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For $A\subseteq\omega$ we define the upper density by $$d_u(A) = \lim\sup_{n\to\infty}\frac{|A\cap n|}{n+1}.$$ For $y\in \omega$ we set $A - y:= \{|a\setminus y|:a\in A\}.$ Note that $|a\setminus y|$ equals the difference of $a$ and $y$ if $a\geq y$, and $0$ otherwise. The upper Banach density is defined by $$d_{uB}(A) = \lim_{n\to\infty}\big(\sup \{\frac{|(A-y)\cap n|}{n+1}: y\in\omega\}\big).$$ It is easy to see that $d_{uB}(A) \geq d_{u}(A)$ for all $A\subseteq \omega$.

Question. What is an example of $A\subseteq \omega$ with $d_{uB}(A) > 0$ and $d_u(A) = 0$?

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I think that it is fairly straightforward to get such a set. You can simply get the set as a union of intervals:$\newcommand{\intrvl}[2]{\langle{#1},#2)}\newcommand{\intrvr}[2]{({#1},#2\rangle}\newcommand{\limti}[1]{\lim\limits_{{#1}\to\infty}}$ $$A=\bigcup\limits_{n\in\omega} \intrvr{a_n}{b_n}.$$ And you choose the intervals in a suitable way. You want them long enough to get $d_{uB}(A)=1$. And, at the same time, you want start each interval far enough to keep $d_u(A)=0$. So it suffices to choose them in such way that \begin{align*} \limti n (b_n-a_n)=\infty\\ \limti n \frac{\sum_{k=1}^n (b_k-a_k)}{b_n}=0 \end{align*} The first conditions will help you to get upper Banach density (a.k.a. upper uniform density) equal to one. And the second condition will keep the asymptotic density (natural density) zero.

For example, you can take $a_n=2^n$ and $b_n=2^n+n$.

Several posts on this site are a bit related - and the answers include some references which might be useful:

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