Not knowing elementary number theory well, I ask this one, which is not very clear to answer, rather I am looking for some results around this question or known theorems. The problem is the following:


The set of prime numbers $\mathbb{P}=2,3,5,7,11...$ generates $\mathbb{N}$ by multiplication. Now I am interested in subsets $S$ of $\mathbb{P}$, which generate a positive fraction of all numbers, that means:


There is $\epsilon>0$ which holds the following equation for all $N \in\mathbb{N}$ big enough: $$\frac{\|span(S)\_{\leq N}\|}{N}\geq \epsilon N$$ where $span(S)\_{\leq N}$ is the subset of $\mathbb{N}\_{\leq N}$, which consists of the numbers, whose prime factors are all elements of $S$.


Clearly $S$ has to be infinite to have this property, but what can one say about $S$ more specificely? Is there any known criterions or examples for $S$ being too small to generate a positive ratio of the natural numbers?



Which kind of subsets of primes one needs to generate a positive ratio of the natural numbers?
Let $T=\mathbb P\setminus S$. If $\sum\_{p\in T}1/p=\infty$, then $\prod\_{p\in T}(1-1/p)=0$ and for any $\epsilon>0$, there exist $p\_1,\ldots,p\_n\in T$ such that $\prod\_{i=1}^n(1-1/p\_i)<\epsilon$. Now modulo $P=p\_1\cdots p\_n$, the fraction of integers that have no factor of the form $p\_i$ with $i\le n$ is $\prod(1-1/p\_i)<\epsilon$. Hence in the entire set of natural numbers, those with no factor in $T$ has density less than $\epsilon$. 


Conversely, if $\sum\_{p\in T} 1/p<\infty$, then let $\alpha=\prod\_{p\in T}(1-1/p)$. Then $0<\alpha<1$. Let $M$ be chosen such that $\sum\_{p\in T; p>M} 1/p<\alpha/2$. Then up to $N$, the number of integers that has a factor in $T$ that exceeds $M$ is at most $\sum\_{p\in T; p>M} N/p<\alpha N/2$. For large $N$, the number of $n\le N$ having no factor in $T$ below $M$ is close to $\alpha N$, and hence the density of $n$'s having no factor in $T$ is at least $\alpha/2$. 


