Given a bounded domain $\Omega\subset\mathbb R^n$ ($n\geq2$), how often can a single real number $r>0$ appear as a distance of two points on $\partial\Omega$? We can make any assumptions about the boundary $\partial\Omega$ if needed; I would prefer weak assumptions like finite $(n-1)$-dimensional Hausdorff measure, but smoothness is also ok. Specifically, I would be interested in a reference or a proof for either of the following two claims, which intuitively seem true:

Claim 1: Fix $r>0$ and a bounded domain $\Omega\subset\mathbb R^n$ with sufficiently regular boundary. For $x\in\mathbb R^n$ and $v\in S^{n-1}$, denote the line through $x$ in direction $v$ by $L_{x,v}=x+v\mathbb R$. Then for almost every $(x,v)\in\mathbb R^n\times S^{n-1}$ the set $L_{x,v}\cap\partial\Omega$ does not contain any two points exactly distance $r$ from each other.


Claim 2: Fix $r>0$ and a bounded domain $\Omega\subset\mathbb R^n$ with sufficiently regular boundary. Then for almost all $x,y\in\partial\Omega$ we have $d(x,y)\neq r$. (The measure on $\partial\Omega\times\partial\Omega$ is the $2(n-1)$-dimensional Hausdorff measure.)

Both claims are of the form "a set $A$ is of zero measure". (The set $A$ is different in different claims, a subset of $\mathbb R^n\times S^{n-1}$ or $\partial\Omega\times\partial\Omega$.) The set $A$ need not be discrete. For example, if $n=2$ and $\Omega=\{x;x_1>0,x_2>0,|x|<1\}$, the set $A$ is quite large for $r=1$ but both claims are still true.

  • $\begingroup$ Perhaps Erdös's Distinct Distances Problem is relevant here? Posed in 1946, for $n$ points in $\mathbb{R}^2$, the lowerbound (and perhaps the "right answer") is $\Omega(ne^{\frac{c \log n}{\log \log n}})$. Cf. Research Problems in Discrete Geometry (Springer link). $\endgroup$ – Joseph O'Rourke Feb 24 '15 at 23:59
  • $\begingroup$ @JosephO'Rourke, that does indeed look like the discrete version of my question in the plane, but I don't see how to translate a discrete result into a continuous one here. If we fix $\epsilon>0$ and sample $n$ points on $\partial\Omega$ so that $\log\log n>c/\epsilon$, then that result shows that at most $\sim n^{1+\epsilon}$ out of the $\sim n^2$ pairs have distance $r$. This suggests that the Hausdorff dimension of $A$ in claim 2 in dimension 2 is at most 1, but I don't see how to actually prove it. $\endgroup$ – Joonas Ilmavirta Feb 25 '15 at 2:21

I cannot prove either of the claims, but a similar result is true. Namely, given $r>0$, set of points $(x,y)\in\partial\Omega\times\partial\Omega$ with $d(x,y)=r$ is closed and has no interior points. In particular, the complement of the set is dense. (This does not imply claim 2.)

Closedness follows from continuity of the distance function. Suppose $(x,y)$ was an interior point. Then there would be $\delta>0$ such that $d(z,w)=r$ for all $z\in B(x,\delta)\cap\partial\Omega$ and $w\in B(y,\delta)\cap\partial\Omega$. Fixing $w=y$ in this condition shows that $\partial\Omega$ has to be a part of the sphere $S(y,r)$ near $x$. Similarly, the boundary near $y$ has to coincide with $S(x,r)$. If we take the line joining $x$ and $y$ and take a line parallel to it and so close to it that it hits the spherical parts of the boundary, the intersection points of the new line with these spheres are actually closer to each other than $r$. This contradiction shows that interior points are impossible.

This proof can also be found in lemma 25 of this paper. This result was eventually sufficient for my purposes for now but I'm still interested in the more general question.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.