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Source. S. Korsky, A resolution of the de Bruijn--Erdős consecutive-gap problem, arXiv:2609.07196v2, Section 4: Theorem 4.1 (p. 7), its derivation from Larcher's proof (p. 8) and Lemma 4.2 (p. 8) 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 of Lemma 4.2; not an independent review; changes no status and assigns no tier. Theorem 4.1 is an external input whose stated form is the source's own derivation from G. Larcher, On the star discrepancy of sequences in the unit interval, J. Complexity 31 (2015), 474--485 (arXiv:1407.2094), Section 3. Larcher's paper is not held and the derivation is not checked here; see "The imported input" below.

Definitions

For a finite list $z_1,\ldots,z_L\in[0,1)$, its maximum prefix counting error is

$$H_L(z_1,\ldots,z_L)=\max_{1\le j\le L}\ \sup_{0\le u\le1}\ \Bigl|\#\{i\le j:\ z_i<u\}-ju\Bigr| .$$

Points, $P_n$ and $N_n(\cdot)$ are as on the Lemma 2.1 page; here only integer times $n$ occur, $P_n=\{x_1,\ldots,x_n\}$.

The imported input (Theorem 4.1, p. 7)

Statement as used. There is an absolute integer $L_0$ such that, for every integer $L\ge L_0$ and every list $z_1,\ldots,z_L\in[0,1)$,

$$H_L(z_1,\ldots,z_L)\ \ge\ \frac1{16}\log L .$$

The source's derivation, as stated (p. 8). Section 3 of Larcher's paper (pp. 12--13 of the arXiv preprint, per the source) starts from a finite list of length $N=\lfloor a^h\rfloor$ with $3<a<4$ and $h\in\mathbb N$ and proves $H_N\ge c_a\log N$ with

$$c_a=\frac{(a-2)(8a+3)}{16(1-2a)^2\log a};\qquad a=\tfrac72:\quad c_a=\frac{31}{384\log(7/2)}>0.064>\frac1{16}.$$

Given a list of length $L$, restrict to its prefix of the largest such length $N\le L$; then $H_L\ge H_N$ (a maximum over fewer prefixes) and $\log N=\log L-O_a(1)$, so $H_L\ge c_a\log L-O_a(1)$ with a constant independent of the list, and the strict margin $c_a>1/16$ gives the statement for $L\ge L_0$. (The arithmetic $c_{7/2}=31/(384\log3.5)\approx0.0644$ is checked here.)

What is and is not checked. Whether Larcher's Section 3 proves the finite-list statement $H_N\ge c_a\log N$ for every list of length $N=\lfloor a^h\rfloor$, as opposed to a statement about infinitely many prefixes of an infinite sequence, is exactly what the source asserts and what is not verified here. The numerical constant $1/100$ in the ratio bound depends on $c_a>1/16$. An authored remark, checked here: the qualitative form of Theorem 4.1, $H_L\ge c\log L-1$ for every list with some absolute $c>0$, follows from Schmidt's theorem for planar point sets (W. M. Schmidt, Irregularities of distribution VII, Acta Arith. 21 (1972), 45--50: every $N$-point set in $[0,1]^2$ has a box anchored at the origin whose count differs from $Nuv$ by at least $c\log N$), applied to the set $\{(z_i,i/L)\}$, since the count of that set in $[0,u)\times[0,v]$ is $\#\{i\le j:z_i<u\}$ with $j=\lfloor Lv\rfloor$ and $|ju-Luv|\le1$. That form suffices for the growth statement $r(\mu_r-1)\to\infty$, with a smaller unspecified constant in place of $1/100$.

Statement (Lemma 4.2, p. 8)

Suppose that $B\ge1$, $S\ge2$, and that for all sufficiently large integers $n$,

$$\Bigl|N_n\bigl((x,x+D/n]\bigr)-D\Bigr|\ \le\ B\qquad (x\in\mathbb T,\ 0\le D\le S). \tag{4.1}$$

If $\lfloor S\rfloor\ge L_0$, then

$$B\ \ge\ \frac1{16}\log\lfloor S\rfloor .$$

Proof

Put $L=\lfloor S\rfloor$. Choose $n_0\ge2$ such that (4.1) holds for all $n\ge n_0$, and let $\delta>0$ be the least circular distance between two distinct points of $P_{n_0}$ (positive since the points are distinct). Choose an integer $N>\max\{L,n_0\}$ so large that $L/N<\delta$.

A short interval with exactly $L$ points. The cyclic $L$-spans of $P_N$, the distances from each point to the point $L$ places later, have mean $L/N$ (each gap lies in exactly $L$ of them), so some $L$-span has length $\ell\le L/N$. The half-open arc from its initial point $p$ to $p+\ell$ contains exactly $L$ points of $P_N$, namely the $L$ points after $p$ including $p+\ell$. Translate this arc forward by a small $\varepsilon>0$: the arc $J=(p+\varepsilon,p+\varepsilon+\ell]$ still contains those $L$ points, excludes $p$, admits no new point for small $\varepsilon$, and has both endpoints outside $P_N$. Write $J=(a,a+\ell]$. Since $\ell\le L/N<\delta$, $J$ contains at most one point of $P_{n_0}$.

The list. List the $L$ points of $J$ in their order of insertion and rescale $J$ to $(0,1)$: the $i$-th listed point $x_{m_i}$ becomes $z_i=(x_{m_i}-a)/\ell$, with $m_1<m_2<\cdots<m_L\le N$ and $z_i\in(0,1)$ because the endpoints of $J$ are not points. Fix $1\le j\le L$ and let $n=m_j\le N$ be the insertion time of the $j$-th listed point. Then the points of $P_n$ in $J$ are exactly the first $j$ listed points, so

$$N_n(J)=j,\qquad \#\{i\le j:\ z_i\le u\}=N_n\bigl((a,a+u\ell]\bigr)\quad(0\le u\le1).$$

Early prefixes. If $n<n_0$, all of the first $j$ points lie in $P_{n_0}\cap J$, so $j\le1$; a one-point prefix has counting error $|\mathbf 1[z_1<u]-u|\le1\le B$.

Late prefixes. If $n\ge n_0$, define

$$f(u)=N_n\bigl((a,a+u\ell]\bigr)-n\ell u\qquad(0\le u\le1).$$

The arc $(a,a+u\ell]$ has length $u\ell=D/n$ with $D=nu\ell\le n\ell\le N\ell\le L\le S$, so (4.1) at time $n$ gives $|f(u)|\le B$. The complementary arc $(a+u\ell,a+\ell]$ has length $(1-u)\ell=D'/n$ with $D'=n(1-u)\ell\le S$, and its count is $N_n(J)-N_n((a,a+u\ell])$, so (4.1) gives $|f(1)-f(u)|\le B$. Since $f(1)=j-n\ell$,

$$\#\{i\le j:z_i\le u\}-ju=f(u)+n\ell u-ju=f(u)-uf(1) =(1-u)f(u)-u\bigl(f(1)-f(u)\bigr),$$

and therefore

$$\bigl|\#\{i\le j:z_i\le u\}-ju\bigr|\ \le\ (1-u)B+uB=B .$$

Controlling both complementary arcs is what keeps the bound at $B$ rather than $2B$.

Conclusion. Every prefix of the list has counting error at most $B$ for the convention $z_i\le u$; for $z_i<u$ the count is the left limit, so the supremum over $u$ is the same. Hence $H_L(z_1,\ldots,z_L)\le B$, and Theorem 4.1 with $L\ge L_0$ gives $B\ge\frac1{16}\log L$.

Role in the argument

Proposition 3.1 supplies (4.1) at integer times with $B=3A+C_1A/\Lambda$ and $S=\sqrt{Ar}/\Lambda^2$; the Section 5 proof compares $B\approx\frac3{100}\log r$ with $\frac1{16}\log\lfloor S\rfloor\approx\frac1{32}\log r$.