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Statement

The conjecture refuted (p. 1). The paper quotes Larman and Rogers's Conjecture 1 (Mathematika 19 (1972), 1--24) as: "Suppose that the distance 1 is not realized in a closed subset $S$ of a spherical ball $B$ of radius 1. Then the Lebesgue measure of $S$ is less than $(1/2)^n$ times the Lebesgue measure of $B$." An open ball of radius $1/2$ realizes no distance $1$, so the bound would be tight.

The construction (p. 1). Let $e_1=(1,0,\ldots,0)\in\mathbb{R}^n$ and $a=\tfrac16(1+\sqrt{10})$. Put

$$T_n=\{\,x\in\mathbb{R}^n:\ x_1>\tfrac12,\ \|x-ae_1\|<\tfrac12,\ \|x\|<1\,\}, \qquad S_n=T_n\cup(-T_n),$$

so $T_n$ is the intersection of the open ball of radius $1/2$ about $ae_1$, the open unit ball about the origin and the open halfspace $x_1>1/2$. Then $S_n$ is a measurable subset of the unit ball $B_n$ containing no two points at distance $1$, and for every $n\ge2$

$$\operatorname{vol}S_n>(1/2)^n\operatorname{vol}B_n .$$

The abstract states the result in this form, for each $n\ge2$. Since the conjecture asks for a closed set, the paper's footnote 1 (p. 1) takes closed inner approximations of $S_n$; a closed subset of $S_n$ of volume close enough to $\operatorname{vol}S_n$ still exceeds $(1/2)^n\operatorname{vol}B_n$ and is then a counterexample in every dimension $n\ge2$.

Volumes (pp. 1--2). With $v_n=\pi^{n/2}/\Gamma(1+n/2)$ the volume of $B_n$, the paper writes (p. 1)

$$\operatorname{vol}T_n=\int_{1/2-a}^{a-1/2}v_{n-1}(1/4-x^2)^{(n-1)/2}\,dx +\int_{2a-1/2}^{1}v_{n-1}(1-x^2)^{(n-1)/2}\,dx,$$

and reports $\operatorname{vol}S_2/\operatorname{vol}B_2=0.2848\ldots$ and $\operatorname{vol}S_3/\operatorname{vol}B_3=0.1563\ldots$, against $(1/2)^2=0.25$ and $(1/2)^3=0.125$. On p. 2 it records the asymptotic relation, displayed as (1),

$$\frac{\operatorname{vol}S_n}{\operatorname{vol}B_n}=(2-o(1))(1/2)^n>(1/2)^n, \qquad(1)$$

and states that a suitable choice of the constant in the concentration inequality below gives $\operatorname{vol}S_n/\operatorname{vol}B_n>(1/2)^n$ for all $n\ge15$, the remaining cases being checked directly. The paper does not print the computations for $4\le n\le14$.

Choice of $a$ (Figure 1, p. 2). The caption says that $a$ makes $ae_1$ equidistant from the hyperplane $x_1=1/2$ and from the hyperplane containing the intersection of the spheres $\|x\|=1$ and $\|x-ae_1\|=1/2$, and that this choice maximizes the volume of $T_n$. (That hyperplane is $x_1=2a-1/2$, the breakpoint of the two integrals above.)

Source. F. M. de Oliveira Filho and F. Vallentin, A counterexample to a conjecture of Larman and Rogers on sets avoiding distance 1, Mathematika 65 (2019), 785--787; arXiv:1808.07299. Pages are those of the arXiv version 2 (11 March 2019) identified on the source card: the conjecture, the construction, the volume formula and the values for $n=2,3$ on p. 1; Figure 1, relation (1) and the range $n\ge15$ on p. 2.

Read depth. Claims checked: the quoted conjecture, the definition of $T_n$ and $S_n$, the volume formula, the two numerical ratios, relation (1) and the range $n\ge15$ were read clause by clause on the printed pages. The direct checks for the dimensions below $15$ are not printed and were not checked; nothing here is independently reviewed.

Proof pointer

The paper calls the avoidance property easy to see. In the corpus's words: two points of $T_n$ lie in one open ball of radius $1/2$, so their distance is below $1$; a point $x$ of $T_n$ and a point $y$ of $-T_n$ satisfy $x_1>1/2$ and $y_1<-1/2$, so $\|x-y\|\ge x_1-y_1>1$. The same holds inside $-T_n$ by symmetry.

For the volume bound with $n\ge3$ the paper keeps only the first integral, the part of $T_n$ cut from the small ball by the slab $1/2<x_1<2a-1/2$, and uses the concentration of the volume of a ball near its equator, citing Theorem 2.7 of Blum, Hopcroft and Kannan, Foundations of Data Science: if $n\ge3$ and $c\ge1$, the fraction of $\operatorname{vol}B_n$ lying in the slab $|x_1|\le c/\sqrt{n-1}$ is at least $1-(2/c)e^{-c^2/2}$. Applied to the ball of radius $1/2$ about $ae_1$, almost all of its volume lies in a slab about $x_1=a$ that shrinks with $n$ and so eventually sits inside $1/2<x_1<2a-1/2$; each of $T_n$ and $-T_n$ then has volume $(1-o(1))(1/2)^n\operatorname{vol}B_n$, which is (1). The paper states the consequence without printing these steps.

Dependencies

The concentration inequality for the ball quoted above, cited by the paper from Blum, Hopcroft and Kannan (Theorem 2.7); nothing else.

Bears on

• Problem 1070: indirect, through $m_1$, the supremum of the upper densities of measurable planar sets avoiding distance $1$, which the problem page uses. The paper (p. 2) recalls L. Moser's conjecture, popularized by Erdős and which it calls still open, that every measurable planar set with no two points at distance $1$ has upper density less than $1/4$, upper density being defined in its footnote 3 as the supremum over $p\in\mathbb{R}^n$ of the limsup, as $T\to\infty$, of $\operatorname{vol}(X\cap(p+[-T,T]^n))/\operatorname{vol}[-T,T]^n$. It notes that Larman and Rogers's conjecture would have given only the bound at most $1/4$, so it would not have implied Moser's even if true. The construction refutes the local conjecture for the unit disk ($n=2$, ratio $0.2848\ldots$), so that route to $m_1\le1/4$ fails; it gives no bound on $m_1$ or on the problem's $f(n)$ and settles nothing in Problem 1070. Moser's conjecture has since been proved by Ambrus, Csiszárik, Matolcsi, Varga and Zsámboki ($m_1\le0.247$), as the problem page records.