Home » Discrete Geometry » Equilateral sets in normed spaces: upper bounds

Equilateral sets in normed spaces: upper bounds

A set S in a normed space X is equilateral if any two points of S are at the same distance:

There exists \rho>0 such that \|x-y\|=\rho for all distinct x,y\in S.

From now on we fix \rho=1 by scaling.

If X is n-dimensional, then an equilateral set has at most 2^n elements (Petty 1971, Soltan 1975). The following is a very simple proof. Let P be the convex hull of S. Then the homothets x+\frac12(P-x), x\in S, are all contained in P, and have pairwise disjoint interiors. This follows from the triangle inequality. Now use the fact that the Lebesgue measure of each homothet is (1/2)^n times that of P.

The next proof is by Füredi, Lagarias and Morgan (1993). Let A=\bigcup_{x\in S} B(x,\frac12), the union of all balls of radius \frac12 centered at the points in S. Then A has diameter 2. Now, by the isodiametric inequality (Busemann 1947, Mel’nikov 1963), the unit ball is the set of diameter 2 of largest Lebesgue volume. Again comparing volumes gives the upper bound.

The isodiametric inequality has a simple proof from the Brunn-Minkowski inequality, so it is not surprising that there is also a proof using the latter inequality. Again consider the set A. Since it has diameter 2, its central symmetral A-A:=\{x-y : x,y\in A\} is contained in a ball of radius 2. By the Brunn-Minkowski inequality, \mu(A-A)^{1/n}\geq\mu(A)^{1/n}+\mu(-A)^{1/n}, and the 2^n upper bound follows.


  1. H. Busemann, Intrinsic area, Ann. Math. 48 (1947), 234-267. [It is quite a challenge to find the isodiametric inequality in this paper. It’s equation (2.2) on p. 241.]
  2. Z. Füredi, J. C. Lagarias, and F. Morgan, Singularities of minimal surfaces and networks and related extremal problems in Minkowski space, Discrete and computational geometry (New Brunswick, NJ, 1989/1990), DIMACS Ser. Discrete Math. Theoret. Comput. Sci., vol. 6, Amer. Math. Soc., Providence, RI, 1991. pp. 95-109.
  3. M. S. Mel’nikov, Dependence of volume and diameter of sets in an n-dimensional Banach space (Russian), Uspehi Mat. Nauk 18 (1963), 165-170.
  4. C. M. Petty, Equilateral sets in Minkowski spaces, Proc. Amer. Math. Soc. 29 (1971), 369-374.
  5. P. S. Soltan, Analogues of regular simplexes in normed spaces (Russian), Dokl. Akad. Nauk SSSR 222 (1975), 1303-1305. English translation: Soviet Math. Dokl. 16 (1975), 787-789.

5 thoughts on “Equilateral sets in normed spaces: upper bounds

  1. Pingback: Equilateral sets in normed spaces II: lower bounds « Konrad Swanepoel’s blog

  2. Pingback: Equidistancia en R^n (parte I) « Spin Foam

  3. What do you think of the following approach to proving Kusner’s conjecture that an upper bound for the size of an equilateral set in \ell_1^n is 2n?

    Let v_2 : \mathbb{Q} \longrightarrow \mathbb{Z} \cup \{\infty\} denote the 2-adic valuation, R = \{a \in \mathbb{Q} : v_2(a) \geq 0\} the corresponding local ring, M = \{a \in \mathbb{Q} : v_2(a) > 0\} its unique maximal ideal and \underline{a} \in \mathbb{F}_2 the reduction of a \in R modulo M. One can prove that there are no equilateral sets with respect to the \ell_1 norm with \rho = 1 in R^n as follows:
    Since R/M = \mathbb{F}_2, \underline{a} = \underline{-a} and hence \underline{a} = \underline{|a|} for any a \in R. Therefore, if (a_1,\dots,a_n),(b_1\dots,b_n) \in R^n are points in an equilateral subset of R^n, the image modulo M of a_1-b_1 + \dots + a_n-b_n equals the image modulo M of |a_1-b_1| + \dots + |a_n-b_n|, which equals 1 when (a_1,\dots,a_n) and (b_1\dots,b_n) are distinct, and 0 otherwise.
    Therefore, if (a_{1,1},\dots,a_{1,n}),\dots,(a_{m,1}\dots,a_{m,n}) \in R^n are the points in an equilateral set of size m in R^n, then the m \times m-matrix \sum_{1 \leq i \leq n}((\underline{a_{1,i}},\dots,\underline{a_{m,i}})^T (1,\dots,1) - (1,\dots,1)^T (\underline{a_{1,i}},\dots,\underline{a_{m,i}}))
    has (i,j)-entry \underline{a_{i,1}}-\underline{a_{j,1}} + \dots+ \underline{a_{i,n}}-\underline{a_{j,n}} and is hence equal to the matrix with main diagonal entries all equal to 0 and all other entries equal to 1. Since this matrix has rank m and is equal to the sum of the 2n rank-1-matrices (\underline{a_{1,i}},\dots,\underline{a_{m,i}})^T (1,\dots,1), - (1,\dots,1)^T (\underline{a_{1,i}},\dots,\underline{a_{m,i}}), we have that m \leq 2n.

    For a proof by contradiction of Kusner’s conjecture, it would therefore suffice to show that the existence of an equilateral set (with respect to the \ell_1 norm) of size m in \mathbb{R}^n implies the existence of an equilateral set (with respect to the \ell_1 norm) of size m in R^n. This sounds like the sort of thing tropical geometry was invented for, but I don’t understand much tropical geometry, so I’m sort of stuck here. I apologize for dumping this text on you, but I’m just a math-obsessed pupil and not (yet) a professional mathematician, so I just don’t know what to do with this observation.

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