If $A\subset \mathbb{N}$ we say that the set $A$ has a natural density if the limit $$\lim_{n \to \infty}\frac{|A\cap [1,n]|}{n}$$ exists and we denote it by $d(A)$. Also for every $k\in \mathbb{N}$ and $\ell\in {0,1,\dots,k-1}$ we define the set $A_{k\mathbb{N}+\ell}$ to be $$A_{k\mathbb{N}+\ell}=\{\alpha\in A: \textrm{there} \ \textrm{exist} \ m\in\mathbb{N} \ \textrm{with} \ \alpha=km+\ell\}.$$ My question is the following:
Does there exist a subset $A\subset\mathbb{N}$ with natural density $0<d(A)<1$ and the property that for every $k\in \mathbb{N}$ and $\ell\in {0,1,\dots,k-1}$ one has $$d(A_{k\mathbb{N}+\ell})=\frac{d(A)}{k}?$$
Edit: Let me give you some motivation about the aforementioned question.
This question arose when I tried to prove another property of the natural density. More specifically, given a set $A\subset \mathbb{N}$ with natural density $d(A)>0$ and a number $0<b<d(A)$ I wanted to prove that there exists a $B\subset A$ with natural density $d(B)=b$.
Although this seems very natural, the proof isn't easy at all. My proof was based on a constructive argument. I approached the requested density by eliminating terms that belonged in appropriate arithmetic progressions. In this way, as I noticed, I was changing the "proportion" of terms in these specific arithmetic progressions. So, I thought that maybe there is a way to do that without influence the "proportion". Starting from $\mathbb{N}$ I tried to construct sets with positive densities with the property that their terms have in some sense "uniform distribution" on arithmetic progressions. So finally I formulated this problem in my mind.
When $d(A)=0$ or $d(A)=1$ the answer is yes and it is trivial. The problem I encountered was that when you try to "distribute" terms, for examble, on $2\mathbb{N}$ and $2\mathbb{N}+1$ you have already affect the "proportion" of terms on $4\mathbb{N}+\ell$. So, I thought that the key was to do it for the progressions of the form $p\mathbb{N}+\ell$ where p is prime. But I am stack here. My intuition says that there must be a way. Maybe my previous attempt on the initial problem have me "binded" and I can't think in a different more sophisticated way.
In case if anyone is interested, I would really like to have some thoughts on this problem.
Thanks in regards.