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Statement

Definition (p. 848). A sequence $A=\{a_i\}$ is pseudorational, in the terminology the paper attributes to Buck and Volkmann, if for every $\varepsilon$ there are two sequences $S_1$ and $S_2$, each a finite union of arithmetic progressions (a rational sequence, in Volkmann's term, the name printed "Volkman" here), such that for large $n$

$$S_1\subset A\subset S_2,\qquad N(S_2-S_1,n)<\varepsilon n,$$

where $N(S_2-S_1,n)$ counts the integers up to $n$ in $S_2$ but not in $S_1$.

Theorem (p. 851, unnumbered). There exist two pseudorational sequences whose sum, the set of all $a+b$ with $a$ in the first and $b$ in the second, is not pseudorational. The paper states it as the proof of a conjecture (p. 848, footnote 5, "Volkman, ibid.") and announces it on p. 851 as: the Schnirelmann sum of two pseudorational sequences does not have to be pseudorational. It notes that the sum of two rational sequences is again rational.

Key step (pp. 851-852). If $A$ is a pseudorational set of density $0$ such that every integer is of the form $a+a'$ with $a,a'\in A$, then some subset $B$ of $A$ has $B+B$ of upper density $1$ and lower density $0$. Such a $B+B$ is not pseudorational, while $B$ is pseudorational as a subset of a pseudorational set of density $0$.

Source. P. Erdős, Some results on additive number theory, Proc. Amer. Math. Soc. 5 (1954), 847-853: the definition and the conjecture on p. 848, the theorem and its proof on pp. 851-852, the further remarks on pp. 852-853. The edition read is identified on the source card.

Read depth. Claims checked: the definition, the theorem and the key step were read clause by clause on the printed pages. The proof (pp. 851-852) was read but not checked step by step; the further remarks below have no proofs in the paper. Nothing here is independently reviewed.

Proof pointer

Pages 851-852. The squares form a pseudorational set of density $0$. If the set $S_1$ of integers $x^2+y^2$ is not pseudorational, the squares already give the example. Otherwise (the paper reports that Ruchte proved $S_1$ pseudorational) $S_1$ has density $0$ and $S_1+S_1$ contains every integer by the four-squares theorem, so the key step applies to $A=S_1$. For the key step, take $n_1<n_2<\cdots$ with $n_{k+1}/n_k\to\infty$ and $n_kN(A,n_{k+1})/n_{k+1}\to0$ (display (9)), and let $B$ consist of the elements of $A$ in the intervals $(n_{2k-1},n_{2k})$; displays (10)-(12) show that $N(B+B,n_{2k+1})/n_{2k+1}\to0$ while $N(B+B,n_{2k})=n_{2k}-o(n_{2k})$. The paper adds, with the proof left to the reader, that the integers of the form $x^2+y^2$ in the intervals $(2^{2^{2k}},2^{2^{2k+1}})$ already give such a $B$.

Further remarks in the paper

Stated without proof on pp. 852-853. For a sequence $S$ of density $0$, let $u^{(k)}_1,\ldots,u^{(k)}_{s_k}$ be the residues modulo $k!$ that contain an element of $S$; $S$ is pseudorational if and only if $s_k/k!\to0$. If this system of residues contains a branching subsystem (an infinite $k_1<k_2<\cdots$ and $2^r$ residues modulo $k_r!$, each congruent modulo $k_r!$ to two of the residues modulo $k_{r+1}!$), then some pseudorational $B$ of density $0$ has $S+B$ of upper density $1$ and lower density $0$; if it contains none, $S$ is pseudorational and $S+B$ is pseudorational for every pseudorational $B$. The powers $2^k$ give a branching subsystem and the factorials $k!$ do not. Finally, for almost every sequence $S$ (in the binary-digit measure), $S+B$ has a density for every $B$.

Dependencies

R. C. Buck, Amer. J. Math. 68 (1946), 560-580, and E. F. Buck and R. C. Buck, ibid. 69, 413-420, for the setting; the four-squares theorem; Ruchte's result that the sums of two squares form a pseudorational set, which the paper does not cite further and says follows from the characterization of sums of two squares.

Bears on

No Erdős problem in the corpus is recorded as concerning this result.