Special generating sets of the additive group of rational numbers

Question

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 $(\mathbb{Q},+)$.

One can find more information in this and this.

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.

Answer 1

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.

Answer 2

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.