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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: Lemma 1 and its proof on p. 8.

Read depth. Claims checked: the statement was read clause by clause on the page image, and the proof (p. 8) was followed step by step. Nothing here is independently reviewed.

Statement

Notation as on the Theorem's page: $A_{\alpha\beta}$ is the set of the integer parts $[2^n\alpha]$ and $[2^n\beta]$, $n\ge0$, and $P(A_{\alpha\beta})$ is the set of its finite sums of distinct elements. The lemma sits in the proof of the Theorem after the reduction to $\alpha\ge1$ (p. 8).

Lemma 1 (p. 8, quoted). "Let $A_\alpha=\{[\alpha],\ldots,[2^n\alpha],\ldots\}=\{a_0<a_1<\ldots\}$ and $A_\beta=\{[\beta],\ldots,[2^n\beta],\ldots\}=\{b_0<b_1<\ldots\}$. If there exist $p$ and $i$ such that $[k,a_p]\subset P(A_{\alpha\beta})$ and

$$(1)\qquad k<\min\{a_p-b_i,b_i-a_{p-1}\}$$

then $A_{\alpha\beta}$ is complete."

Here $[k,a_p]$ is the set of integers from $k$ to $a_p$. Condition (1) places $b_i$ strictly between $a_{p-1}$ and $a_p$, more than $k$ from each. The proof establishes the explicit form: every integer $m\ge k$ lies in $P(A_{\alpha\beta})$. It uses $k\ge1$ (in the steps $2k>1$ and $2k-1\ge k$) and the doubling relations $a_{n+1}\in\{2a_n,2a_n+1\}$ and $b_{n+1}\in\{2b_n,2b_n+1\}$, which hold for the integer parts of $2^n x$.

Proof pointer

P. 8. An induction on $p$ and $i$ together: from $[k,a_p]\subset P(A_{\alpha\beta})$ and (1) the paper shows $[k,a_{p+1}]\subset P(A_{\alpha\beta})$ and that (1) still holds with $p+1$, $i+1$ in place of $p$, $i$. Adding $a_p$ to the sums in $[k,a_p)$ covers the integers from $k+a_p$ up to $2a_p$, exclusive; adding $b_i$ to the sums in $[k,b_i)$, which use neither $b_i$ nor $a_p$, covers the gap between $a_p$ and $k+a_p$; adding both $a_p$ and $b_i$ to them reaches $2a_p$. Since $a_{p+1}$ is $2a_p$ or $2a_p+1$ and is itself a term, this gives $[k,a_{p+1}]$. Condition (1) propagates because each of the two gaps at least doubles, less one, at each step.

Dependencies

None beyond the base-two relation $a_{n+1}=2a_n+\varepsilon_{n+1}(\alpha)$ recorded on p. 8.

Bears on

• Problem 354: the lemma gives a sufficient condition for completeness of the base-$2$ sequences of the first question, one finite check: an interval of subset sums together with a term $b_i$ placed as in (1). The paper uses it to prove that completeness persists under small changes of $\beta$ (pp. 8--9), for the Theorem. It does not by itself decide any pair $(\alpha,\beta)$.