Product-free subsets of $(0,1)$

Abstract: The third problem in Ben Green's collection of 100 open problems asks whether an open subset of $(0,1)$ that does not contain $x,y,z$ with $xy=z$ must have measure at most 1/3. We give an affirmative answer to this question. As part of the proof we obtain a result of independent interest that gives a lower bound for the size of the sumset and the difference set of a set of reals in terms not just of its size but also of a parameter that measures how far it is from being an interval.

• The paper proves that every open product-free subset of (0,1) has Lebesgue measure strictly less than 1/3, a sharp bound approached by periodic constructions.
• The authors establish the discrepancy inequality |A+A|, |A-A| ≥ 4|A| − 2d(A), yielding quantitative inverse results, stability near small doubling, and continuous analogues of Freiman-type theorems.
• The proof transforms multiplication into addition via −log x and derives the key inequality f(u−1)+f(u)+f(u+1) ≤ u for sum-free sets, with elementary balanced-interval arguments enforcing period-3 extremal structure.

The problem and the main result

The paper by Franchi, Gowers and Yip resolves a question posed by Ben Green in his collection of 100 open problems: if $A$ is an open subset of $(0,1)$ containing no triple $x,y,z$ with $xy=z$, must $|A|\le 1/3$? The authors answer affirmatively. Their main theorem states that every open product-free subset of $(0,1)$ has Lebesgue measure strictly less than $1/3$. The bound is best possible up to the strictness of the inequality: sets of the form $$\bigcup_{n\in\mathbb Z}(\alpha^{3n+1},\alpha^{3n+2})$$, intersected with $(0,1)$ for $\alpha>1$ close to $(0,1)$0, are product-free with measure arbitrarily close to $(0,1)$1.

The problem is a continuous analogue of Erdős's 1965 question on sum-free subsets of sets of integers, answered asymptotically by Eberhard, Green and Manners in 2014. Under the change of variables $(0,1)$2, multiplication becomes addition and Lebesgue measure on $(0,1)$3 becomes the exponential measure $(0,1)$4 on $(0,1)$5; Green's question is thus equivalent to showing that every sum-free measurable subset of $(0,1)$6 has exponential measure at most $(0,1)$7. The paper works with closed sets (which dominate open sets under approximation) and reduces the main theorem to a key inequality about distribution functions.

A quantitative inverse theorem via interval discrepancy

The principal technical tool is a new inequality relating the size of a sumset or difference set to how far the set deviates from being an interval. The relevant deviation is measured by the positive interval discrepancy

$(0,1)$8

the supremum over intervals $(0,1)$9, which equals $x,y,z$0 for the maximizing interval when $x,y,z$1 is compact. The second main theorem asserts that for every Borel set $x,y,z$2 of finite measure,

$x,y,z$3

The bound interpolates between two known regimes. When $x,y,z$4 is an interval, $x,y,z$5 and one recovers the sharp Brunn–Minkowski/Kneser bound $x,y,z$6. At the other extreme, if $x,y,z$7 is small relative to $x,y,z$8, the theorem forces near-doubling only when $x,y,z$9 occupies a large fraction of some interval. Equality holds not only for intervals but also for thickened arithmetic progressions $xy=z$0 and for sets of the form $xy=z$1 where $xy=z$2 is small enough that both the maximizing interval is $xy=z$3 and $xy=z$4; the class of sharp examples is therefore substantial rather than degenerate.

Reformulated contrapositively, if $xy=z$5 then there exists an interval $xy=z$6 with $xy=z$7. For $xy=z$8 this recovers the classical fact that a set with minimal doubling equals its convex hull almost everywhere; for $xy=z$9 it yields the sharp symmetric stability theory for one-dimensional Brunn–Minkowski, i.e. the continuous counterpart of Freiman's $|A|\le 1/3$0 theorem. Indeed, the authors derive from their theorem, via an amplification argument using long strings of translates to dilute discrepancy, the Figalli–Jerison diagonal inequality

$|A|\le 1/3$1

previously known through work of Ruzsa, Freiman, Figalli–Jerison and de Roton. For $|A|\le 1/3$2 close to $|A|\le 1/3$3, the result gives a sharp version of Theorem 6.2 of Eberhard–Green–Manners, with sharpness witnessed by $|A|\le 1/3$4.

The proof of the discrepancy theorem is elementary — no Fourier analysis or regularity lemmas are used — and proceeds by induction on the number of component intervals, organized around the notion of balanced intervals: intervals $|A|\le 1/3$5 such that $|A|\le 1/3$6 for every subinterval $|A|\le 1/3$7 of $|A|\le 1/3$8. Maximal balanced intervals partition the set, and a central lemma shows that balanced intervals of equal discrepancy in two sets force the corresponding restricted sumsets to fill entire intervals. This machinery also yields off-diagonal versions ($|A|\le 1/3$9 when $(0,1)$0) and, after a fractional-part thinning argument to equalize discrepancies, the general asymmetric corollary

$(0,1)$1

Two further consequences deserve mention. First, combining the theorem with simple observations (a set disjoint from its translate by $(0,1)$2 has discrepancy at most $(0,1)$3; a set meeting no unit-length interval in more than half its length has discrepancy at most $(0,1)$4) gives clean lower bounds of the form $(0,1)$5 for sumsets and difference sets under local sparsity hypotheses. Second, a discretization argument produces a purely combinatorial statement: if $(0,1)$6 satisfies $(0,1)$7, then $(0,1)$8, which is sharp for thickened even progressions and is in fact equivalent to the continuous corollary. The authors note that existing inverse theorems near doubling threshold $(0,1)$9 (Eberhard–Green–Manners, van Hintum–Keevash, Jing–Mudgal) do not appear strong enough to yield these bounds directly; the direct elementary route seems essential.

From product-freeness to the key distributional inequality

The reduction to the main theorem passes through a strikingly clean statement. Let $1/3$0 be a measurable sum-free subset of $1/3$1 with $1/3$2, and let $1/3$3. Then for every $1/3$4,

$1/3$5

This inequality is sharp: for $1/3$6 equality holds identically. Its consequence for the original problem follows from a short integration-by-parts argument: since $1/3$7,

$1/3$8

so $1/3$9, with strict inequality because $$\bigcup_{n\in\mathbb Z}(\alpha^{3n+1},\alpha^{3n+2})$$0.

Before introducing this idea, the authors show that cruder methods stall: optimizing elementary bounds involving $$\bigcup_{n\in\mathbb Z}(\alpha^{3n+1},\alpha^{3n+2})$$1 and the discrepancy estimate $$\bigcup_{n\in\mathbb Z}(\alpha^{3n+1},\alpha^{3n+2})$$2 yields an upper bound of roughly $$\bigcup_{n\in\mathbb Z}(\alpha^{3n+1},\alpha^{3n+2})$$3, but reaching $$\bigcup_{n\in\mathbb Z}(\alpha^{3n+1},\alpha^{3n+2})$$4 "does not seem possible without a further idea." The key lemma supplies that idea, and its proof is the technical core of the paper.

Structure of the proof of the key lemma

The proof is a rigidity-by-contradiction argument modeled on the period-3 extremal example. Suppose the inequality fails at some point. A preliminary reduction, based on the absolute continuity of $$\bigcup_{n\in\mathbb Z}(\alpha^{3n+1},\alpha^{3n+2})$$5, produces a failure point $$\bigcup_{n\in\mathbb Z}(\alpha^{3n+1},\alpha^{3n+2})$$6 with $$\bigcup_{n\in\mathbb Z}(\alpha^{3n+1},\alpha^{3n+2})$$7: the local configuration already resembles the extremal example, with $$\bigcup_{n\in\mathbb Z}(\alpha^{3n+1},\alpha^{3n+2})$$8 lying in a gap flanked by occupied blocks.

One then fixes a discrepancy-maximizing interval $$\bigcup_{n\in\mathbb Z}(\alpha^{3n+1},\alpha^{3n+2})$$9 of length $(0,1)$0 (maximal intervals for a sum-free set have length below $(0,1)$1) and discrepancy $(0,1)$2, and writes $(0,1)$3 for the position of the block. A sequence of lemmas progressively forces period-3 geometry:

• Excess mass in $(0,1)$4 (quantified as $(0,1)$5) forces $(0,1)$6.
• The maximizing interval is pushed leftward in stages: first $(0,1)$7, then $(0,1)$8, then $(0,1)$9, then $\alpha>1$0, each step combining difference-set estimates from the balanced-interval machinery with sum-freeness.
• The gap around $\alpha>1$1 has length at least $\alpha>1$2; subsequently a gap of length at least $\alpha>1$3 around $\alpha>1$4 is established, which forces the crucial density estimate $\alpha>1$5.
• Bootstrapping confines $\alpha>1$6 to the narrow window $\alpha>1$7, so the left endpoint of $\alpha>1$8 sits within $\alpha>1$9 of $(0,1)$00, exactly as in the model configuration.
• The same constraints propagate to the beginning of the set: the point $(0,1)$01 lies in a gap of size at least $(0,1)$02, where $(0,1)$03 is a slack parameter tracking deviation from exact rigidity.
• Finally one proves $(0,1)$04 (where $(0,1)$05 is the last point of $(0,1)$06 before $(0,1)$07), so the gaps around $(0,1)$08 and around $(0,1)$09 coincide, and a three-case analysis combining Brunn–Minkowski estimates with the accumulated gap and density constraints yields $(0,1)$10, contradicting the choice of $(0,1)$11.

The argument is delicate precisely because the inequalities employed are essentially sharp in the extremal configuration, leaving no room for loss; each estimate must be tight up to the controlled parameters $(0,1)$12.

Limitations and open questions

Several boundaries of the results are worth noting. The constant $(0,1)$13 is attained only in the limit by periodic constructions, and the theorem gives strict inequality for genuine subsets of $(0,1)$14; whether a structural stability statement accompanies the bound — describing product-free sets of measure close to $(0,1)$15 — is not addressed. The discrepancy theorem's sharp examples include thickened progressions and punctured intervals, but a complete characterization of equality cases is not given. The authors also record a natural strengthening that fails: conjecturing $(0,1)$16 in terms of one-sided discrepancies alone is false, with $(0,1)$17 a counterexample (found with AI assistance, though the proofs themselves are entirely due to the authors). Whether the discrete corollary $(0,1)$18 admits a direct combinatorial proof independent of the continuum is left implicit. More broadly, the analogous question for other measures or higher-dimensional analogues of the discrepancy framework is untouched.

Conclusion

The paper settles Green's third open problem with a sharp $(0,1)$19 bound for product-free open subsets of $(0,1)$20, via a reduction to a three-term distributional inequality for sum-free sets that is itself exactly sharp. Along the way it establishes a quantitative inverse theorem for sumsets and difference sets in terms of interval discrepancy, recovering and sharpening several strands of the one-dimensional inverse theory for small doubling, including the continuous $(0,1)$21 stability result, all by elementary means. The combination of a clean new geometric parameter, an induction-friendly balanced-interval toolkit, and a rigid contradiction argument calibrated against the period-3 extremal example makes the paper a self-contained contribution to both additive combinatorics and geometric measure theory.