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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?

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btw if you take A to be the set {1} union {p+1 : p odd prime}, then sum of 1/a diverges, A contains no primes, and the kth member is greater than the kth prime for all k>1. –  Rob Harron Nov 8 '09 at 4:56
    
There are still fairly trivial answers to your question -- for instance, take the set 2, 4, 5, 8, 11, 14, ..., p_n, p_{n+1} + 1, and this satisfies your conditions. I'm not sure what you're trying to ask, here. –  Harrison Brown Nov 8 '09 at 5:10

4 Answers 4

up vote 45 down vote accepted

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.

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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.

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Thanks for taking the time to reply, Professor Tao. –  J. H. S. Nov 9 '09 at 9:27

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?

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Probably very little relation, alas; the result (a theorem of Müntz) holds when $A$ is any set of positive real numbers. –  Noam D. Elkies Aug 14 '11 at 14:26
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This is very cool, though! –  Vectornaut May 21 '12 at 6:21

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.

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