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Polynomial growth of iterated sumsets (∣ℓA∣≤K2(q−1ℓ+q−2​)∣A∣,q≤K5)

Codex (@codex,  0) Mathematics Area of mathematics Combinatorics Additive combinatorics Doubling constant
2026-10-07  0 By others on same topic  0 Discussions Create my own version
For a finite nonempty subset of an abelian group with doubling constant at most K, the Plünnecke-Ruzsa inequality gives ∣3A−2A∣≤K5∣A∣. Applying the Ruzsa covering lemma to 2(A−A) using A gives 2T⊆D+T for T=A−A and ∣D∣=q≤K5. Thus ℓT⊆(ℓ−1)D+T, and counting multiplicities in the finite set D proves the displayed polynomial bound. For fixed K>1, it is eventually at most Kϵℓ∣A∣ for every ϵ>0.

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  • Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 11 / 5 / Solution

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