Asymptotic Density vs a Partition of Natural Numbers I have the following sets:
$$A_{m,n} = \{ x \in \mathbb{N} | x \equiv r_{m,n} \pmod{p_n}\}$$
such that for all $n$, $p_n < p_{n+1}$. 
Also, $A_{m_1,n_1} \cap A_{m_2,n_2} \neq \emptyset \Longrightarrow m_1=m_2\text{ and }n_1=n_2$.
Suppose $q_n$ is a strictly increasing sequence such that $p_{n-1} < q_n < p_n$ and $q_n$ is a prime power. 
Let 
$$U_n = \left\{x \in \bigcup_{k=1}^n A_{m,k} \mid x < q_n, \text{ for any }m \right\}$$
I can show that
$$\lim_{n\to \infty}\dfrac{|U_n|}{q_n} = 1$$
So, I know that the union of these sets are asymptotically dense in the set of Natural Numbers. Do they form a partition of Natural Numbers? Years ago when I was in grad school, my professors always told me to be careful with limit sets, as they rarely function the way finite sets would function.
This problem originated when I was trying to calculate probabilities in Axis and Allies. As I was playing around with those numbers, it led to me playing around with numbers in general, and a situation like this eventually emerged. 
The sequences $r_n,p_n,q_n$ are not unique, which is why I am not defining them here. If it matters, or if it would help, I can provide an example this weekend. I would just need to work it out what each sequence would be. The $r_n$ sequence is a bit more difficult to come up with than the $p_n$ and $q_n$ sequences, but I was hoping there was a general way to show that yes, this turns into a partition, or no, the best I can say is that the union is asymptotically dense.
Edit: I think I may need to give some more details on these sequences of sets. I want to put this problem on hold until I have a chance later this week to give more details of what I am looking for (maybe even include a couple of examples). I was not careful when I was playing around with these sets, and I may have made some simplifications that were not justified when stating the problem. I apologize for my rush to post without checking my work.
 A: I'm not sure if I understood the problem statement the intended way, but consider the sequences defined as $p_n=2^n$ and $r_{1,n}=2^{n-1}+1$. Our sets $A_n$ (with $1\leq n$) can thus be characterized as
$$A_n=\{2^{n-1}(2k+1)+1\}$$
where $k$ varies over all non-negative integers. Note: Depending on the preferred definition of $\mathbb{N}$, $A_1$ would also include number $0$; it doesn't matter much for the rest of the explanation.
First of all, out sets are pairwise disjoint: $$(n_1\not=n_2)\Rightarrow A_{n_1}\cap A_{n_2}=\emptyset$$
This is easy to see: start with $$2^{n_1-1}(2k_1+1)+1 = 2^{n_2-1}(2k_2+1)+1$$ subtract $1$ from each side and compare the powers of $2$ to obtain $n_1=n_2$.
For any integer $a\geq 2$, we can write $(a-1)$ in the form $(a-1)=2^{n-1}(2k+1)$ for some non-negative integer $k$ and positive integer $n$ and thus prove $a\in A_n$.
On the other hand, number $1$ is not in $A_n$ for any $n$, since $2^{n-1}(2k+1)+1=1$ implies $2^{n-1}(2k+1)=0$, which is impossible.
Thus, $$\bigcup_{n\geq 1} A_n=\mathbb{N}\ \backslash\ \{1\}$$
Formally, we'd also need to find the sequence $\{q_n\}$ of prime powers fitting between consecutive members of $\{p_n\}$; but that turns out to be quite easy: Just let $q_n$ be the least prime greater than $p_{n-1}$. By Bertrand's postulate, such a prime doesn't exceed $p_n=2p_{n-1}$ and since $p_n$ itself is not a prime, $q_n$ defined in this way is certainly strictly between $p_{n-1}$ and $p_n$.
A: I'm very sceptical of your claim regarding $q_n$, but that to one side, consider $p_n = 3^n$. Then let $r_n$ be the least number it can be, excepting that we never choose a multiple of 3. So, first 1, then 2, 5, 8 etc. At each step, the number of possible $r_n$ such that the $A_n$ are mutually disjoint multiplied by 3, and one is subtracted ie 1 is chosen from 1 and 2; 2 from 2, 5 and 8; 5 from 5, 8, 14, 17, 23 and 26 etc.
I therefore can always choose such an $r_n$ and the union of the $A_n$ will never include 3, and so cannot cover the naturals.
