# Special generating sets of the additive group of rational numbers

We are looking for two classes of subsets $$A$$ of $$(\mathbb{Q},+)$$ such that:

Class 1):

(i) $$1\in A$$

(ii) $$A-A\neq \mathbb{Q}$$ but $$(A-A)+\cdots +(A-A)=\mathbb{Q}$$ (i.e., some finite sumation of $$A-A$$ is equal to $$\mathbb{Q}$$)

(iii) $$A$$ is neither co-finite nor bounded (i.e., $$\mathbb{Q}\setminus A$$ is an infinite set and $$A$$ is unbounded)

Class 2):

(i) $$1\in A$$

(ii) $$(A-A)+\cdots +(A-A)\neq \mathbb{Q}$$ (i.e., every finite sumation of $$A-A$$ is not equal to $$\mathbb{Q}$$)

(iii) $$\langle A\rangle=\mathbb{Q}$$

(iv) $$A$$ is neither co-finite nor bounded

Note that $$A-A:=\{ a-\alpha:a,\alpha\in A\}$$ and $$B_1+\cdots +B_k:=\{ b_1+\cdots + b_k:b_1,\cdots,b_k\in B\}$$. It is interesting to know that for every subset $$A$$, either a finite summation $$(A-A)+\cdots +(A-A)$$ or the infinite summation $$(A-A)+(A-A)\cdots$$ is a subgroup of $$\mathbb{Q}$$. Also, all $$A$$ from both classes are generating sets of $$(\mathbb{Q},+)$$.

Can one give some explicit forms (or examples) of subsets $$A$$ of Class 1 and Class 2?

Thank you so much for your attention and participation.

For any rational number $$a$$, write $$a = \pm\sum_{i = -\infty}^{+\infty} a_i 2^i$$ its expansion in base $$2$$, with $$a_i = 0$$ for $$i \gg 0$$ (this adapts to any base).

Class $$1$$: Take $$A$$ to be the set of rationals $$a$$ such that $$a_1 = a_2 = a_3 = 0$$. We do have $$1 \in A$$, $$A + A + A + A = \Bbb Q$$ (the details are a bit tedious to write, but the core idea is that $$A+A$$ contains $$2 \times A$$), but $$A - A \neq \Bbb Q$$ (a difference of elements of $$A$$ cannot make $$a_2$$ nonzero). $$A$$ is neither cofinite nor bounded.

Class 2: Take $$A$$ to be the set of rationals $$a$$ such that $$a_i = 0$$ for all $$i >0$$ which are not a power of $$2$$. The subset $$A$$ contains $$[-1,1] \cap \Bbb Q$$, which ensures $$\langle A \rangle = \Bbb Q$$, but no finite summation of $$A$$ yields $$\Bbb A$$ as a whole (because of the massive gaps between the nonzero digits). Again, $$A$$ is neither cofinite nor bounded.

This works for $$\Bbb R$$ as well.

Class 1: $$A = ([0,1] \cap \mathbb Q) \cup 10\mathbb Z$$. This is clearly neither cofinite nor bounded, and $$A-A$$ does not equal $$\mathbb Q$$, but the $$5$$-fold sum of $$(A-A)$$ does.

Class 2: $$A = ([0,1] \cap \mathbb Q) \cup \{2^n : n \in \mathbb N\}$$. Similar reasoning as above. Of course the $$\{2^n\}$$ is just to ensure that the set is unbounded.

Another Class 2 example: $$A = \{$$reciprocals of prime powers$$\} \cup \mathbb Z$$. The summation of $$n$$ copies of $$(A-A)$$ can only contain fractions with at most $$2n$$ distinct primes in the denominator, so no such finite summation equals $$\mathbb Q$$. And yet $$A$$ does generate $$\mathbb Q$$, by a version of partial fractions decomposition.