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Source. S. Korsky, A resolution of the de Bruijn--Erdős consecutive-gap problem, arXiv:2609.07196v2, Section 6, hypothesis (6.1) and Lemma 6.1 (p. 10) of the retained PDF, read in the canonical conversion and checked against the text layer; held by its library card, Korsky 2026, resolution.

Standing. Author-recorded reconstruction; not an independent review; changes no status and assigns no tier. The source is an unrefereed preprint.

Definitions

Points, $P_t$, $r$-spans $S_i(t)$ of $P_t$ and the moves by $kr$ places are as on the Lemma 2.1 page. $M_n^{(r)}$ and $m_n^{(r)}$ are the largest and smallest $r$-spans at the integer time $n$. For $p\in P_t$ let $L_{t,k}(p)$ be the clockwise distance from $p$ to the point $kr$ places after it in the cyclic order of $P_t$; here $r$ and $k$ are fixed and $t$ is large enough that $kr<|P_t|$.

Hypothesis (6.1). A number $A\ge1$ is fixed, and one of the two alternatives

$$nM_n^{(r)}-r\ \le\ A\qquad\text{or}\qquad r-nm_n^{(r)}\ \le\ A$$

holds for every sufficiently large integer $n$; which alternative holds is fixed throughout.

Statement (Lemma 6.1, p. 10)

Under either alternative in (6.1), for all sufficiently large $t$,

$$\sum_{p\in P_t}\Bigl|L_{t,k}(p)-\frac{kr}t\Bigr|\ \le\ 2kA+\frac{kr}t . \tag{6.2}$$

The estimate is uniform when $t$ ranges over a fixed multiplicative interval.

Proof

The zero-sum identity at an integer time. Let $n$ be a large integer and $S_1,\ldots,S_n$ the $r$-spans of $P_n$, one from each point. Each gap lies in exactly $r$ spans, so $\sum_iS_i=r$ and

$$\sum_{i=1}^n\Bigl(S_i-\frac rn\Bigr)=0 .$$

Under the first alternative every summand is at most $M_n^{(r)}-r/n\le A/n$, so the sum of the positive summands is at most $n\cdot A/n=A$; by the zero-sum identity the sum of the negative parts equals the sum of the positive parts, so

$$\sum_{i=1}^n\Bigl|S_i-\frac rn\Bigr|\ \le\ 2A . \tag{6.3}$$

Under the second alternative every summand is at least $m_n^{(r)}-r/n\ge-A/n$, the negative parts sum to at most $A$, and the same identity gives (6.3).

From $r/n$ to $r/t$. Let $n=\lfloor t\rfloor$, so $P_t=P_n$ and the spans of $P_t$ are $S_1,\ldots,S_n$. Replacing $r/n$ by $r/t$ in (6.3) changes the left side by at most

$$n\Bigl|\frac rn-\frac rt\Bigr|=\frac{r(t-n)}t\ \le\ \frac rt ,$$

so $\sum_i|S_i-r/t|\le2A+r(t-n)/t\le2A+r/t$.

From $r$-spans to $kr$-spans. If $p$ is the $i$-th point of $P_t$, then $L_{t,k}(p)=S_i+S_{i+r}+\cdots+S_{i+(k-1)r}$, the sum of $k$ consecutive $r$-spans starting at $p$, and by the triangle inequality

$$\Bigl|L_{t,k}(p)-\frac{kr}t\Bigr|\ \le\ \sum_{j=0}^{k-1} \Bigl|S_{i+jr}-\frac rt\Bigr| .$$

Summing over $i$, each $S_m$ occurs once for each of the $k$ values of $j$, so

$$\sum_{p\in P_t}\Bigl|L_{t,k}(p)-\frac{kr}t\Bigr|\ \le\ k\sum_{m=1}^n \Bigl|S_m-\frac rt\Bigr|\ \le\ 2kA+\frac{kr}t ,$$

which is (6.2). The only requirement on $t$ is that (6.1) hold at $\lfloor t\rfloor$ and $kr<n$, so the bound is uniform for $t$ in any fixed multiplicative interval once its lower end is large.

Role in the argument

The one-sided hypothesis gives no pointwise bound in the other direction, so Lemma 2.1 does not apply; (6.2) is the substitute that Lemma 6.2 uses to control, in spatial $L^1$, the error of the cyclic-walk comparison, and its intermediate bound $\sum_i|S_i-r/t|\le2A+r(t-n)/t$ is what Lemma 6.3 uses at the scale $r$.