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Source. S. Korsky, A resolution of the de Bruijn--Erdős consecutive-gap problem, arXiv:2609.07196v2, Section 2 and Lemma 2.1 (pp. 4--5) of the retained PDF, read in the canonical conversion beside the PDF and checked against the text layer at the displayed constants; 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, AI-assisted preprint registered as a proof claim for Problem 1221.

Definitions

Let $(x_n)_{n\ge1}$ be distinct points of $\mathbb T=\mathbb R/\mathbb Z$ and fix $r\in\mathbb N$. For real $t\ge1$ put

$$P_t=\{x_1,\ldots,x_{\lfloor t\rfloor}\},\qquad N_t(I)=\#(P_t\cap I).$$

The sets are nested: $P_s\subseteq P_t$ for $s\le t$. The $r$-spans of $P_t$ are the clockwise distances from each point of $P_t$ to the point $r$ places after it in cyclic order, written $S_i(t)$; they are the $r$-spans at time $\lfloor t\rfloor$. Intervals are oriented half-open arcs $(x,x+\ell]$ of length $\ell<1$; lifts to $\mathbb R$ are used to translate endpoints and to measure displacements. For $D>0$,

$$U_t(D)=\frac1D\sup_{x\in\mathbb T}N_t\bigl((x,x+D/t]\bigr),\qquad V_t(D)=\frac1D\inf_{x\in\mathbb T}N_t\bigl((x,x+D/t]\bigr),$$

so that $U_t(D)$ and $V_t(D)$ compare the largest and smallest counts in intervals of length $D/t$ with $D$, the count a perfectly spread set would give. Write $(z)_+=\max(z,0)$.

Hypothesis (2.1). A number $A\ge1$ is fixed, and for every sufficiently large $t$ there are $a_t,b_t\ge0$ with $a_t+b_t\le A$ such that every $r$-span of $P_t$ satisfies

$$\frac{r-a_t}t\ \le\ S_i(t)\ \le\ \frac{r+b_t}t .$$

Cyclic moves. For a time $s$ with $kr<|P_s|$, let $F_s$ be the map sending each point of $P_s$ to the point $kr$ places after it in the cyclic order of $P_s$, and $B_s=F_s^{-1}$ the move by $kr$ places backward. Both are bijections of $P_s$. The clockwise displacement of $F_s$ at a point is the sum of $k$ consecutive $r$-spans of $P_s$ (the $kr$ gaps after the point, grouped in $k$ runs of $r$), so under (2.1) it lies in

$$\Bigl[\frac{kr-ka_s}s,\ \frac{kr+kb_s}s\Bigr].$$

The counterclockwise displacement of $B_s$ at a point $p'$ is the clockwise displacement of $F_s$ at $B_s(p')$, so it lies in the same range. All times below are large enough that (2.1) holds, that $kr<|P_s|$, and that every interval used has length less than $1$.

Statement (Lemma 2.1, p. 4)

Fix $D,E>0$ and an integer $k\ge1$, and put $q=E/(kr)$. If $q<1$, then for all sufficiently large $t$,

$$U_t(D)\ \le\ \Bigl(1+\frac{3kA}D\Bigr)(1+q)\,U_{(1+q)t}(E), \tag{2.2}$$ $$V_t(D)\ \ge\ \Bigl(1-q-\frac{kA}D(3-q)\Bigr)_+V_{(1-q)t}(E). \tag{2.3}$$

For fixed $E$ and $k$ the time threshold can be chosen uniformly for $D$ in any bounded range.

Proof of (2.2)

Put $t_+=(1+q)t$ and $\ell=E/t_+$. For $0\le u\le\ell$ let $s=s(u)$ be the solution of

$$\frac{kr}t-\frac{kr}s=u,\qquad\text{that is,}\qquad s=\frac{t}{1-ut/(kr)} .$$

As $u$ runs from $0$ to $\ell$, $ut/(kr)$ runs from $0$ to $Et/(t_+kr)=q/(1+q)$, so $s$ runs from $t$ to $t/(1-q/(1+q))=t_+$; thus $t\le s\le t_+$.

The injection. Consider

$$T_u:\ P_t\xrightarrow{\ F_t\ }P_t\hookrightarrow P_s \xrightarrow{\ B_s\ }P_s\hookrightarrow P_{t_+}.$$

Each arrow is injective, so $T_u$ is an injection of $P_t$ into $P_{t_+}$. Its clockwise displacement at $p$ is the displacement of $F_t$ at $p$, in $[k(r-a_t)/t,k(r+b_t)/t]$, minus the counterclockwise displacement of $B_s$ at $F_t(p)$, in $[k(r-a_s)/s,k(r+b_s)/s]$. Using $kr/t-kr/s=u$, the displacement minus $u$ lies in

$$\Bigl[-\frac{ka_t}t-\frac{kb_s}s,\ \frac{kb_t}t+\frac{ka_s}s\Bigr] \ \subseteq\ \frac kt\bigl[-a_t-A,\ b_t+A\bigr],$$

because $s\ge t$ and $a_s,b_s\le A$.

Enlarging the interval. Let $I=(x,x+D/t]$. Extend $I$ to the left by $k(a_t+A)/t$ and to the right by $k(b_t+A)/t$, obtaining

$$J=\Bigl(x-\frac{k(a_t+A)}t,\ x+\frac Dt+\frac{k(b_t+A)}t\Bigr],\qquad |J|=\frac{D+k(a_t+b_t+2A)}t\ \le\ \frac{D+3kA}t .$$

If $p\in P_t\cap I$, then $T_u(p)-u$ lies within $k(a_t+A)/t$ to the left and $k(b_t+A)/t$ to the right of $p$ (the left bound strict in the sense that $T_u(p)-u\ge p-k(a_t+A)/t>x-k(a_t+A)/t$), so $T_u(p)\in J+u$. As $T_u$ is injective into $P_{t_+}$,

$$N_t(I)\ \le\ N_{t_+}(J+u)\qquad(0\le u\le\ell).$$

Averaging. Integrate over $u\in[0,\ell]$. For each point $p$ of $P_{t_+}$, the set of $u\in[0,\ell]$ with $p\in J+u$ has the same measure as the set of $v\in J$ with $p\in(v,v+\ell]$ (both are the set of $v=p-u$ in $J\cap[p-\ell,p)$, up to endpoints), so

$$N_t(I)\,\ell\ \le\ \int_0^\ell N_{t_+}(J+u)\,du =\int_JN_{t_+}\bigl((v,v+\ell]\bigr)\,dv\ \le\ |J|\,E\,U_{t_+}(E),$$

the last step because $\ell=E/t_+$, so each $N_{t_+}((v,v+\ell])\le E\,U_{t_+}(E)$ by the definition of $U$. Dividing by $D\ell=DE/t_+$,

$$\frac{N_t(I)}D\ \le\ \frac{|J|\,t_+}D\,U_{t_+}(E)\ \le\ \Bigl(1+\frac{3kA}D\Bigr)(1+q)\,U_{t_+}(E),$$

and the supremum over $x$ gives (2.2). The half-open endpoint conventions affect none of the integrals.

Proof of (2.3)

Put $t_-=(1-q)t$ and $\ell=E/t_-$. For $0\le u\le\ell$ let $s=s(u)$ solve

$$\frac{kr}s-\frac{kr}t=u,\qquad\text{that is,}\qquad s=\frac t{1+ut/(kr)};$$

as $u$ runs from $0$ to $\ell$, $ut/(kr)$ runs from $0$ to $Et/(t_-kr)=q/(1-q)$ and $s$ from $t$ down to $t(1-q)=t_-$, so $t_-\le s\le t$.

The injection. This time use

$$T'_u:\ P_{t_-}\hookrightarrow P_s\xrightarrow{\ B_s\ }P_s \hookrightarrow P_t\xrightarrow{\ F_t\ }P_t ,$$

an injection of $P_{t_-}$ into $P_t$. Its clockwise displacement is the displacement of $F_t$, in $[k(r-a_t)/t,k(r+b_t)/t]$, minus the counterclockwise displacement of $B_s$, in $[k(r-a_s)/s,k(r+b_s)/s]$. Using $kr/s-kr/t=u$, the displacement minus $(-u)$ lies in

$$\Bigl[-\frac{ka_t}t-\frac{kb_s}s,\ \frac{kb_t}t+\frac{ka_s}s\Bigr] \ \subseteq\ \Bigl[-\frac{ka_t}t-\frac{kA}{t_-},\ \frac{kb_t}t+\frac{kA}{t_-} \Bigr],$$

because $s\ge t_-$.

Shrinking the interval. Let $I=(x,x+D/t]$. Move its left endpoint to the right by $ka_t/t+kA/t_-$ and its right endpoint to the left by $kb_t/t+kA/t_-$; call the result $J$, empty if its length is not positive. Then

$$|J|\ \ge\ \Bigl(\frac Dt-\frac{k(a_t+b_t)}t-\frac{2kA}{t_-}\Bigr)_+ \ \ge\ \Bigl(\frac Dt-\frac{kA}t-\frac{2kA}{t_-}\Bigr)_+ .$$

If $p\in P_{t_-}\cap(J+u)$, then $T'_u(p)=p-u+\eta$ with $\eta$ in the range above, so $T'_u(p)>x$ and $T'_u(p)\le x+D/t$: the point $T'_u(p)$ lies in $I$. Injectivity gives

$$N_{t_-}(J+u)\ \le\ N_t(I)\qquad(0\le u\le\ell).$$

Averaging. As before,

$$N_t(I)\,\ell\ \ge\ \int_0^\ell N_{t_-}(J+u)\,du =\int_JN_{t_-}\bigl((v,v+\ell]\bigr)\,dv\ \ge\ |J|\,E\,V_{t_-}(E),$$

since $\ell=E/t_-$ makes each $N_{t_-}((v,v+\ell])\ge E\,V_{t_-}(E)$. Dividing by $D\ell=DE/t_-$, the coefficient of $V_{t_-}(E)$ is $|J|t_-/D$, which is at least

$$\Bigl(\frac{t_-}t-\frac{kA}D\Bigl(\frac{t_-}t+2\Bigr)\Bigr)_+ =\Bigl(1-q-\frac{kA}D(3-q)\Bigr)_+ ,$$

using $t_-/t=1-q$. The infimum over $x$ gives (2.3).

Uniformity

Both constructions use only inclusions from an earlier point set into a later one and the moves $F_t$, $B_s$, $F_t$; the comparison times $s(u)$ range over $[t,t_+]$ or $[t_-,t]$ and depend on $E$, $k$ and $t$ but not on $D$. The threshold on $t$ must make (2.1) hold at all these times, make $kr<|P_{t_-}|$, and make the intervals $J$, $J+u$ shorter than $1$; for $D$ in a bounded range one threshold does all of this.

Role in the argument

Iterated along a chain of doubling scales from $r\pm A$ down to $\sqrt{Ar}$ and then applied once more, the lemma gives the short-interval counting bound of Proposition 3.1. Its averaged form, with pointwise span control replaced by $L^1$ control, is Lemma 6.2.