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Statement

Notation (p. 101): $A(x)$ and $A_3(x)$ count the elements of $A$ and of the threefold sumset $3A=A+A+A$ below $x$; a basis is a basis of some order $h$, a set of natural numbers such that every sufficiently large integer is a sum of at most $h$ of its elements.

Theorem 2 (p. 102, quoted). "If $A$ is a basis and $A(x)=o(x)$, then $A_3(3x)/A(x)\to\infty$."

The paper calls this "something more modest" than its Conjecture 1 (p. 102), which asks the same of $A_2(2x)/A(x)$. Both compare a sumset counted up to a multiple of $x$ with $A$ counted up to $x$; neither speaks of $A_2(x)/A(x)$, the ratio of the Erdős--Graham conjecture that Theorem 1 disproves.

Source. I. Z. Ruzsa and S. Turjányi, A note on additive bases of integers, Publ. Math. Debrecen 32 (1985), 101--104; the statement in Section 3 on p. 102 and its proof in Section 5 on p. 103, read on the page images of the copy identified on the source card.

Read depth. Claims checked: the statement was read clause by clause on the page image. The proof was read but not checked step by step. Nothing here is independently reviewed.

Proof pointer

P. 103, Section 5. Let $A$ be a basis of order $h$ and put $X=A\cap[1,x]$, so that $\lvert X\rvert=A(x)$, $\lvert3X\rvert\le A_3(3x)$, and $hX$ contains all but a bounded number of the integers below $x$, giving $\lvert hX\rvert\ge x-c$ with $c$ independent of $x$. Theorem 3 with $s=\lvert3X\rvert/\lvert X\rvert$ and $k=h$ gives $x-c\le A(x)\,(A_3(3x)/A(x))^h$. The final display prints the resulting lower bound for $A_3(3x)/A(x)$ as $(x-c)/A(x)$ in parentheses with no exponent; the preceding inequality gives it with the exponent $1/h$, which still tends to infinity because $A(x)=o(x)$ (an observation of this page).

Dependencies

Theorem 3 of the same paper.

Bears on

• Problem 337: the problem asks whether $\lvert(A+A)\cap\{1,\ldots,N\}\rvert/\lvert A\cap\{1,\ldots,N\}\rvert\to\infty$ for every basis $A$ with $\lvert A\cap\{1,\ldots,N\}\rvert=o(N)$. The theorem does not decide that question, which Theorem 1 answers in the negative; it proves a variant with the threefold sumset counted up to $3x$ in place of the twofold sumset counted up to $x$.