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\((m, n)\)-equidistant sets in \(\mathbb{R}^{k},\mathbb{S}^{k}\), and \(\mathbb P^k\). (English) Zbl 1158.54012

For a metric space \((X,d)\) and a subset \(A\) of \(X\) with at least two points, the equidistance set \(E(A)\) for \(A\) consists of all \(x\in X\) with the property that \(d(x,a)=d(x,b)\) whenever \(\{a,b\}\in[A]^2\), where \([A]^2\) is the set of all subsets of \(A\) with cardinality 2. Note that \(E(A)\subset X\setminus A\). For \(m\geq 2\), the author calls a metric space \(X(m,n)\)-equidistant if \([X]^m\neq 0\) and \(E(A)\in[X]^n\) whenever \(A\in[X]^m\). Obviously, \(\mathbb{R}^1\) is \((2,1)\)-equidistant, but \(\mathbb{R}^2\) is not. However, \(\mathbb{R}^2\) contains a subset which is \((2, 1)\)-equidistant. This observation is generalized as follows.
Theorem 1. If \(k \geq 2\), then \(\mathbb{R}^k\) contains an \((m,1)\)-equidistant subset if and only if \(k\geq m\).
Theorem 2. If \(m\geq\max\{3,k\}\) and \(n\geq 3\), then \(\mathbb{R}^k\) contains no \((m,n)\)-equidistant subset.
Theorem 3. \(\mathbb{R}^4\) contains no \((4,2)\)-equidistant subset. Obviously, the circle \(\mathbb{S}^1\subset\mathbb{R}^2\) is \((2,2)\)-equidistant. Similarly, the following holds.
Theorem 4. The sphere \(\mathbb{S}^2\subset\mathbb{R}^3\) is \((3,2)\)-equidistant. However, \(\mathbb{S}^3\subset\mathbb{R}^4\) contains no \((4,2)\)-equidistant subset.
Theorem 5. If \(k\geq 1\), then \(\mathbb{S}^k\) contains an \((m,1)\)-equidistant subset if and only if \(k+1\geq m\). Moreover, if \(k+1\geq m\), then the projective space \(\mathbb{P}^k\) contains an \((m,1)\)-equidistant subset.
The author conjectures that, if \(X\) is a \((3,2)\)-equidistant continuum, then \(X\) is homeomorphic to \(\mathbb{S}^2\subset \mathbb{R}^3\).

MSC:

54E35 Metric spaces, metrizability
51K99 Distance geometry
51D20 Combinatorial geometries and geometric closure systems
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