Is the following true?


For any $c\in (0,1)$ there exists $f(c)>0$ such that for any subset $A\subset \{1,2,\dots,n\}$ of cardinality $|A|\geq cn$, the set 
 $$B=\left\{ k \in \{1,2,\dots,n!\} \colon \text{ there is } a \in A \text{ that divides } k\right\}$$ 
of numbers having at least one divisor in $A$ satisfies $$|B|\geq f(c) n!.$$


Of course, $n!$ may be replaced to the any common multiple of elements of $A$, or we may ask about densities or probabilities.


I do not know the answer even for $A=\{ cn\ $consecutive integers$\}$



numbers divisible by at least one  of many numbers
The asymptotic version is certainly false. Let $\varepsilon (x,y)$ be the density of numbers having a divisor in the interval $[x,y]$, then Besicovitch proved in "On the density of certain sequences of integers" that $$\liminf\_{x\to \infty} \varepsilon (x,2x)=0.$$
Later, Erdos improved this to $$\varepsilon(x,2x)\sim(\log x)^{-\delta +o(1)} , \delta \approx 0.086.$$ Denoting $H(x,y,z)$ to be the number of elements in $\{1,2,\dots,x\}$ which have a divisor in the interval $[y,z]$, some sharper estimates are given in [this](http://arxiv.org/abs/math.NT/0401223) paper of Kevin Ford.


