Timeline for Additivity of upper densities with respect to arithmetic progressions of integers
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 18, 2015 at 20:51 | vote | accept | Paolo Leonetti | ||
| Dec 18, 2015 at 16:22 | answer | added | Salvo Tringali | timeline score: 2 | |
| Aug 21, 2015 at 9:56 | comment | added | Salvo Tringali | I've just realized that the answer to my previous question is negative: Looking at the case of the upper asymptotic density, ${\sf d}^\ast$, on $\mathbf N^+$, fix $\alpha \in {]0,1[}$ and consider as $X$ the set $\bigcup_{n \ge 1} [\![\alpha(2n-1)! + (1-\alpha)(2n)!+1, (2n)!]\!]$; then recall from mathoverflow.net/questions/207522 that ${\sf d}^\ast(X) = \alpha$. Edit: I see we added a comment almost at the same time. | |
| Aug 21, 2015 at 9:56 | comment | added | Paolo Leonetti | I am almost sure that what you are asking holds if and only if $X$ belongs to the domain of the Buck's density, which holds if and only if $X$ belongs to the topology generated by the set of arithmetic progressions $\mathscr{A}$, or differ by these by a finite set of elements. Anyway, for a negative answer to your question, it sufficient to provide a counterexample (which holds for all $\mu^\star)$. Set $X=\cup_{n\ge 1}\{(2n)!,\ldots,(2n+1)!\}$, then it contains no AP, and the smallest AP containing $X$ is $\mathbf{N}^+$ itself. | |
| Aug 21, 2015 at 9:43 | comment | added | Salvo Tringali | I didn't ask you to agree with me (on something); I asked if you have a counterexample (to something). Anyway, here is an idea: Let $\mathcal A^\sharp$ be the collection of all subsets of $\mathbf N^+$ that can be expressed as a finite union of sets of arithmetic progressions of ${\bf N}^+$, or which differ from these by finitely many integers. Given $X\subseteq{\bf N}^+$, can you show that there exists a nonincreasing (resp., nondecreasing) sequence $(U_n)_{n \ge 1}$ of $\mathcal A^\sharp$ such that $X\subseteq U_n$ (resp., $U_n\subseteq X$) for each $n$ and $\lim_n\mu^\ast(U_n)=\mu^\ast(X)$? | |
| Aug 20, 2015 at 8:38 | comment | added | Paolo Leonetti | I agree that your question would be natural (and obviously stronger); the reason why I asked in this form is that, on the one hand, I guess this is not simpler from the general case, and, on the other hand, this would be enough to prove that also its associated lower density $\mu_\star \colon \mathcal{P}(\mathbf{N}^+)\to \mathbf{R}\colon X\mapsto 1-\mu^\star(X^c)$ satisfies (F4) too.. | |
| Aug 18, 2015 at 12:01 | comment | added | Salvo Tringali | Do you have any counterexample to the stronger (and more natural?) statement that $\mu^\ast(X \cup Y) = \mu^\ast(X) + \mu^\ast(Y)$ whenever $X,Y \subseteq \mathbf N^+$, $X \cap Y = \emptyset$, and $\mu^\ast(Y) + \mu^\ast(Y^c) = 1$, i.e. $Y$ belongs to the domain of the density induced by $\mu^\ast$? | |
| Aug 16, 2015 at 14:47 | history | edited | Paolo Leonetti | CC BY-SA 3.0 |
added 33 characters in body
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| Aug 16, 2015 at 14:07 | history | asked | Paolo Leonetti | CC BY-SA 3.0 |