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Statement

Printed p. 205 (PDF p. 1 of the retained scan, page image). With 0<a1<a2<⋯<an0<a_1<a_2<\cdots<a_n arbitrary real numbers and f(t)f(t) the number of solutions of

∑i=1nεiai=t;εi=0 oder 1(1)\sum_{i=1}^n\varepsilon_ia_i=t;\qquad\varepsilon_i=0\text{ oder }1 \tag{1}

the paper recalls that Erdős and Moser proved max⁡0≤t<+∞f(t)<c12nn3/2log⁡3/2n\max_{0\le t<+\infty}f(t)<c_1\frac{2^n}{n^{3/2}}\log^{3/2}n and conjectured max⁡0≤t<+∞f(t)<c22nn3/2\max_{0\le t<+\infty}f(t)<c_2\frac{2^n}{n^{3/2}}, adding that for a1=1,a2=2,…,an=na_1=1,a_2=2,\ldots,a_n=n one has max⁡t=0,1,2,…,n2f(t)>c3(2n/n3/2)\max_{t=0,1,2,\ldots,n^2}f(t)>c_3(2^n/n^{3/2}). Then: "SATZ. Es sei ε>0\varepsilon>0 eine beliebige Zahl. Dann ist für n>n0(ε)n>n_0(\varepsilon)

max⁡0≤t<+∞f(t)<(1+ε)8π⋅2nn3/2.\max_{0\le t<+\infty}f(t)<(1+\varepsilon)\frac8{\sqrt\pi}\cdot\frac{2^n}{n^{3/2}}.

"

In English: for every ε>0\varepsilon>0 and every n>n0(ε)n>n_0(\varepsilon), for any nn distinct positive reals, every real tt is the sum of fewer than (1+ε)(8/π)2n/n3/2(1+\varepsilon)(8/\sqrt\pi)2^n/n^{3/2} of the 2n2^n subsets. Positive integers are positive reals, so the bound covers the sets A⊆NA\subseteq\mathbb N of Problem 362; the constants c1,c2,…c_1,c_2,\ldots are positive constants (p. 205).

Source. A. Sárközy and E. Szemerédi, Über ein Problem von Erdös und Moser, Acta Arith. 11 (1965), no. 2, 205--208, DOI 10.4064/aa-11-2-205-208; printed p. 205, read on the page image (the scan has no text layer). Library home: sarkozi_1965_uber_ein_problem_von_erdos_und.

Read depth. Claims checked: the definitions, the recalled bounds and the Satz were read clause by clause on the page image. The proof (pp. 205--208) was read for structure and not checked; nothing here is independently reviewed.

Proof pointer

Pages 205--208, indirect. The Lemma (pp. 205--206, "eine modifizierte und schwächere Gestalt eines Satzes von Katona", proved on p. 206 from Sperner's theorem): if A=B∪CA=B\cup C with B∩C=∅B\cap C=\emptyset, ∣B∣=b|B|=b, ∣C∣=c|C|=c, and M1,…,MlM_1,\ldots,M_l are subsets of AA with l≥2b(c[c/2])+1l\ge2^b\binom c{[c/2]}+1, then two of them satisfy Mu∩B=Mv∩BM_u\cap B=M_v\cap B (2) and Mu∩C⊂Mv∩CM_u\cap C\subset M_v\cap C (3). Assuming f(t)≥(1+ε)8π2nn3/2f(t)\ge(1+\varepsilon)\frac8{\sqrt\pi}\frac{2^n}{n^{3/2}} for some tt (4), BB is the set of the [n/2][n/2] smallest and CC the set of the remaining aia_i; the solution sets AiA_i with more than n4⋅1+ε/31+2ε/3\frac n4\cdot\frac{1+\varepsilon/3}{1+2\varepsilon/3} elements in BB (5) number f1(t)>(1+2ε/3)8π2nn3/2f_1(t)>(1+2\varepsilon/3)\frac8{\sqrt\pi}\frac{2^n}{n^{3/2}} (6); removing one element of BB from each in all ways gives more than 2[n/2](n−[n/2][(n−[n/2])/2])+12^{[n/2]}\binom{n-[n/2]}{[(n-[n/2])/2]}+1 distinct sets DijD_i^j (the central binomial asymptotic enters here, for m=n−[n/2]m=n-[n/2] in the form (m[m/2])∼2π2mn\binom{m}{[m/2]}\sim\frac2{\sqrt\pi}\frac{2^m}{\sqrt n}, p. 207, where the print's exponent reads n/2−[n/2]n/2-[n/2]), so the Lemma yields two of them with (7) and (8)--(11), which force a[n/2]≥a[n/2]+1a_{[n/2]}\ge a_{[n/2]+1}, a contradiction (p. 208).

Dependencies

Sperner's theorem (the paper's [2]); the Lemma modifies a theorem of Katona (the paper's [1], "im Druck" in 1965).

Bears on

  • Problem 362: the status-defining source of the first question, #{S⊆A:∑S=t}≪2N/N3/2\#\{S\subseteq A:\sum S=t\}\ll2^N/N^{3/2} for ∣A∣=N|A|=N, with an absolute implied constant; the introduction's report of the Erdős--Moser bound is the site's "with an additional factor of (log⁡n)3/2(\log n)^{3/2}".