Thinking about the question [Four polynomials representing all integers modulo m](https://mathoverflow.net/questions/45697/four-polynomials-representing-all-integers-modulo-m/45709#45709) lead me to the following complementary question:


If $S$ is a set of positive integers, say that a positive integer $m$ is *covered* by $S$ if every congruence class $\bmod m$ has a representative in $S$. Denote by $C(S)$ the set of positive integers covered by $S$. If $x>0$ let $S(x) = \{ k \in S : k \le x \}$ and the *lower density* of $S$, $\ell(S) := \lim \inf\_{x \rightarrow \infty} |S(x)|/x$. My question: is there a non-trivial lower bound on $\ell(C(S))$ in terms of $\ell(S)$? That is, is there a continuous function $f : [0,1] \rightarrow [0,1]$, not identically 0, such that $\ell(C(S)) \ge f(\ell(S))$.


The set in the question I referred to has density 0, so my question wouldn't apply to it. However, I wondered if there were a simple argument in the case of positive lower density. This has the smell of the kind of question that Erdos would ask, so I wouldn't be surprised to see it there.



Density of congruence classes covered by a set
Denote by $P$ the set of prime powers not covered by $S$ (for each prime $p$, take only the smallest non-covered its power). If $\sum\_{x\in P} 1/x=+\infty$, then $\prod\_{x\in P} (1-1/x)$ is 0, so the product over some finite subset is arbitrarily small. But this product is a density of numbers without forbidden remainders modulo respective prime powers. So, $S$ has density 0. A contradiction. Hence $a=\prod\_{x\in P} (1-1/x)$ is positive and $\ell(S)\leq a$. But then complement of $C(S)$ is the set of numbers divisible by at least one element of $P$. Density of such numbers equals $1-a$ (this is rather technical, but standard). So, we get that $\ell(C(S))\geq \ell(S)$. 


