Which sets of positive rationals are closed under addition?

Question

This question evolved because I was interested in generalizing power series so the exponents were rational numbers instead of integers, i.e., $\sum_{i=1}^{\infty} a_n x^{r_n}$, with the $a_i$ real and the $r_i$ non-negative rationals.

The theorem I wanted to hold was that if the product of two such series was zero, then one of the series had to be zero.

I quickly realized that for the series to be meaningful, the $r_i$ could not have any accumulation point.

This means that there had to be a positive real $c$ such that $r_{i+1} \ge r_i+c$ for all $i$.

Also, for the ordinary Cauchy product of these series to have the same exponents, the series had to be closed under addition, so that, for each pair of exponents $r_i$ and $r_j$, there is an exponent $r_k$ such that $r_i+r_j = r_k$.

An obvious set of such rationals is $(\frac{i}{n})_{i=0}^{\infty}$ for $n$ a positive integer.

My question is this:

Are there any other sets of non-negative rationals without any accumulation point which are closed under addition?

Answer 1

This might not be very satisfying considering what you're looking for, but you could remove part of the beginning part of the set and retain this property. These are called numerical semigroups, if $0$ is included. For example, $$\{0,3,5,6,8,9,10,\ldots\}=\mathbb{N}\setminus\{1,2,4,7\}$$ (and of course you can divide this by any denominator you like).

A perhaps more appealing solution is to take an additive subsemigroup of $\mathbb{N}$ that is not a numerical semigroup, i.e. that has a common divisor, and put it over a relatively prime denominator. For example, $$\left\{\frac{2n}{7}:n\in \mathbb{N}\right\}$$

Answer 2

Fleshing out my comments into an answer: the 'meaningfulness' (which I take to mean 'having a positive convergence radius') and additive-closed nature of the series are not enough in and of themselves to guarantee a lower bound on $r_{n+1}-r_n$; there is an example (in fact, many examples) of a sequence of exponents $r_n$ such that series of the form $a_nx^{r_n}$ with $a_n\in\Theta(1)$ have a positive radius of convergence and with $r_n$ additively closed but $\lim_{n\to\infty}(r_{n+1}-r_n)=0$.

For the simplest (to me) example, consider $r_n$ defined such that $r_{2^i+j}=i+\frac{j}{2^i}$ if $j\lt2^i$; i.e., $r_1=0$, $r_2=1$, $r_3=1\!\frac12$, $r_4=2$, $r_5=2\!\frac14$, etc. Then it's easy to see that $r_n$ is additively closed; $r_a+r_b\ (b\gt a)$ can't have a 'dyadic denominator' larger than that of $r_b$, and all such numbers $\geq r_b$ are in the sequence. What's more, the series $\sum_{n=0}^\infty x^{r_n}$, the prototypical example of a 'rational power series' (the analogue to the geometric series), converges for $0\leq x\lt\frac12$; for we can break it into a double sum $\displaystyle\sum_{i=0}^\infty\sum_{j=0}^{2^i-1}x^{(i+j2^{-i})}$. Each term in the inner sum is $\leq x^i$ if $x\lt1$, so the entire inner sum is bounded by $2^ix^i$ and the sum as a whole is bounded by $\sum_{i=0}^\infty 2^ix^i$, which clearly converges for $x\lt\frac12$.

Picking other denominators for the sequence of $r$s will give plenty of other examples of additively-closed sequences under the looser conditions I'm using here, and you can combine this with Vadim's suggestion of removing initial subsets to provide many more examples. On the other hand, all of these techniques only provide countably many examples of sequences $r_n$, and I suspect that the set of all such sequences (additively closed and 'exponentially bounded', which I believe is the right accumulation condition to imply - it's the statement that there's some $C$ s.t. for all $n$ there are only $C^n$ members of the sequence $r_i$ less than $n$) is in fact actually countable, but I can't see any easy proof for that.