---

Statement

Theorem 4 (p. 338, quoted). "For all $\varepsilon>0$ there is a Sidon set $\mathcal A$ and a positive integer $i_0$ such that the sum set $\mathcal S_{\mathcal A}=\mathcal A+\mathcal A=\{s_1,s_2,\ldots\}$ satisfies"

$$s_{i+1}-s_i<s_i^{1/2}(\log s_i)^{(3/2)+\varepsilon} \qquad(10.1)$$

"for $i>i_0$."

The paper introduces it (p. 338) as a slightly weaker result for infinite Sidon sets, following the finite Theorem 3. The authors remark (p. 338) that the right-hand side of (10.1) can probably be replaced by $s_i^{\varepsilon}$, but that proving this seems hopeless.

Source. P. Erdős, A. Sárközy, V. T. Sós, On Sum Sets of Sidon Sets, I, J. Number Theory 47 (1994), 329--347, doi:10.1006/jnth.1994.1040; the statement on p. 338, the proof with Lemmas 1 (p. 339) and 2 (p. 340) on pp. 339--342. The edition read is identified on the source card.

Read depth. Claims checked: the statement and the remark were read clause by clause on the page images of the journal print. The proof was read but not checked step by step.

Proof pointer

Pp. 339--342, adapting the probabilistic method of Erdős and Rényi in the setting of Halberstam and Roth's Sequences. Each $n$ is put in a random set independently with probability $\alpha_n=n^{-3/4}(\log(n+3))^{-(1+\varepsilon)/4}$ (10.2). Lemma 1 (p. 339): almost surely every large $n$ has at most one representation $n=b+b'$, $b\le b'$, by Borel--Cantelli. Lemma 2 (p. 340): with $u_1=1000$, $v_n=[\frac16u_n^{1/2}(\log u_n)^{(3/2)+\varepsilon}]$ and $u_{n+1}=u_n+2v_n$, almost surely every large $n$ has $b<b'$ in the set with $[u_n/10]\le b$ and $u_n\le b+b'<u_{n+1}$, again by Borel--Cantelli. Removing the elements below the point from which Lemma 1 holds leaves a Sidon set, and Lemma 2 places a sum of two of its elements in every window $[u_n,u_{n+1})$ with $n$ large (p. 342).

Dependencies

The Erdős--Rényi probability space as set out in Halberstam and Roth (the paper's reference [5], Theorem 13, p. 142) and the Borel--Cantelli lemma.

Bears on

No Erdős problem page of the corpus consumes this theorem.