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Source. N. Hegyvári, On complete sequences, Ann. Univ. Sci. Budapest. Eötvös Sect. Math. 34 (1991), 7--10, identified on the source card: the setting and the unnumbered Theorem on p. 7, the proof on pp. 8--9, with Lemma 1 (p. 8) and Lemma 2 (p. 9).

Read depth. Claims checked: the definitions, the conjectures of p. 7, the Theorem and Lemma 2 were read clause by clause on the page images. The proof (pp. 8--9) was read through but not checked step by step. Nothing here is independently reviewed.

Statement

Setting (p. 7). A sequence $A=\{a_1<a_2<\cdots\}$ is "called complete, if every sufficiently large integer can be expressed as the sum of distinct elements of $A$". $P(A)$ is the set of finite sums of distinct terms of $A$, so completeness means that $\mathbb N\setminus P(A)$ is finite. For $\alpha,\beta>0$ the paper writes

$$A_{\alpha\beta}=\{[\alpha],[\beta],\ldots,[2^n\alpha],[2^n\beta],\ldots\},$$

with $[x]$ the integer part, and for each fixed $\alpha$ sets

$$X_\alpha=\{\beta:A_{\alpha\beta}\text{ is incomplete}\},$$

a subset of $(0,\infty)$ (p. 8 writes its complement as $(0,\infty)-X_\alpha$). $\mu$ is Lebesgue measure. Completeness is for the set $A_{\alpha\beta}$ written as an increasing sequence, so each value is used at most once in a sum.

Theorem (p. 7, quoted). "$X_\alpha$ is measurable and either $\mu(X_\alpha)=0$ or $\mu(X_\alpha)=\infty$."

Context on p. 7. The paper places the Theorem below three conjectures, each implied by the one before it:

• Erdős and Graham's, from their 1980 monograph, p. 58: if $\alpha/\beta$ is irrational then $A_{\alpha\beta}$ is complete.
• A weaker form, posed by the paper: for every fixed $\alpha$ the set $X_\alpha$ is countable.
• An even weaker one: $\mu(X_\alpha)=0$.

The paper also restates, from the author's 1989 paper (Acta Math. Hung. 53, 149--154), that the Erdős--Graham conjecture holds when $\alpha$ is a finite dyadic fraction and $\beta$ an infinite one, and the stronger conjecture stated there: if $\beta/\alpha\ne2^m$ ($m\in\mathbb Z$) and $\alpha$ is an infinite dyadic fraction, then $A_{\alpha\beta}$ is complete. The Theorem proves neither the countability nor the measure-zero form; it shows that if the measure-zero form fails for some $\alpha$, then $X_\alpha$ has infinite measure.

Proof pointer

Pp. 8--9. The paper reduces to $\alpha\ge1$ ("It is clear", p. 8) and uses the binary digits of $\alpha$, with which $a_{n+1}=2a_n$ or $2a_n+1$ for $a_n=[2^n\alpha]$. Measurability is the main step. With $X'=(0,\infty)-X_\alpha$ the set of $\beta$ giving completeness and $M=\{2^m\alpha:m\in\mathbb Z,\ 2^m\alpha\ge1\}$, the paper shows that $X'-M$ is open: for $\beta$ in it, every integer from some $k$ on is representable, and Lemma 1 transfers completeness to every $\gamma$ close enough to $\beta$, because the first terms $[2^n\gamma]$ agree with $[2^n\beta]$ up to an index where the gap condition of the lemma holds. The paper says this makes $X'$ open as well; for measurability it suffices that $X'$ differs from the open set $X'-M$ by a subset of the countable set $M$ (an observation of this page). The dichotomy comes from Lemma 2 (p. 9): if $A_{\alpha,\delta}$ is incomplete then so is $A_{\alpha,2\delta}$, since $A_{\alpha,2\delta}\subset A_{\alpha,\delta}$. Hence $2X_\alpha\subset X_\alpha$, so $\mu(X_\alpha)\ge\mu(2X_\alpha)=2\mu(X_\alpha)$, which forces $\mu(X_\alpha)$ to be $0$ or $\infty$. The paper's set $2X_\alpha$ is printed as $\{2\delta:\delta\in X\}$, with $X$ in place of $X_\alpha$.

Dependencies

Lemma 1 (p. 8) and Lemma 2 (p. 9) of the same paper; nothing from other sources.

Bears on

• Problem 354: the first question asks whether the doubling floor sequences of $\alpha$ and $\beta$ are complete whenever $\alpha/\beta$ is irrational. The Theorem is a measure-theoretic statement about the exceptional set for a fixed $\alpha$, in the paper's set reading of completeness: $X_\alpha$ is Lebesgue measurable with measure $0$ or $\infty$. It does not say which alternative holds, decides no individual pair, and concerns base $2$ only, not the second question's bases $\gamma\in(1,2)$.