Mathematics > General Topology
[Submitted on 18 Oct 2018 (v1), last revised 6 Nov 2018 (this version, v4)]
Title:The normality and bounded growth of balleans
View PDFAbstract: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.
Submission history
From: Taras Banakh [view email][v1] Thu, 18 Oct 2018 10:15:20 UTC (28 KB)
[v2] Sun, 21 Oct 2018 19:29:56 UTC (31 KB)
[v3] Mon, 5 Nov 2018 13:19:28 UTC (33 KB)
[v4] Tue, 6 Nov 2018 17:47:12 UTC (33 KB)
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