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Source. S. Korsky, A resolution of the de Bruijn--Erdős consecutive-gap problem, arXiv:2609.07196v2, Lemma 6.3 (p. 12) 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

Notation as on the Lemma 6.2 page: $\Delta_t(x,D)=N_t((x,x+D/t])-D$, $Z_t(D)=\int_{\mathbb T}(\Delta_t)_+$, hypothesis (6.1) with constant $A\ge1$, and the $r$-spans $S_i(t)$ of $P_t$.

Statement (Lemma 6.3, p. 12)

Under (6.1), for every sufficiently large $t$,

$$Z_t(r)\ \le\ A .$$

Proof

Let $n=\lfloor t\rfloor$ and let $y_1,\ldots,y_n$ be the points of $P_t$ in cyclic order, indices mod $n$; let $S_{i-r}(t)=y_i-y_{i-r}$ be the $r$-span ending at $y_i$ (the clockwise distance from $y_{i-r}$ to $y_i$), and take $t$ large enough that $r<n$ and every span is shorter than $1$.

Two decompositions. For all $x$ outside the finite set of endpoints,

$$N_t\bigl((x,x+r/t]\bigr)=\sum_{i=1}^n\mathbf 1_{(y_i-r/t,\,y_i]}(x),$$

since $y_i\in(x,x+r/t]$ exactly when $x\in[y_i-r/t,y_i)$. On the other hand, the $r$-span arcs $(y_{i-r},y_i]$ cover every point of the circle, apart from endpoints, exactly $r$ times: $x$ lies in $(y_{i-r},y_i]$ exactly when $y_i$ is one of the $r$ points following $x$. So

$$r=\sum_{i=1}^n\mathbf 1_{(y_{i-r},\,y_i]}(x).$$

Pairing arcs with the same right endpoint. Subtracting,

$$\Delta_t(x,r)=\sum_{i=1}^n\Bigl(\mathbf 1_{(y_i-r/t,\,y_i]}(x) -\mathbf 1_{(y_{i-r},\,y_i]}(x)\Bigr).$$

The two arcs in the $i$-th term share the right endpoint $y_i$ and have lengths $r/t$ and $S_{i-r}(t)$, so their indicators differ on an arc of length $|S_{i-r}(t)-r/t|$. By the triangle inequality and the bound $\sum_i|S_i(t)-r/t|\le2A+r(t-n)/t$ from the proof of Lemma 6.1,

$$\int_{\mathbb T}\bigl|\Delta_t(x,r)\bigr|\,dx\ \le\ \sum_{i=1}^n \Bigl|S_{i-r}(t)-\frac rt\Bigr|\ \le\ 2A+\frac{r(t-n)}t .$$

The mean. Each point lies in $(x,x+r/t]$ for $x$ in a set of measure $r/t$, so

$$\int_{\mathbb T}\Delta_t(x,r)\,dx=\frac{rn}t-r=-\frac{r(t-n)}t .$$

Conclusion. The positive part of a function is half the sum of its absolute value and the function itself, so

$$Z_t(r)=\frac12\Bigl(\int_{\mathbb T}|\Delta_t(x,r)|\,dx +\int_{\mathbb T}\Delta_t(x,r)\,dx\Bigr) \ \le\ \frac12\Bigl(2A+\frac{r(t-n)}t-\frac{r(t-n)}t\Bigr)=A .$$

Role in the argument

This is the starting value $z_t(r)=Z_t(r)/r\le A/r=\theta^2$ of the scale iteration in Proposition 6.4.