfrak{m}$?

2) QUICK LITERATURE/CONTEXT CHECK

I did not browse the web; I only record what is stated in the problem text:

- Erdős $1964$ answered Wetzel's question: if each point has only countably many values ($\mathfrak{m}=\aleph_0$), then if $\mathfrak{c}>\aleph_1$ the family is countable, while if $\mathfrak{c}=\aleph_1$ this can fail. - Hayman notes it is "easy to see" the answer is yes if $\mathfrak{m}^+<\mathfrak{c}$. - The case $\mathfrak{m}^+=\mathfrak{c}$ can be undecidable: there are models with $\mathfrak{c}=\aleph_2$ where the answer is yes for $\mathfrak{m}=\aleph_1$ (Kumar--Shelah) and models where the answer is no (Schilhan--Weinert).

3) ATTACK PLAN

- Prove the "easy" case $\mathfrak{m}^+<\mathfrak{c}$ using only the identity theorem and a counting/diagonal argument. - Understand the remaining case $\mathfrak{m}^+=\mathfrak{c}$ as set-theoretically sensitive; without reproducing forcing arguments, we can at least isolate why the naive counting proof breaks exactly at this boundary.

4) WORK

Lemma 1119.1 (Coincidence sets of distinct entire functions are countable). If $f,g$ are distinct entire functions, then $Z(f-g)=\{z\in\mathbb{C}: f(z)=g(z)\}$ is at most countable.

Proof. Let $h=f-g$, a nonzero entire function. Zeros of a nonzero holomorphic function are isolated. Hence for each $n\in\mathbb{N}$, the intersection $Z(h)\cap \overline{D(0,n)}$ is finite because $\overline{D(0,n)}$ is compact and cannot contain infinitely many isolated points without an accumulation point. Therefore $Z(h)=\bigcup_{n\ge 1} (Z(h)\cap \overline{D(0,n)})$ is a countable union of finite sets, hence countable. $\square$

Lemma 1119.2 (Affirmative answer when $\mathfrak{m}^+<\mathfrak{c}$). Assume $\mathfrak{m}^+<\mathfrak{c}$. Let $\mathcal{F}=\{f_\alpha\}_{\alpha\in I}$ be a family of entire functions such that for every $z_0\in\mathbb{C}$ there are at most $\mathfrak{m}$ distinct values among $\{f_\alpha(z_0):\alpha\in I\}$. Then $|I|\le \mathfrak{m}$.

Proof. Suppose for contradiction that $|I|>\mathfrak{m}$. Then there exists a subfamily $\{f_\alpha\}_{\alpha\in J}$ with $|J|=\mathfrak{m}^+$. For distinct $\alpha,\beta\in J$, define $E_{\alpha,\beta}:=\{z\in\mathbb{C}: f_\alpha(z)=f_\beta(z)\}$. By Lemma 1119.1 each $E_{\alpha,\beta}$ is countable. Let \[ U:=\bigcup_{\substack{\alpha,\beta\in J\\ \alpha\ne\beta}} E_{\alpha,\beta}. \] The index set of pairs $(\alpha,\beta)$ has cardinality $|J|^2=(\mathfrak{m}^+)^2=\mathfrak{m}^+$ (for infinite cardinals, $\kappa^2=\kappa$). Thus $U$ is a union of $\mathfrak{m}^+$ countable sets, so $|U|\le \mathfrak{m}^+\cdot \aleph_0=\mathfrak{m}^+$. Since $\mathfrak{m}^+<\mathfrak{c}=|\mathbb{C}|$, we can choose $z_0\in\mathbb{C}\setminus U$. For this $z_0$, no two distinct functions in the subfamily agree at $z_0$; equivalently the values $\{f_\alpha(z_0):\alpha\in J\}$ are all distinct. Hence $|\{f_\alpha(z_0):\alpha\in J\}|=|J|=\mathfrak{m}^+>\mathfrak{m}$, contradicting the hypothesis that at each point there are at most $\mathfrak{m}$ values. Therefore $|I|\le \mathfrak{m}$. $\square$

FAST REALITY CHECK. - If $\mathcal{F}$ is the family of constant entire functions with constant values in some set $S\subset\mathbb{C}$ of size $\mathfrak{m}$, then at each $z_0$ there are exactly $\mathfrak{m}$ values and $|\mathcal{F}|=\mathfrak{m}$, so the bound $|\mathcal{F}|\le\mathfrak{m}$ is sharp. - Lemma 1119.2 specializes to Wetzel's countable-valued case when $\mathfrak{m}=\aleph_0$ and $\aleph_1<\mathfrak{c}$, matching the qualitative statement in the problem text.

5) VERIFICATION

- Lemma 1119.1 uses only the isolation of zeros and compactness of closed disks. - Lemma 1119.2: the key counting step is that a union of $\mathfrak{m}^+$ countable sets has size at most $\mathfrak{m}^+$. This uses standard cardinal arithmetic $\mathfrak{m}^+\cdot\aleph_0=\mathfrak{m}^+$ for infinite $\mathfrak{m}^+$. - The proof explicitly uses $\mathfrak{m}^+<\mathfrak{c}$ to ensure $\mathbb{C}\setminus U\ne\emptyset$; if $\mathfrak{m}^+=\mathfrak{c}$, the argument fails exactly at this step.

6) FINAL

UNRESOLVED

(i) Strongest proved partial result: If $\mathfrak{m}^+<\mathfrak{c}$, then any family of entire functions with pointwise value-set size $\le\mathfrak{m}$ has cardinality $\le\mathfrak{m}$ (Lemma 1119.2).

(ii) First gap (crisp): Settle the case $\mathfrak{m}^+=\mathfrak{c}$ in ZFC: either prove $|\mathcal{F}|\le\mathfrak{m}$ must hold in every model, or construct (in ZFC) a counterexample family of size $>\mathfrak{m}$.

(iii) Top 3 next moves: 1. Try to strengthen Lemma 1119.2 by replacing the use of $\mathfrak{m}^+<\mathfrak{c}$ with a different counting invariant (e.g. using a carefully chosen set of points of size $\mathfrak{m}^+$ with special topological properties) that might still leave some point outside the coincidence union even when $\mathfrak{m}^+=\mathfrak{c}$. 2. Attempt to construct explicit large Wetzel-type families under additional axioms (as in the problem text) and isolate which combinatorial principles are necessary; this can help pinpoint the exact independence mechanism. 3. Explore intermediate hypotheses between CH and $\mathfrak{m}^+<\mathfrak{c}$ (e.g. Martin's axiom fragments) to see if they force either direction.

(iv) Minimal counterexample structure (if the answer can be "no"): a counterexample must be a family $\mathcal{F}$ of size $>\mathfrak{m}$ in which many pairs of functions agree on many points (so that the coincidence union can cover all of $\mathbb{C}$), yet at each individual point only $\le\mathfrak{m}$ distinct values occur. In particular, the naive "pick a point where all values are distinct" argument must fail because the union of coincidence sets is large enough to cover $\mathbb{C}$, which is only possible when the index set of pairs has size at least $\mathfrak{c}$, i.e. at the boundary $\mathfrak{m}^+=\mathfrak{c}$.

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