On the independence of lower and upper asymptotic and Banach densities

Question

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.

Answer 1

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").

Answer 2

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 $$ Y=\bigcup_{n\text{ even}}\left(S_\beta\cap [2^{2^n},2^{2^{n+1}})\right) \cup \bigcup_{n\text{ odd}}\left(S_\gamma\cap [2^{2^n},2^{2^{n+1}})\right). $$