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Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

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Statement

Notation as on the Théorème 1 page: A\mathcal A is admissible when two subsets of different cardinalities never have the same sum, P(A,k)P(\mathcal A,k) is the number of integers that are sums of exactly kk distinct elements of A\mathcal A, and F(N)F(N) is the largest size of an admissible subset of NN={1,…,N}\mathbb N_N=\{1,\ldots,N\}.

Lemme 1 (printed p. 56). Let A\mathcal A be a finite set and k∈Nk\in\mathbb N with k≤∣A∣k\le|\mathcal A|. Then

P(A,k)≥k(∣A∣−k)+1.P(\mathcal A,k)\ge k(|\mathcal A|-k)+1.

"C'est le théorème 2 de Straus [2], qui se démontre facilement par récurrence sur ∣A∣|\mathcal A|": stated with this pointer, no proof printed.

Lemme 2 (printed p. 57, display (4)). For N∈NN\in\mathbb N,

F(N)<43N1/2+1.F(N)<\frac{4}{\sqrt3}N^{1/2}+1.

"Nous utiliserons le théorème 4 de Straus sous la forme suivante" (p. 56, the sentence introducing the lemma); the proof opens "Nous suivons la preuve de Straus" (p. 57).

Source. P. Erdős, J.-L. Nicolas and A. Sárközy, Sommes de sous-ensembles, Sém. Théor. Nombres Bordeaux (2) 3 (1991), no. 1, 55–72; Numdam file, printed p. nn on PDF p. n−53n-53. Lemme 1 on printed p. 56 (PDF p. 3) and Lemme 2 with its proof on printed p. 57 (PDF p. 4), read on the page images.

Read depth. Claims checked for both lemmas (statements read clause by clause on the page images). The half-page proof of Lemme 2 was read line by line and its two displayed computations were followed here; this is a reading by the compiler, not an independent review, and Lemme 1, which the proof uses, is not proved in the paper.

Proof pointer

The proof of Lemme 2 (p. 57): for admissible A⊆NN\mathcal A\subseteq\mathbb N_N the sets P(A,k)\mathcal P(\mathcal A,k), k≤xk\le x, are disjoint subsets of [1,xN][1,xN], so by Lemme 1

xN≥∑k=1xP(A,k)≥∑k=1x(k(∣A∣−k)+1)=∣A∣x(x+1)2−x(x+1)(2x+1)6+x;xN\ge\sum_{k=1}^xP(\mathcal A,k)\ge\sum_{k=1}^x\bigl(k(|\mathcal A|-k)+1\bigr) =|\mathcal A|\frac{x(x+1)}2-\frac{x(x+1)(2x+1)}6+x;

with x=[3∣A∣/4]x=[3|\mathcal A|/4] this gives (5) 6(N−1)>98∣A∣(∣A∣−2)−16(N-1)>\tfrac98|\mathcal A|(|\mathcal A|-2)-1, and ∣A∣≥43N1/2+1|\mathcal A|\ge\frac4{\sqrt3}N^{1/2}+1 would make the right side at least 6N−98−1>6(N−1)6N-\tfrac98-1>6(N-1), a contradiction.

Dependencies

Straus's Theorem 2 (Lemme 1) and Theorem 4, from E. G. Straus, On a problem in combinatorial number theory, J. Math. Sci. 1 (1966), 77–80 (not held; a zbMATH record for it was read).

Bears on

  • Problem 874: the site's "Straus [St66] ... proved that lim sup⁡k(N)/N1/2≤4/3\limsup k(N)/N^{1/2}\le4/\sqrt3"; Lemme 2 is that bound with an explicit constant term, proved in a refereed paper.
  • Problem 789: the site's "Straus [St66] proved h(n)≪n1/2h(n)\ll n^{1/2}". Every admissible subset of A={1,…,n}A=\{1,\ldots,n\} has at most F(n)F(n) elements, so h(n)≤F(n)<43n1/2+1h(n)\le F(n)<\frac4{\sqrt3}n^{1/2}+1 for the problem's h(n)h(n), a one-line deduction made here from Lemme 2.