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Statement

Setting (p. 115). For reals $\alpha,\beta>0$ the paper writes $A_{\alpha\beta}=\{[2^n\alpha],[2^n\beta]\mid n\in\mathbb N\}$ and, for a sequence $A$, $P(A)={\sum\varepsilon_ia_i\mid a_i\in A;\ \varepsilon_i=0 \text{ or } 1}$, the set of its subset sums. $A$ is complete when every sufficiently large integer lies in $P(A)$. The printed definition of the weaker notion reads: "An infinite sequence of integers is said to be subcomplete if it contains an infinite arithmetic progression." The paper uses it for $P(A)$: its remark on p. 116 that $A_\alpha$ is subcomplete if and only if $\alpha$ is a finite dyadic fraction concerns $P(A_\alpha)$, since $\{[2^n\alpha]\}$ itself grows geometrically, and the proof of Theorem 1 shows that $P(A_{\alpha\beta})$ contains no infinite arithmetic progression.

Theorem 1 (p. 116, quoted). "There are continuum many pairs of $(\alpha,\beta)$ for which $A_{\alpha\beta}$ is not subcomplete."

It sharpens Theorem A, recalled on p. 116 from the author's earlier paper (his reference [3], Acta Math. Hungar. 53, 149--154): there are continuum many pairs $(\alpha,\beta)$ for which $A_{\alpha\beta}$ is not complete.

The pairs built (pp. 117--118). Every pair in the proof has $\alpha\ge2$ and $\beta=2^n\alpha$ for a natural number $n$, the setting of Lemma 1; the continuum comes from the free choice of infinitely many binary digits of $\alpha$. The ratio $\beta/\alpha$ is therefore a power of $2$ in every case.

Source. N. Hegyvári, On sumset of certain sets, Publ. Math. Debrecen 45 (1994), no. 1--2, 115--122: the definitions on p. 115, Theorem A and the statement on p. 116, the proof in Section 3, pp. 117--118.

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

Proof pointer

Section 3, pp. 117--118. With $a_n=[2^n\alpha]$ and $x_n=a_0+\cdots+a_n+1$, Lemma 1 puts every $x_n$ outside $P(A_{\alpha\beta})$, so it suffices to choose continuum many $\alpha$ for which the $x_m$ meet every residue class modulo every $n$. The digits of $\alpha$ are fixed block by block, the moduli taken in nondecreasing order. A first lemma (the first Lemma 2, p. 117) writes $x_{N+m}-x_{N-1}$ through $a_N$ and the digits $\varepsilon_{N+1}(\alpha),\ldots,\varepsilon_{N+m}(\alpha)$, using $a_{n+1}=2a_n+\varepsilon_{n+1}(\alpha)$. Taking $m=n^3$, the proof finds a $u$ prime to $n$ with $2^x-1\equiv u\pmod n$ for infinitely many $x$, and sets suitable digits of the block to $1$ so that $x_{N+m}$ falls in the wanted class. The digit $\varepsilon_N(\alpha)$ before each block is free, which gives continuum many $\alpha$.

Bears on

• Problem 354: the problem assumes $\alpha/\beta$ irrational. Every pair the proof builds has $\beta=2^n\alpha$, a rational ratio, so the theorem decides no case of the problem; it shows only that for such ratios completeness can fail in the stronger sense that no infinite arithmetic progression lies in the subset sums.