Timeline for Conditions on $R\subseteq \mathbb{N}$ so that $\{\{xr\}:r\in R\}$ is dense in $[0,1]$ for all irrational $x$
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| Aug 7 at 16:07 | comment | added | John Griesmer | It's worth observing that "goodness," in your sense, lacks a compactness property satisfied by "adequacy" (in the sense of the linked question): a set $R\subset \mathbb Z$ is adequate if and only if for all $\epsilon>0$ and all $K\in \mathbb N$, there is a finite $R'\subset R$ such that for all $k\in \mathbb Z$ with $|k|< K$, there is an $r\in R'$ such that $|\{(r+k)x\}|<\epsilon$. In contrast, $2\mathbb Z+1$ is good, but no finite subset $R'\subset 2\mathbb Z+1$ approximates goodness: there is always an irrational $x$ such that $|\{rx\}|>1/4$ for all $r\in R'$ (take $x$ close to $1/2$). | |
| Jul 29 at 22:12 | comment | added | Varun Vejalla | @Asaf You don't necessarily have to enumerate those elements in increasing order though. I'm just saying that you'd need a sequence $(r_1,r_2,\cdots)$ such that $r_i\in R$ for all $i$ and $(r_n.x)$ is equidistributed. Does that make sense? The sequence can be different for different values of $x$. | |
| Jul 29 at 14:05 | comment | added | Asaf | @VarunVejalla , I don't understand your term "equidistributed sequence" in the set. For equidistribution (at least in the standard meaning in number theory), we fix a sequence and then a base point, i.e. the sequence $n^{2}.x$ is equidistributed for all irrational $x$. If you enumerate the elements of $\{2^{n}3^{m}\}$ in increasing order, call the resulting sequence $r_{n}$, there are irrationals $x$ such that $\{r_{n}.x\}$ is not equidistributed. Obviously for almost all $x$ (Lebesgue), the sequence $\{r_{n}.x\}$ will be equidistributed, by ergodicity of $\times 2,\times 3$. | |
| Jul 27 at 3:43 | comment | added | Varun Vejalla | @Asaf But there would be an equidistributed sequence with elements all in that set, right? | |
| Jul 23 at 19:30 | comment | added | Wojowu | Positive (lower) density is not enough - for instance the set $R=\{n\in\mathbb N\mid \{n\sqrt{2}\}\in[0,1/2]\}$ has density $1/2$ (by Weyl) but by construction the set $R\sqrt{2}$ is not dense mod $1$. On the other hand, if $R$ has (upper?) density $1$, it should hold, as again by Weyl you can argue it will visit every interval infinitely often. | |
| Jul 23 at 19:07 | comment | added | Asaf | I am not sure about your answer on the other site, but it is a landmark result of Furstenberg that for any irrational $x$, the set $\{2^{n}3^{m}.x\}$ is dense mod $1$, but is not (necessarily) equidistributed, i.e. by choosing a proper Liouville number. | |
| Jul 23 at 18:10 | history | edited | Varun Vejalla | CC BY-SA 4.0 |
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| Jul 23 at 17:55 | history | edited | YCor |
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| S Jul 23 at 17:01 | review | First questions | |||
| Jul 23 at 17:25 | |||||
| S Jul 23 at 17:01 | history | asked | Varun Vejalla | CC BY-SA 4.0 |