Growth of measurable sets avoiding integer quotients
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How fast can the measure of a measurable set S⊂(0,∞) grow if no quotient of two distinct points is an integer? For nondecreasing F→∞, the target asks when |S∩(0,x)|≥F(x) eventually can coexist with infinite measure.
Why it matters and what counts as progress
Why it matters: The integral criterion describes exactly which measure-growth profiles are compatible with excluding integer ratios.
What counts as progress: Prove the sharp criterion ∫₁∞F(x)/x² dx<∞ or construct sets achieving admissible growth profiles.
How the problem developed
- 1969 · Open problem [1]
W. M. Schmidt Asked whether an infinite-measure measurable set can avoid having an integer quotient between any two distinct points; Erdős later recorded the question and asked how fast its measure can grow.
- 1970–1971 · Existence result [1]
Haight and Szemerédi Independently constructed infinite-measure sets avoiding integer quotients, answering the existence part while leaving the sharp growth question open.
- April 2026 · Reported claim [1]
Boon Suan Ho and GPT-5.4 Pro Proved a sharp criterion: a nondecreasing target F tending to infinity is eventually attainable exactly when the integral of F(x)/x² from 1 to infinity is finite.
AI contribution
Erdős’s ratio-growth problem and a quantitative Haight-type theorem
For nondecreasing F tending to infinity, an infinite-measure measurable set S of positive reals with no distinct-point integer quotient and |S∩(0,x)|≥F(x) eventually exists exactly when ∫₁∞ F(x)/x² dx is finite.
Sources
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Publication dates and editorial changes.
AI-assisted work gives a sharp growth criterion for Erdős #1195