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Statement

Setting (p. 116). For α>0\alpha>0, AαA_\alpha is the sequence {[2nα]}\{[2^n\alpha]\} (the paper uses an=[2nα]a_n=[2^n\alpha] in the proof) and P(Aα)P(A_\alpha) its set of subset sums, as in Theorem 1. Definition 3 prints

fα(x)=max⁡{L∣∃y≤x−L for which ∀t, 1≤t≤1, y+t∉P(Aα)},f_\alpha(x)=\max\{L\mid\exists y\le x-L\text{ for which }\forall t,\ 1\le t\le1,\ y+t\notin P(A_\alpha)\},

with "1≤t≤11\le t\le1" [sic], and glosses it: fα(x)f_\alpha(x) is the biggest gap in P(Aα)∩[1,x]P(A_\alpha)\cap[1,x]. The same page states that AαA_\alpha is subcomplete if and only if α\alpha is a finite dyadic fraction.

Theorem 3 (pp. 116--117).

  1. lim sup⁡x→∞fα(x)/log⁡2x≤1\limsup_{x\to\infty}f_\alpha(x)/\log_2x\le1.

  2. As printed: "For almost all α\alpha we have lim⁡x→∞fα(x)=1/2\lim_{x\to\infty}f_\alpha(x)=1/2." [sic] The proof (p. 121) concludes instead that for almost all α\alpha (Lebesgue measure), fα(x)/log⁡2x→1/2f_\alpha(x)/\log_2x\to1/2 as x→∞x\to\infty, and that is the statement proved.

  3. For a parameter AA let

    GA(η,x)=G(η,x)={α∣A−1≤α<AG_A(\eta,x)=G(\eta,x)=\{\alpha\mid A-1\le\alpha<A and (fα(x)−log⁡2x2)/(log⁡x/2)≤η}(f_\alpha(x)-\tfrac{\log_2x}{2})/(\sqrt{\log x}/2)\le\eta\}.

    Then lim⁡x→∞μ(G(η,x))=Φ(η)\lim_{x\to\infty}\mu(G(\eta,x))=\Phi(\eta), where μ\mu is Lebesgue measure and Φ(η)=12π∫−∞ηe−t2/2 dt\Phi(\eta)=\frac1{\sqrt{2\pi}}\int_{-\infty}^{\eta}e^{-t^2/2}\,dt.

In part 3 the denominator is printed log⁡x/2\sqrt{\log x}/2, with no base on the logarithm; the proof (p. 121) works with the condition fα(x)≤log⁡x/2+ηlog⁡2x/2f_\alpha(x)\le\log x/2+\eta\sqrt{\log_2x}/2 and with n=log⁡2x+O(1)n=\log_2x+O(1) binary digits, which matches the base-22 normalization. The paper does not state the range of AA or of η\eta.

Source. N. Hegyvári, On sumset of certain sets, Publ. Math. Debrecen 45 (1994), no. 1--2, 115--122: Definition 3 and the statement on pp. 116--117, the proof in Section 4, pp. 120--122.

Read depth. Claims checked: the definition and the three parts were read clause by clause on the journal print. The proof was read but not checked step by step.

Proof pointer

Section 4, pp. 120--122. Lemma 3.1 (p. 120): the biggest gap in P(Aα)∩[1,an]P(A_\alpha)\cap[1,a_n] is the interval [∑i<nai+1,an)[\sum_{i<n}a_i+1,a_n), of length ∑i=1nεi(α)+a0−1\sum_{i=1}^{n}\varepsilon_i(\alpha)+a_0-1, proved by induction from an+1=2an+εn+1(α)a_{n+1}=2a_n+\varepsilon_{n+1}(\alpha). For an≤x<an+1a_n\le x<a_{n+1} this gives part 1, since n≤log⁡2xn\le\log_2x. Part 2 follows because in almost every number the binary digit 11 has frequency 1/21/2 (Lemma 3.2, p. 121, a special case of Theorem 148 of Hardy and Wright). Part 3 compares G(η,x)G(\eta,x) with the sets of α\alpha in [A−1,A)[A-1,A) whose first nn digits sum to less than n/2+(η+δ)n/2n/2+(\eta+\delta)\sqrt n/2, and similarly with η−δ\eta-\delta, and applies the normal approximation to the binomial distribution (p. 122).

Bears on

No Erdős problem page in this corpus cites this theorem. It concerns the subset sums of the single sequence {[2nα]}\{[2^n\alpha]\}, not the pairs of Problem 354.