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Statement

Setting (p. 117). For $\alpha>0$ put $a_n=[2^n\alpha]$ and $x_n=a_0+a_1+\cdots+a_n+1$. $A_{\alpha\beta}$ and $P(\cdot)$ are as in Theorem 1.

Lemma 1 (p. 117, quoted). "Let $\alpha\ge2$ and $\beta=2^n\alpha$ ($n\in\mathbb N$). Then $x_n\notin P(A_{\alpha\beta})$ for every $n\in\mathbb N$."

The letter $n$ serves twice as printed: once for the fixed exponent in $\beta=2^n\alpha$ and once for the index of $x_n$, which ranges over all of $\mathbb N$. The paper does not say whether $0\in\mathbb N$. It writes $A_{\alpha\beta}$ with set braces and does not say whether a value $[2^j\beta]=[2^{j+n}\alpha]$ is counted once or twice in $P(A_{\alpha\beta})$. Since $x_n\to\infty$, the lemma implies that $A_{\alpha\beta}$ is not complete.

Source. N. Hegyvári, On sumset of certain sets, Publ. Math. Debrecen 45 (1994), no. 1--2, 115--122, p. 117. The paper gives no proof; it refers to the proof of Theorem 2 in the author's earlier paper (its reference [3]: N. Hegyvári, Some remarks on a problem of Erdős and Graham, Acta Math. Hungar. 53 (1989), 149--154).

Read depth. Claims checked: the statement was read clause by clause on the journal print. No proof is given in this paper, and the 1989 proof was not read.

Proof pointer

None in this paper (p. 117: "See the proof of Theorem 2 in [3]."). The lemma is the input to Theorem 1.

Bears on

• Problem 354: the lemma concerns $\beta/\alpha=2^n$, a rational ratio outside the problem's hypothesis that $\alpha/\beta$ be irrational, so it decides no case of the problem. It is the incompleteness for $\alpha\ge2$ and $\beta=2^k\alpha$ that the problem page records from the 1989 paper, here restated without proof.