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The normality and bounded growth of balleans

Published 18 Oct 2018 in math.GN, math.CT, and math.MG | (1810.07979v4)

Abstract: By a ballean we understand a set XX endowed with a family of entourages which is a base of some coarse structure on XX. Given two unbounded ballean X,YX,Y with normal product X×YX\times Y, we prove that the balleans X,YX,Y have bounded growth and the bornology of X×YX\times Y has a linearly ordered base. A ballean (X,EX)(X,\mathcal E_X) is defined to have bounded growth if there exists a function GG assigning to each point x∈Xx\in X a bounded subset G[x]⊂XG[x]\subset X so that for any bounded set B⊂XB\subset X the union ⋃x∈BG[x]\bigcup_{x\in B}G[x] is bounded and for any entourage E∈EXE\in\mathcal E_X there exists a bounded set B⊂XB\subset X such that E[x]⊂G[x]E[x]\subset G[x] for all x∈X∖Bx\in X\setminus B. We prove that the product X×YX\times Y of two balleans has bounded growth if and only if XX and YY have bounded growth and the bornology of the product X×YX\times Y has a linearly ordered base. Also we prove that a ballean XX has bounded growth (and the bornology of XX has a linearly ordered base) if its symmetric square [X]<sup>≤</sup>2[X]<sup>{\le</sup> 2} is normal (and the ballean XX is not ultranormal). A ballean XX has bounded growth and its bornology has a linearly ordered base if for some n≥3n\ge 3 and some subgroup G⊂SnG\subset S_n the GG-symmetric nn-th power [X]<sup>nG[X]<sup>n_G of XX is normal. On the other hand, we prove that for any ultranormal discrete ballean XX and every n≥2n\ge 2 the power X<sup>nX<sup>n is not normal but the hypersymmetric power [X]<sup>≤</sup>n[X]<sup>{\le</sup> n} of XX is normal. Also we prove that the finitary ballean of a group is normal if and only if it has bounded growth if and only if the group is countable.

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