Covering Systems of infinite sets of residue classes mod primes Take an infinite set of distinct primes and a (edit: or 2 , etc.) residue class for every prime. For exammple you can take all the primes bigger than some prime or the primes of a specific form (i.e. of some arithmetic progression or whatelse)
The question is under what circumstances every natural number is of some of this residue classes?
Of course there are conditions that the answer is easy(for example if you take as residue class for a prime on the set to be  $1\bmod{p_1}$, $2\bmod{p_2}$), but do we have any well known general results on this direction? (On infinite sets of residue classes etc.)
Note:if you think that the question is not very specific please ask me to be more.
EDIT: to be more precise i will give an example : Take all the primes from a point and after, split the natural numbers on intervals of a given length, say A, and for each prime is not allowed to "hit" a number that is in lower interval than itself, (i.e take all the primes greater or equal than 7 and $A=100$)
 A: If you insist on prime moduli you will need infinitely many classes. If you don't insist on prime moduli then you can get away with finitely many, these are called covering systems and the linked article looks good. 
Let $(r,m)=\lbrace r+mj \mid j \ge 0 \rbrace$ and $p_1<p_2<p_3<\cdots$ be a list of some or all the primes. You want a list $(r_1,p_1),(r_2,p_2),\cdots$ whose union is the natural numbers (optinally with some natural numbers allowed to be missed). One method to do this is to greedily define $r_i$ to be the smallest integer not in $\cup_{j<i}(r_j,p_j)$. Is this (almost) optimal? I don't know but I will wildly guess that it is and from now on only discuss the greedy choice. If we use all the primes and try to get all the natural numbers starting with 2 then (making the greedy choice) $r_j=p_j$. This becomes $r_j=p_j-1$ (or $r_j=p_j-2$) if we also want to get 1 ( or 0 and 1). In any case the greedy algorithm gives $\frac{r_i}{p_i} \approx 1$ (Well in the very first we could as well have $r_i=0$)
If we use all the primes except $2$ and try to get all the natural numbers except powers of $2$ (and 0) then again $r_i=p_i$. If we try to get all the natural numbers then it would appear that $\frac{r_i}{p_i} \approx \frac{1}{2}$.
Based on this I will say that there is little hope that you can exclude the primes 2,3,5 (or even just 2) and keep $p_i-r_i<100$. I suspect that for each prime $q$ there is a constant $C_q$ so that using the greedy choice to cover all the natural numbers using primes $q$ and greater results in $\frac{r_i}{p_i} \approx C_q$. It looks as if perhaps $C_7$ is something like 0.43.
