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Source. Y. Yu and K. Chen, Erdős Problem 354(i): Strong Completeness of Two Dyadic Floor Sequences, manuscript of 13 September 2026, Lemma 2.2 with its display (2.2) and the consequence stated after it, physical p. 3, in the seventeen-page PDF held by its library source card, Yu and Chen (2026).

Standing. This is an author-recorded reconstruction. It is not an independent review, changes no status and assigns no tier.

Definitions

For a finite set $W\subseteq\mathbb Z$ with at least two elements, listed as $w_0<w_1<\cdots<w_m$,

$$\operatorname{span}(W)=w_m-w_0,\qquad \operatorname{gap}(W)=\max_{0\le j<m}(w_{j+1}-w_j).$$

A gap of $k$ means that at most $k-1$ consecutive integers of the interval $[w_0,w_m]$ are missing from $W$. The hull of $W$ is the real interval $[w_0,w_m]$. For an integer $c$, $W+c$ is the translate.

Statement

Lemma 2.2. If $\operatorname{span}(W)\ge c>0$ and $\operatorname{gap}(W)\le k$, then

$$\operatorname{gap}\bigl(W\cup(W+c)\bigr)\le k,\qquad \operatorname{span}\bigl(W\cup(W+c)\bigr)=\operatorname{span}(W)+c.$$

Consequence. Let $c_1\le c_2\le\cdots$ be positive integers with $c_{i+1}\le2c_i$ for every $i$, and let $W_0$ have $\operatorname{gap}(W_0)\le k$ and $\operatorname{span}(W_0)\ge c_1$. Put $W_i=W_{i-1}\cup(W_{i-1}+c_i)$. Then every $W_i$ has gap at most $k$, $\min W_i=\min W_0$, and $\operatorname{span}(W_i)=\operatorname{span}(W_0)+c_1+\cdots+c_i$.

Proof

Write $w_0=\min W$ and $w_m=\max W$. The hull of $W+c$ is $[w_0+c,w_m+c]$. Since $c\le\operatorname{span}(W)=w_m-w_0$, we have $w_0+c\le w_m$: the two hulls intersect or touch, and the union of the hulls is the single interval $[w_0,w_m+c]$. The minimum of $W\cup(W+c)$ is $w_0$ and the maximum is $w_m+c$, which gives the span identity. All four hull endpoints $w_0$, $w_m$, $w_0+c$, $w_m+c$ belong to the union.

Let $w<w'$ be consecutive elements of $W\cup(W+c)$; we show $w'-w\le k$. The open interval $(w,w')$ contains no element of the union, hence no hull endpoint. There are three cases.

If $w'\le w_m$, both points lie in the hull of $W$. Let $w_-$ be the largest element of $W$ with $w_-\le w$ (it exists because $w\ge w_0$) and $w_+$ the smallest element of $W$ with $w_+\ge w'$ (it exists because $w'\le w_m$). No element of $W$ lies in $(w_-,w]$, by the choice of $w_-$; none lies in $(w,w')$, by consecutiveness in the union; none lies in $[w',w_+)$, by the choice of $w_+$. So $w_-$ and $w_+$ are consecutive in $W$, and $w'-w\le w_+-w_-\le k$.

If $w\ge w_0+c$, both points lie in the hull of $W+c$, and the same argument applied to $W+c$ (whose gap is also at most $k$) gives $w'-w\le k$.

Otherwise $w<w_0+c$ and $w'>w_m$. Since $w_0+c\le w_m<w'$, the union point $w_0+c$ lies in $(w,w')$, contradicting consecutiveness. So this case does not occur.

For the consequence, induct on $i$. Given $\operatorname{gap}(W_{i-1})\le k$ and $\operatorname{span}(W_{i-1})\ge c_i$, the lemma gives $\operatorname{gap}(W_i)\le k$, $\min W_i=\min W_{i-1}$ and $\operatorname{span}(W_i)=\operatorname{span}(W_{i-1})+c_i\ge2c_i\ge c_{i+1}$, which is the hypothesis for the next step. The base case is the assumption on $W_0$.