gpt pro 5.2 attempt on Math-689
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Contents
Erdős Problem #143 (Integer dilation approximation)
1) FORMAL RESTATEMENT
Let $\mathcal A\subset (1,\infty)$ be a countably infinite set. Assume the integer-dilation separation condition \begin{equation} \label{wsref-0-65-71-3a-31-34-33-2d-73-65-70} \forall x\neq y\in \mathcal A\ \forall k\in\mathbb Z_{\ge 1}:\quad |kx-y|\ge 1. \end{equation} Define the truncated harmonic sum \[ H_{\mathcal A}(X):=\sum_{\substack{x\in\mathcal A\\ x\le X}}\frac1x\qquad (X\ge 1), \] and the "primitive-set analogue" sum \[ P_{\mathcal A}:=\sum_{x\in\mathcal A}\frac{1}{x\log x}, \] where $\log$ is the natural logarithm (so $x>1$ ensures $\log x>0$).The question asks whether (eq:143-sep) forces $\mathcal A$ to be sparse, in particular whether it implies either \begin{align*} \text{(Q1)}\quad & P_{\mathcal A}<\infty,\\ \text{(Q2)}\quad & H_{\mathcal A}(X)=o(\log X)\quad\text{as }X\to\infty. \end{align*}
2) QUICK LITERATURE / CONTEXT CHECK
- If $\mathcal A\subset\mathbb N$ consists of integers, then (eq:143-sep) is equivalent to $\mathcal A$ being primitive (no element divides another): for integers, $|kx-y|\ge 1$ always holds unless $kx=y$. For primitive integer sets, Erdős proved $\sum_{n\in\mathcal A}\frac{1}{n\log n}<\infty$, and later works of Behrend and Erdős--Sárkőzy--Szemerédi gave quantitative bounds for $\sum_{n\le X,\,n\in\mathcal A}\frac1n$.
- For general real sets $\mathcal A\subset(1,\infty)$, the second implication $Q2$ is now known to hold: Koukoulopoulos--Lamzouri--Lichtman $2025$ proved that if $\limsup_{X\to\infty}\frac{1}{\log X}\sum_{x\in\mathcal A\cap[1,X]}\frac1x>0$, then for every $\varepsilon>0$ there exist infinitely many distinct $\alpha,\beta\in\mathcal A$ and a positive integer $n$ with $|n\alpha-\beta|<\varepsilon$. The contrapositive yields $H_{\mathcal A}(X)=o(\log X)$ under (eq:143-sep).
- The convergence question $Q1$ for general real sets satisfying (eq:143-sep) appears to remain open.
3) ATTACK PLAN
Proof track.- Extract immediate structural consequences of (eq:143-sep) (e.g. $1$-separation from the case $k=1$) to obtain baseline bounds.
- Use the strongest known theorem (Koukoulopoulos--Lamzouri--Lichtman) as a black box to deduce $Q2$ rigorously.
- For $Q1$, try to adapt "primitive set" arguments: one would like to partition $\mathcal A$ into classes where "multiplicative structure" forces a convergent majorant, but in the real setting there is no prime factorization. A plausible route is to convert (eq:143-sep) into a graph/Diophantine approximation statement and seek a "density increment" or "energy" inequality strong enough to upgrade $Q2$ to $Q1$.
Disproof track.
- Attempt to construct $\mathcal A$ with relatively large local density (e.g. about one point per interval of length $\log\log x$) while arranging fractional parts so that no integer dilation approximates another element within $1$.
- Heuristic: to make $\sum \frac{1}{x\log x}$ diverge, one needs $\mathcal A$ to have counting function $A(X)=|\mathcal A\cap[1,X]|$ decaying slowly enough, e.g. $A(X)\asymp X/\log\log X$ would already force divergence of $\int \frac{A'(t)}{t\log t}\,dt$. So a counterexample would need to be much denser than the known examples (like primes) but still satisfy (eq:143-sep).
4) WORK
Phase 1 sanity checks (tiny examples).
- Integer example: If $\mathcal A$ is the set of prime numbers, then (eq:143-sep) holds (no prime is a multiple of another prime), and one has $H_{\mathcal A}(X)=\sum_{p\le X}\frac1p\sim\log\log X$ (hence $o(\log X)$) and $\sum_{p}\frac{1}{p\log p}<\infty$.
- Trivial upper bound: Any $\mathcal A$ satisfying (eq:143-sep) is $1$-separated, hence has at most one element per unit interval.
Lemma (Unit separation): If $\mathcal A$ satisfies (eq:143-sep), then for all distinct $x,y\in\mathcal A$ we have $|x-y|\ge 1$.
Proof: Take $k=1$ in (eq:143-sep). ∎
Corollary (Baseline harmonic bound): For all $X\ge 2$, \[
H_{\mathcal A}(X)\le 1+\sum_{m=2}^{\lfloor X\rfloor}\frac1m\le 1+\log X.
\] In particular $H_{\mathcal A}(X)=O(\log X)$.
Proof: By Lemma Lemma (Unit separation), $\mathcal A$ contains at most one element in each interval $[m,m+1)$ with integer $m\ge 2$. If $x\in\mathcal A\cap[m,m+1)$ then $1/x\le 1/m$. Summing over $m\le X$ gives the first inequality; the second is the standard bound on harmonic numbers. ∎
Known theorem (used as a black box).
We record the 2025 theorem in a form tailored to this problem.Theorem (Koukoulopoulos--Lamzouri--Lichtman $2025$, contrapositive form): Let $\mathcal A\subset\mathbb R_{\ge 1}$ be countable and fix $\varepsilon>0$. Assume that there do not exist infinitely many pairs $(\alpha,\beta)\in\mathcal A^2$ with $\alpha\ne\beta$ and a positive integer $n$ such that $|n\alpha-\beta|<\varepsilon$. Then \[
\limsup_{X\to\infty}\frac{1}{\log X}\sum_{\alpha\in\mathcal A\cap[1,X]}\frac{1}{\alpha}=0,
\] i.e. $H_{\mathcal A}(X)=o(\log X)$.
Proof: This is exactly the contrapositive of the statement proved in the cited work: if the displayed limsup were $>0$, then for every $\varepsilon>0$ there would be infinitely many pairs $(\alpha,\beta)$ with $|n\alpha-\beta|<\varepsilon$ for some $n\in\mathbb Z_{\ge 1}$. ∎
Corollary (Resolution of $Q2$): If $\mathcal A\subset(1,\infty)$ satisfies (eq:143-sep), then \[
H_{\mathcal A}(X)=o(\log X)\qquad (X\to\infty).
\]
Proof: Under (eq:143-sep), for every distinct $\alpha,\beta\in\mathcal A$ and every $n\in\mathbb Z_{\ge 1}$ we have $|n\alpha-\beta|\ge 1$. Hence the hypothesis of Theorem thm:KLL-contrapos holds with (say) $\varepsilon=1$. The conclusion is exactly $Q2$. ∎
Status of $Q1$.
The argument above settles $Q2$ but does not appear to imply $Q1$. Indeed, there exist abstract sequences $(a_n)$ with $\sum_{a_n\le X} \frac1{a_n}=o(\log X)$ but $\sum_n \frac{1}{a_n\log a_n}=\infty$ (for instance, choosing $a_n$ so that $|\{a_n\le X\}|\asymp X/\log\log X$). The open difficulty is to show that (eq:143-sep) forbids such "near $X/\log\log X$" densities.5) VERIFICATION
- Quantifiers: Corollary cor:143-Q2 matches $Q2$ exactly: $\lim_{X\to\infty} H_{\mathcal A}(X)/\log X=0$.
- Edge cases: $x>1$ is used only to ensure $\log x>0$ in $Q1$. The $Q2$ deduction works already for $\mathcal A\subset[1,\infty)$.
- Self-pairs: Theorem thm:KLL-contrapos requires $\alpha\ne\beta$, matching (eq:143-sep).
6) FINAL
UNRESOLVED. The stronger convergence claim $Q1$ remains open (to the best of my current knowledge), but $Q2$ is resolved affirmatively.- (i) Strongest fully proved partial result obtained:
- $Q2$ holds: $\sum_{x\in\mathcal A,\,x\le X}\frac1x=o(\log X)$ (Corollary cor:143-Q2).
- Trivial baseline: $H_{\mathcal A}(X)\le 1+\log X$ for $X\ge 2$ (Corollary Corollary (Baseline harmonic bound)).
- (ii) Exact first gap: I do not have a proof upgrading $Q2$ to $\sum_{x\in\mathcal A}\frac{1}{x\log x}<\infty$, nor an explicit construction of $\mathcal A$ satisfying (eq:143-sep) for which that sum diverges.
- (iii) Top 3 next moves:
- Try to refine the KLL "GCD graph" machinery to obtain a quantitative bound on the counting function $A(X)=|\mathcal A\cap[1,X]|$, strong enough to force $\int \frac{dA(t)}{t\log t}<\infty$.
- Attempt a transference principle: discretize $\mathcal A$ by integer parts and fractional parts, and show that (eq:143-sep) enforces an "approximate primitivity" property that controls $\sum 1/(x\log x)$.
- Explore constructive disproof: build $\mathcal A$ by choosing one point in many unit intervals via a Lovász local lemma / entropy-compression argument, arranging to avoid all forbidden neighborhoods $|kx-y|<1$.
- (iv) What a minimal counterexample would likely look like: a set with counting function roughly $A(X)\approx X/(\log\log X)$ (or slightly smaller), spread across unit intervals, with carefully chosen fractional parts so that for each $x\in\mathcal A$ the nearest multiple $kx$ stays at distance $\ge 1$ from every other point. Such a set would make $\sum \frac{1}{x\log x}$ diverge while remaining consistent with $H_{\mathcal A}(X)=o(\log X)$.