On the series 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + ... It is well-known that A: The series of the reciprocals of the primes diverges
My question is whether property A is in some sense a truth strongly tied to the nature of the prime numbers.
Property A tells us that the primes are a rather fat subset of $\mathbb{N}$. Is there a way to define a topology on $\mathbb{N}$ such that every dense subset of $\mathbb{N}$ (under this topology) corresponds to a fat subset of the natural numbers?
What do you think about this?
A: Yes, it's possible.  Define the closed sets to be the sets the sum of whose reciprocals converges, together with $\mathbb{N}$.  This collection of subsets is closed under arbitrary intersection and finite union, so it does form the closed sets of a topology.  
A subset of $\mathbb{N}$ is dense in this topology if its closure is $\mathbb{N}$, in other words, if it is not contained in any smaller closed set -- in other words, if it is not contained in any set the sum of whose reciprocals converges.  This is equivalent to the sum of its reciprocals not converging.   
A: By Fubini's theorem, the sum of the reciprocals of the primes is equal to $\int_1^\infty \frac{\pi(x)}{x^2}\ dx$, where $\pi(x)$ is the number of primes less than x.  The prime number theorem tells us that $\pi(x) \sim x/\log x$ for large x, which implies the divergence of this integral.  (One does not need the full strength of the PNT here; the more elementary fact that $\pi(x)$ is bounded from below by a constant multiple of $x / \log x$ would suffice.)  A variant of this argument shows that $\sum_{p \leq x} 1/p = \log \log x + O(1)$ (again, this can also be shown by more elementary means - see Mertens' theorem).
The same argument shows that slightly thinner sets than the primes would also have this property, e.g. any set for which the analogue of $\pi(x)$ is asymptotically larger than $x / \log x \log \log x$, or $x / \log x \log \log x \log \log \log x$, would still diverge.  On the other hand, if the analogue of $\pi(x)$ is $O( x / \log^{1+\varepsilon} x )$ for some $\varepsilon > 0$ then one will have convergence.  So the primes are close to the edge of the sparsest set with this property (as one could already guess from the double-logarithmic growth of the sum).
For instance, sieve theory tells us that the number of twin primes less than x is $O( x /\log^2 x)$, which implies Brun's theorem that the sum of reciprocals of twin primes converges.
A: There is also this result...
$\displaystyle\sum_{a \in A} \frac{1}{a}$ diverges if and only if the span of $\{x^a | a \in A\}$ is dense in the continuous functions on an interval.  (I guess you have to include the constant $1$).  There is probably no relation to the primes?  Or is there?
A: Also, it is possible to prove that for any numbers a and b for
which gcd(a,b)=1, the sum of all the 1/p's for p prime that
satisfy p = a (mod b) diverges.
