# Which kind of subsets of primes one needs to generate a positive ratio of the natural numbers?

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?

• Note that the set is related to the totatives of finite products of members outside of S. In particular, if the complement of S is a finite set T, the product of the members of T is then called n, then S is related to the totatives of n, and your epsilon can be chosen not much smaller than a constant times 1/log(log(n)). The next thing to look at would be S having only those primes congruent to b mod d for some integers b coprime to d. Gerhard "Ask Me About Erik Westzynthius" Paseman, 2015.03.17 – Gerhard Paseman Mar 17 '15 at 15:38
• It may develop that you are after zero-density subsets of the primes, where there is some already developed notion of density. You might search this forum for the word density with some number theory tag attached. Gerhard "The Answer May Be Close" Paseman, 2015.03.17 – Gerhard Paseman Mar 17 '15 at 15:42
• So... Who is Erik Westzynthius? – Vincent Mar 17 '15 at 15:43
• If $S$ consists of primes congruent to 0 or 1 mod 4, then the density is 0. This is because every element of span$(S)$ is a sum of two squares, and the set of positive integers that are a sum of two squares has density 0. In fact, we can throw into $S$ the squares of all primes congruent to 3 mod 4, and the density is still 0. – Richard Stanley Mar 17 '15 at 16:07
• I think you're right. The condition should be the sum of the reciprocals of $\mathbb P\setminus S$ should be finite. I strongly suspect this is necessary and sufficient. – Anthony Quas Mar 17 '15 at 17:07

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 $\alpha/2$.

• Thanks, two questions to that: So now we have a simple characterization criterion for a subseries of the harmonic sum to converge/diverge, especially a very easy way to see the divergence of the sum of reciprocels of $\mathbb{P}$. Why this criterion is not the usual one to teach and learn, has it a name? And does it give some possibility to reformulate or simplify the proof of the existence of brun's constant? – Slamice Mar 17 '15 at 20:07
• To add more information how this question came up: I noticed that Erdös used only property of $\mathbb{P}$ in his proof of the divergence of the sum of the reciprocals of $\mathbb{P}$, so the generalization was quite near. Is this criterion ever used in any context, or is it just to trivial to be useful? – Slamice Mar 17 '15 at 20:07
• My guess is that so crude that it's unlikely there is really fine information hidden there. I'm not at all an analytic number theorist so can't really comment further. – Anthony Quas Mar 17 '15 at 21:51
• Ok, not a really fine information, but it's much simpler than most of the proofs of the (of course already simple) classical statement $\sum_{p\in \mathbb{P}}1/p=\infty$, isnt it? And: It reveals the "real reason" for this statement, whre the pure manipluation analytics arguments with logarithms does not. Also it should be not too hard to understand, that the span of the prime twins set has zero densitiy ... I expect that this would be an easier approach too bruns theorem (without the estimating of course, but only the convergence). – Slamice Mar 18 '15 at 8:44
• @Slamice: I don't think you can prove along these lines that the sum of reciprocals of twin primes is finite. That is, I don't think there is a direct proof that the span of twin primes has zero density. Brun's proof estimates from above the number of twin primes up to $x$ (which is a finer information than the convergence of the sum of reciprocals), and I don't think it gets simpler than that. In fact it is not too natural to look at the reciprocal sum of twin primes or their span. What is important is the density of twin primes among all primes, and this is what Brun addressed. – GH from MO Mar 18 '15 at 17:00