2018/10/18 by Taras Banakh, Banakh, Taras, Igor Protasov +1
Mathematics · #20F65 #54D15 #Category Theory (math.CT) #FOS: Mathematics #General Topology (math.GN) #Metric Geometry (math.MG) #math.CT #math.GN #math.MG #msc:20F65 #msc:54D15
paper · pdf · doi:10.48550/arxiv.1810.07979
28 pages
arxiv created 2018/11/06 · arxiv updated 2018/11/07
By a ballean we understand a set X endowed with a family of entourages which is a base of some coarse structure on X. Given two unbounded ballean X,Y with normal product X× Y, we prove that the balleans X,Y have bounded growth and the bornology of X× Y has a linearly ordered base. A ballean (X,\mathcal EX) is defined to have bounded growth if there exists a function G assigning to each point x∈ X a bounded subset G[x]⊂ X so that for any bounded set B⊂ X the union \bigcupx∈ BG[x] is bounded and for any entourage E∈\mathcal EX there exists a bounded set B⊂ X such that E[x]⊂ G[x] for all x∈ X∖ B. We prove that the product X× Y of two balleans has bounded growth if and only if X and Y have bounded growth and the bornology of the product X× Y has a linearly ordered base. Also we prove that a ballean X has bounded growth (and the bornology of X has a linearly ordered base) if its symmetric square [X]≤ 2 is normal (and the ballean X is not ultranormal). A ballean X has bounded growth and its bornology has a linearly ordered base if for some n≥ 3 and some subgroup G⊂ Sn the G-symmetric n-th power [X]nG of X is normal. On the other hand, we prove that for any ultranormal discrete ballean X and every n≥ 2 the power Xn is not normal but the hypersymmetric power [X]≤ n of X 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.