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arxiv: 1810.07979 · v4 · pith:7NPKGOV6new · submitted 2018-10-18 · 🧮 math.GN · math.CT· math.MG

The normality and bounded growth of balleans

classification 🧮 math.GN math.CTmath.MG
keywords boundedballeangrowthnormalsubsetbaseprovebornology
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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 $X\times Y$, we prove that the balleans $X,Y$ have bounded growth and the bornology of $X\times Y$ has a linearly ordered base. A ballean $(X,\mathcal E_X)$ is defined to have bounded growth if there exists a function $G$ assigning to each point $x\in X$ a bounded subset $G[x]\subset X$ so that for any bounded set $B\subset X$ the union $\bigcup_{x\in B}G[x]$ is bounded and for any entourage $E\in\mathcal E_X$ there exists a bounded set $B\subset X$ such that $E[x]\subset G[x]$ for all $x\in X\setminus B$. We prove that the product $X\times Y$ of two balleans has bounded growth if and only if $X$ and $Y$ have bounded growth and the bornology of the product $X\times Y$ 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 $[X]^{\le 2}$ 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 $n\ge 3$ and some subgroup $G\subset S_n$ the $G$-symmetric $n$-th power $[X]^n_G$ of $X$ is normal. On the other hand, we prove that for any ultranormal discrete ballean $X$ and every $n\ge 2$ the power $X^n$ is not normal but the hypersymmetric power $[X]^{\le n}$ 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.

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