Timeline for Upper density of subsets of an amenable group
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Jul 7, 2022 at 8:25 | history | edited | YCor | CC BY-SA 4.0 |
formatting
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| May 6, 2020 at 19:59 | history | became hot network question | |||
| May 6, 2020 at 13:57 | comment | added | Gabe Conant | More generally, in $\mathbb{Z}$ or $\mathbb{N}$, $\sup m(X)=1$ if and only if $X$ contains arbitrarily large intervals, which is equivalent to the notion of a "thick" set (as defined in my answer below). | |
| May 6, 2020 at 13:41 | comment | added | YCor | By the way I forgot to say, but we can always consider $\sup m(X)$ when $m$ ranges over invariant means. Probably it means the same as the supremum over all Følner sequences of the upper densities wrt the Følner sequence. Nevertheless it's frustrating that it maps $\bigcup [2^{2^n},2^{2^n}+n]$ to 1. | |
| May 6, 2020 at 13:28 | answer | added | R W | timeline score: 1 | |
| May 6, 2020 at 12:40 | history | edited | Martin Sleziak |
added the (amenability) tag
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| May 6, 2020 at 12:37 | comment | added | YCor | The problem with this is that you have disjoint ones, i.e. two Følner sequences and two disjoint subsets which have density 1 for one and zero for the other; in particular the density will never be independent of the Følner sequence apart from being in $\{0,1\}$. The notion "subset that has density 0 for every Følner sequence" is thus quite restrictive. Still it seems not void, i.e., not reduced to finite subsets (excludes $\bigcup [2^n,2^n+n]$, but seems to include $\bigcup\{2^n\}$). | |
| May 6, 2020 at 12:34 | answer | added | Gabe Conant | timeline score: 3 | |
| May 6, 2020 at 12:31 | comment | added | Otto | So maybe one way is to look over \emph{all} Folner sequences? | |
| May 6, 2020 at 12:28 | comment | added | YCor | One difficulty is the dependence on the Følner sequence. For instance, take $X=\bigcup_n [2^n,2^n+3n]$, and $F_n=\bigcup_{k=0}^n[2^k+n,2^k+2n[$. Then $F_n$ has cardinal $n^2$ and boundary of cardinal $2n$, so is Følner. But $F_n\subset X$ while $X$ has natural density zero. Only forcing $F_n$ to cover won't be enough to fix the issue (change $F_n$ to $F_n\cup [-\sqrt{n},\sqrt{n}]$ if necessary). In $\mathbf{Z}^2$ one has less artificial examples with $F_n=[-n,n]\times [-n^2,n^2]$ and $X=\{(x,y):|x|\le |y|\}$ which has density zero w.r.t this choice. | |
| May 6, 2020 at 11:57 | history | asked | Otto | CC BY-SA 4.0 |