12
$\begingroup$

Given a set $X \subseteq \mathbf N^+$, denote by $\mathsf{d}_\ast(X)$ and $\mathsf{d}^\ast(X)$, respectively, the lower and upper asymptotic (or natural) density of $X$, viz. $$\mathsf{d}_\ast(X) := \liminf_{n \to \infty} \frac{|X \cap [1, n]|}{n}$$ and $$ \mathsf{d}^\ast(X) := \limsup_{n \to \infty} \frac{|X \cap [1, n ]|}{n},$$ and by $\mathsf{bd}_\ast(X)$ and $\mathsf{bd}^\ast(X)$, respectively, the lower and upper Banach (or uniform) density of $X$, i.e. $$\mathsf{bd}_\ast(X) := \lim_{n \to \infty} \min_{m \ge 1} \frac{|X \cap [m+1, m+n ]|}{n}$$ and $$ \mathsf{bd}^\ast(X) := \lim_{n \to \infty} \max_{m \ge 1} \frac{|X \cap [m+1, m+n]|}{n}.$$ It is mentioned in @Martin Sleziak's answer to Question 66191: Density of a set of natural numbers whose differences are not bounded that for all $\alpha, \beta, \gamma, \delta \in [0,1]$ with $\alpha \le \beta \le \gamma \le \delta$ there exists a set $X \subseteq \mathbf N^+$ such that $\mathsf{bd}_\ast(X) = \alpha$, $\mathsf{d}_\ast(X) = \beta$, $\mathsf{d}^\ast(X) = \gamma$, and $\mathsf{bd}^\ast(X) = \delta$.

Q. Do you know of a reference for the claim above?

Notice that Martin is not sure about the claim (he writes, "If I remember correctly, I have seen the result etc."), and from the comments to his own answer it seems he could not indeed retrieve a reference.

I am aware of results along these lines in the case where the pair $(\mathsf{d}^\ast, \mathsf{bd}^\ast)$ and its "dual" $(\mathsf{bd}_\ast, \mathsf{d}_\ast)$ are replaced, respectively, with the pair $(\mathsf{ld}^\ast, \mathsf{d}^\ast)$ and its dual $(\mathsf{d}_\ast, \mathsf{ld}_\ast)$, where $\mathsf{ld}_\ast$ and $\mathsf{ld}^\ast$ are, respectively, the lower and upper logarithmic density on $\mathbf N^+$, namely $$\mathsf{ld}_\ast(X) := \liminf_{n \to \infty} \frac{\sum_{x \in X \cap [1,n]} \frac{1}{x}}{\log n}.$$ and $$\mathsf{ld}^\ast(X) := \limsup_{n \to \infty} \frac{\sum_{x \in X \cap [1,n]} \frac{1}{x}}{\log n}.$$ In fact, this can even be proven in the general case where the pair $(\mathsf{ld}^\ast, \mathsf{d}^\ast)$ is replaced by a pair $(\mu^\ast, \nu^\ast)$ of $\mathfrak{m}$-weighted upper densities on an arithmetical semigroup $\mathbb{A} = (A, \cdot\,, |\cdot|)$ and $(\mathsf{d}_\ast, \mathsf{ld}^\ast)$ by the dual pair of $(\mu^\ast, \nu^\ast)$, under the assumption that $\mu^\ast(X) \le \nu^*(X)$ for all $X \subseteq A$ and $$\sum_{a \in A,\, |a| \le x} \mathfrak{m}(a) \sim x^s$$ for some $s > 0$ as $x \to \infty$, see

F. Luca, C. Pomerance, and Š. Porubský, Sets with prescribed arithmetic densities, Unif. Distr. Theory 3 (2008), No. 2, 67-80

for further details. Yet, I could not find a reference for the case I am asking for.

$\endgroup$
  • 2
    $\begingroup$ It might be worth adding that Martin Sleziak presents the claim with the caveat that he might not recall it correctly and mix it up with similar results. $\endgroup$ – user9072 May 16 '15 at 18:07
  • 1
    $\begingroup$ I just want to confirm that I cannot find a reference you ask for and I have probably seen result for some other kind of densities. (Which means the my answer back then was partially misleading. Although, this was probably not the most important part of that answer.) $\endgroup$ – Martin Sleziak May 16 '15 at 19:08
  • 1
    $\begingroup$ Here is an old thread where a more difficult question is addressed: mathoverflow.net/questions/103111/…. The thread has also the sketch of an answer by @Anthony Quas (mathoverflow.net/a/103127/16537), but there is no reference to the question stated in the OP there. $\endgroup$ – Salvo Tringali May 20 '15 at 16:18
  • $\begingroup$ A related question: mathoverflow.net/questions/225400. $\endgroup$ – Salvo Tringali Dec 19 '15 at 12:41
2
$\begingroup$

I don't know a reference for the claim, but I think I know a fairly easy construction which gives such a set $X$. I hope it's OK if I give this as an answer even if it doesn't answer your question about a reference.

EDIT: By MartinSleziak's comment, we can assume WLOG $\delta=1$. Writing out the construction for general $0 \leq \alpha \leq \beta \leq \gamma \leq 1$ would still be too technical, so I'll illustrate with some examples: First take $\alpha=1/4$, $\beta=1/3$ and $\gamma=1/2$. Then $$X=\{1,2,6,10,14,18,19,20,21,22,23,24,25,29,33,37,\dots\}.$$

I.e. we start with 1, then we add every fourth number until the total density is at most $1/3$ (which is the case at $18$, since $6/18=1/3$), then we add every number until the total density is at least $1/2$ (which is the case at $24$, since $12/24=1/2$), then we add every fourth number again until the total density is at most $1/3$ (which will be the case at $81$, where $27/81=1/3$) and so on.

This set then clearly will have uniform densities $1/4$ and $1$ and natural densities $1/3$ and $1/2$ and one can check that this construction always works if $\alpha < \beta < \gamma < 1$ (if e.g. $\alpha>0$ isn't a unit fraction, we have to consider progressions $m+\lfloor k/\alpha\rfloor$, $k=0,\dots,N$; if $\alpha=0$, the corresponding progressions are replaced by suitably long "holes").

What if $\beta=\gamma$? E.g. $\alpha=1/4$, $\beta=\gamma=1/3$. Then $$X=\{1,2,6,7,10,13,16,17,18,19,20,23,26,29,\dots,44,45,49,53,57,58,61,64,\dots\}.$$ So we start with $1$ and then alternate between arithmetic progressions of distance $3$ and arithmetic progressions of distance $1$ or $4$, with the lengths of the former growing quadratically and the lengths of the latter growing linearly. Both natural densities are clearly $1/3$ here, as the latter arithmetic progressions are "drowned out" by the former, but the uniform densities are $1/4$ and $1$, as one can find arbitrary long arithmetic progressions of distance $1$ or $4$ in the set. Again this always works if $\beta=\gamma$.

Finally what if $\alpha=\beta$? If $\gamma<1$, I illustrate with $\alpha=\beta=1/3$, $\gamma=1/2$. Then $$X=\{1,2,5,6,7,10,11,12,15,18,19,20,23,\dots\}.$$ We start with $1$. Each run of consecutive numbers is exactly as long as it needs to be to make the total density at least $1/2$ and is followed by an arithmetic progression of distance $3$, which is exactly as long as it needs to be to make the total density $\leq 1/3+1/(n+2)$, where $n$ counts the number of arithmetic progressions of distance $3$, separated by runs of consecutive numbers, occuring in the set so far. This set then clearly has natural densities $1/3$ and $1/2$ and uniform densities $1/3$ and $1$. Again this will work in general and a similar construction works for $\gamma=1$ and $\beta > \alpha$.

If $\gamma=1$ and e.g. $\alpha=\beta=1/2$, consider $$X=\{1,2,3,5,7,9,10,11,\dots,25,26,28,\dots,534,536,537,538,\dots\}.$$ Here we begin with $1$, $2$ and alternate between runs of consecutive numbers and arithmetic progressions of distance $2$, each of which has length equal to the length of the preceding progression squared. This delivers a set with natural and uniform densities $1/2$ and $1$. The same construction works in general for $\gamma=1$ and $\alpha=\beta$ (if $\alpha=\beta=0$, replace the corresponding progressions by "holes").

$\endgroup$
  • 2
    $\begingroup$ Changing a set $A$ to $B=\{\lfloor ca \rfloor; cA\}$ for $c\in(0,1)$ changes all four densities with factor $c$. So it is sufficient to prove the case $\delta=1$. (Which is not much of a simplification, but perhaps it helps at least a bit.) $\endgroup$ – Martin Sleziak May 17 '15 at 8:50
  • $\begingroup$ @Gabriel. I've at least two issues with your answer. First, as you too have remarked, it doesn't answer my question. Second, I'm not so convinced that, after filling in the technical details that you're alluding to, the construction you are suggesting will still be so fairly easy: I might mention a number of situations where constructions involving densities are relatively easy (or even trivial to some degree) as long as the "relevant parameters" are rational, but get significantly more complicated otherwise. In any case, thanks for your contribution to the discussion. $\endgroup$ – Salvo Tringali May 17 '15 at 9:41
  • $\begingroup$ I've edited my answer to include @MartinSleziak's comment and fix a mistake (the first case requires $\beta > \alpha$, not $\beta > 0$, in order to work; this required adjusting the following cases). I know that this answer is probably not very convincing; unfortunately I haven't been able to write down the general construction without getting bogged down in too many technical details. Anyway, I hope it does at least somewhat illuminate the problem; if it's only confusing, I will gladly delete it. $\endgroup$ – Gabriel Dill May 17 '15 at 10:10
2
$\begingroup$

So here is an explicit construction that works for any $\alpha\le\beta\le\gamma\le\delta$.

I first claim that if you can produce a set $Y$ such that bd$_*(Y)=d_*(Y)=\beta$, bd$^*(Y)=d^*(Y)=\gamma$, then it's easy to modify it to produce a set $X$ with the required properties.

First, notice that by unique ergodicity of the circle rotation $x\mapsto x+\pi\pmod 1$, for any $\rho\in [0,1]$, $S_\rho=\{n\colon \langle n\pi\rangle\le \rho\}$ has upper and lower Banach density equal to $\rho$ (where $\langle x\rangle$ is the fractional part of $x$). Now given $Y$, modify it on $[2^{2n},2^{2n}+2n]$ to match $S_\alpha$ and on $[2^{2n+1},2^{2n+1}+2n+1]$ to match $S_\delta$. That is:

$$ X=\left(Y\setminus\bigcup_n[2^n,2^n+n]\right)\cup\bigcup_{n\text{ even}} (S_\alpha\cap[2^n,2^n+n])\cup \bigcup_{n\text{ odd}}(S_\delta\cap[2^n,2^n+n]). $$ Clearly $d_*(X)=d_*(Y)$ and $d^*(X)=d^*(Y)$ since the set has been modified on a vanishingly small proportion of the integers. Let $A=\mathbb N\setminus\bigcup_n[2^n,2^n+n]$.

By the unique ergodicity, it is clear that bd$_*(X)\le\alpha$ and bd$^*(X)\ge\delta$. On the other hand, given a long interval $J$, if $J$ intersects $A$ on a large part sub-interval, $K$, the density $X$ on $K$ is at least $\beta-\epsilon$, while on the other substantial parts the density is at least $\alpha-\epsilon$. This shows bd$_*(X)\ge\alpha$, so that bd$_*(X)=\alpha$ and similarly bd$^*(X)=\delta$.

Hence it suffices to construct a $Y$ with the correct upper and lower densities. Such a $Y$ is given by $$ Y=\bigcup_{n\text{ even}}\left(S_\beta\cap [2^{2^n},2^{2^{n+1}})\right) \cup \bigcup_{n\text{ odd}}\left(S_\gamma\cap [2^{2^n},2^{2^{n+1}})\right). $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.