{
  "id": "1812.10696",
  "version": "v1",
  "published": "2018-12-27T11:11:07.000Z",
  "updated": "2018-12-27T11:11:07.000Z",
  "title": "A new upper bound for the size of $s$-distance sets in boxes",
  "authors": [
    "Gábor Hegedüs"
  ],
  "comment": "7 pages",
  "categories": [
    "math.CO"
  ],
  "abstract": "Let $q,d\\geq 2$ be integers. Define $$ J(q,d):=\\frac 1q \\Big( \\min_{0<x<1} \\frac{1-x^q}{1-x} x^{-\\frac{q-1}{d}}\\Big). $$ Let $\\mbox{$\\cal G$}\\subseteq {\\mathbb R}^n$ be an arbitrary subset. We denote by $d(\\mbox{$\\cal G$})$ the set of (non-zero) distances among points of $\\mbox{$\\cal G$}$: $$ d(\\mbox{$\\cal G$}):=\\{d( p_1, p_2):~ p_1, p_2\\in \\mbox{$\\cal G$}, p_1\\ne p_2\\}. $$ Our main result is a new upper bound for the size of $s$-distance sets in boxes. More concretely, let $A_i\\subseteq \\mathbb R$, $|A_i|=q\\geq 2$ be subsets for each $1\\leq i\\leq n$. Consider the box $\\mbox{$\\cal B$}:=\\prod_{i=1}^n A_i\\subseteq {\\mathbb R}^n$. Suppose that $\\mbox{$\\cal G$}\\subseteq \\mbox{$\\cal B$}$ is a set such that $|d(\\mbox{$\\cal G$})|\\leq s$. Let $d:=\\frac{n(q-1)}{s}$. Then $$|\\mbox{$\\cal G$}|\\leq 2(qJ(q,d))^n.$$ We use Tao's slice rank bounding method in our proof.",
  "revisions": [
    {
      "version": "v1",
      "updated": "2018-12-27T11:11:07.000Z"
    }
  ],
  "analyses": {
    "keywords": [
      "distance sets",
      "upper bound",
      "taos slice rank bounding method",
      "main result",
      "arbitrary subset"
    ],
    "note": {
      "typesetting": "TeX",
      "pages": 7,
      "language": "en",
      "license": "arXiv",
      "status": "editable"
    }
  }
}