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Source. S. Korsky, A resolution of the de Bruijn--Erdős consecutive-gap problem, arXiv:2609.07196v2, Section 3 and Proposition 3.1 (pp. 6--7) of the retained PDF, read in the canonical conversion and checked against the text layer at the displayed constants; held by its library card, Korsky 2026, resolution. The comparison lemma is reconstructed on the Lemma 2.1 page, whose definitions and hypothesis (2.1) are used here.

Standing. Author-recorded reconstruction; not an independent review; changes no status and assigns no tier. The source is an unrefereed preprint. The implied constants of the source's $O(\cdot)$ terms are absolute; the reconstruction keeps them implicit as the source does.

Statement (Proposition 3.1, p. 6)

There are absolute constants $C_0,C_1>0$ with the following property. Suppose that (2.1) holds for all sufficiently large $t$, with $A\ge1$ and $r\ge C_0A$. Put

$$\Lambda=\log(r/A),\qquad S=\frac{\sqrt{Ar}}{\Lambda^2}.$$

Then, for every sufficiently large $t$,

$$\sup_{x\in\mathbb T}\ \sup_{0\le D\le S}\ \Bigl|N_t\bigl((x,x+D/t]\bigr)-D \Bigr|\ \le\ 3A+\frac{C_1A}\Lambda . \tag{3.1}$$

The time threshold may depend on $r$, $A$ and the sequence, but not on $x$ or $D$.

Proof

Put $\theta=\sqrt{A/r}$ and $K=\sqrt{Ar}$, so $A/r=\theta^2$, $K/r=\theta$ and $A/K=\theta$. Take $C_0$ large; in particular $\theta\le1/12$, which also gives $K<r-A$ (equivalent to $\theta<1-\theta^2$), and $\Lambda\ge1$.

Terminal bounds (3.2). An interval $(x,x+(r-A)/t]$ contains at most $r$ points of $P_t$: if it contained $r+1$, the first and the last of them, in cyclic order, would be $r$ places apart in $P_t$ (all points between them lie in the interval), and the $r$-span from the first to the last would have length less than $(r-A)/t\le(r-a_t)/t$, contrary to (2.1). An interval $I=(x,x+(r+A)/t]$ contains at least $r$ points: let $p_0$ be the last point of $P_t$ at or before $x$, and $p_1,\ldots,p_r$ the next $r$ points; then $p_1>x$, and the span bound gives $p_r\le p_0+(r+b_t)/t\le x+(r+A)/t$, so $p_1,\ldots,p_r\in I$. Hence, for all sufficiently large $t$,

$$U_t(r-A)\ \le\ \frac r{r-A},\qquad V_t(r+A)\ \ge\ \frac r{r+A}. \tag{3.2}$$

Scale chains. For the upper estimate take the scales $K=D_0<D_1<\cdots<D_{h_U}=r-A$ with $D_{i+1}=2D_i$ except that the last step is shortened to end at $r-A$; for the lower estimate the same with terminal scale $r+A$ and $h_V$ steps. Adjacent scales $D<E$ satisfy $K\le D\le E\le2D$, and

$$h_U,\ h_V\ \le\ \log_2\frac{r+A}K+1\ \le\ \frac{\Lambda}{2\log2}+2 \ \le\ C\Lambda$$

for an absolute $C$, since $(r+A)/K\le2r/K=2\sqrt{r/A}$.

One comparison step. For adjacent scales $D<E$ apply Lemma 2.1 with $k=\lceil D/K\rceil$. Since $D\ge K$, $D/K\le k\le2D/K$, so

$$q=\frac E{kr}\ \le\ \frac{2D}{(D/K)r}=\frac{2K}r=2\theta,\qquad \frac{3kA}D\ \le\ \frac{6A}K=6\theta,\qquad \frac{kA}D\le2\theta ,$$

and $q\le2\theta\le1/6<1$. The upper multiplier in (2.2) is at most $(1+6\theta)(1+2\theta)=1+8\theta+12\theta^2\le1+9\theta$, using $\theta\le1/12$; the lower multiplier in (2.3) is at least $1-2\theta-2\theta\cdot3=1-8\theta>0$. So

$$U_t(D)\le(1+9\theta)\,U_{(1+q)t}(E),\qquad V_t(D)\ge(1-8\theta)\,V_{(1-q)t}(E).$$

Iteration to scale $K$. Starting from $U_t(K)$ and applying the upper step along the chain, with the time multiplied by $1+q_i$ at the $i$-th step, and ending with (3.2) at the terminal time, gives

$$U_t(K)\ \le\ \frac r{r-A}(1+9\theta)^{h_U},\qquad V_t(K)\ \ge\ \frac r{r+A}(1-8\theta)^{h_V},$$

for all $t$ large enough that every one of the finitely many comparisons and both terminal bounds apply; the terminal times are $t$ times fixed finite products of factors $1\pm q_i$. Now $(1+9\theta)^{h_U}\le\exp(9C\theta\Lambda)=1+O(\theta\Lambda)$ because $\theta\Lambda=\sqrt{A/r}\log(r/A)\to0$ as $r/A\to\infty$; $r/(r-A)=1/(1-\theta^2)=1+O(\theta^2)$; $(1-8\theta)^{h_V}\ge1-8\theta h_V\ge1-O(\theta\Lambda)$ by Bernoulli's inequality; and $r/(r+A)\ge1-\theta^2$. Hence, for all sufficiently large $t$,

$$U_t(K)\ \le\ 1+O(\theta\Lambda),\qquad V_t(K)\ \ge\ 1-O(\theta\Lambda), \tag{3.3}$$

with absolute implied constants.

Descent to short intervals. Apply Lemma 2.1 once more with $E=K$ and $k=1$, so $q=K/r=\theta$. For $D>0$ and $I=(x,x+D/t]$, the proof of (2.2) before the supremum gives $N_t(I)\le|J|\,E\,U_{t_+}(E)/\ell$ with $|J|\le(D+3A)/t$ and $E/\ell=t_+=(1+\theta)t$, and the proof of (2.3) gives $N_t(I)\ge|J|\,E\,V_{t_-}(E)/\ell$ with $|J|\ge(D/t-A/t-2A/t_-)_+$ and $E/\ell=t_-=(1-\theta)t$. Thus

$$N_t(I)\ \le\ (D+3A)(1+\theta)\,U_{(1+\theta)t}(K),\qquad N_t(I)\ \ge\ \bigl(D(1-\theta)-(3-\theta)A\bigr)_+V_{(1-\theta)t}(K).$$

With (3.3) at the times $(1\pm\theta)t$ and $\theta\le\theta\Lambda$, the upper bound is $D+3A+O((D+A)\theta\Lambda)$. For the lower bound, if $D(1-\theta)-(3-\theta)A\le0$ then $D\le(3-\theta)A/(1-\theta)=3A+O(A\theta)$ and the trivial $N_t(I)\ge0$ gives $N_t(I)-D\ge-3A-O(A\theta)$; otherwise $N_t(I)\ge(D(1-\theta)-(3-\theta)A)(1-O(\theta\Lambda))\ge D-3A-O((D+A)\theta\Lambda)$. In all cases

$$\bigl|N_t(I)-D\bigr|\ \le\ 3A+O\bigl((D+A)\theta\Lambda\bigr). \tag{3.4}$$

The choice of $S$. For $0<D\le S=K/\Lambda^2$,

$$D\theta\Lambda\ \le\ \frac{K\theta}\Lambda=\frac A\Lambda,\qquad A\theta\Lambda=\frac A\Lambda\cdot\theta\Lambda^2=O\Bigl(\frac A\Lambda\Bigr),$$

the second because $\theta\Lambda^2=\sqrt{A/r}\log^2(r/A)\to0$ as $r/A\to\infty$ and is bounded once $C_0$ is large. So (3.4) becomes (3.1) after fixing absolute $C_0$ and $C_1$. The case $D=0$ is trivial.

Uniformity of the threshold. The chains use finitely many predetermined comparison times, and the last comparison uses the times $(1\pm\theta)t$, none of which depends on $x$ or $D$; Lemma 2.1's threshold is uniform for $D$ in the bounded range $(0,S]$. So one time threshold serves every $x$ and every $0\le D\le S$.

Role in the argument

Under the ratio hypothesis of Section 5, (2.1) holds with $A=(\log r)/100$; (3.1) then feeds Lemma 4.2, which reads the points of one short interval as a finite list whose prefix discrepancies are all at most $3A+C_1A/\Lambda$. The assembly is on the Theorem 1.1 page.