Some co-finite subsets of rational numbers Is there any $A\subseteq \mathbb{Q}$ such that $A-A$ is a non-trivial co-finite
subset?
Note that $A-A=\{a_1-a_2: a_1,a_2\in A\}$, also see 
A type of integer numbers set.
 A: Let $\Bbb{Q} \cap (1, +\infty) = \{ r_k \}_{k \ge 1}$, then
$$A= \left\{ 1+r_1+\sum_{k=1}^n r_k : n \ge 0\right\} \cup \{ x \in \Bbb{Q} : 0 \le x < 1\}$$
works. With this choice of $A$ you have
$$A-A = \Bbb{Q} \setminus \{ -1, 1 \}$$
WHY DOES $A$ WORK?
For simplicity, let's denote $$B=\left\{ 1+r_1+\sum_{k=1}^n r_k : n \ge 0\right\} = \{ 1+r_1 , 1+2r_1 , 1+2r_1+r_2, 1+2r_1+r_2+r_3 , \dots \}$$, and $C= \Bbb{Q} \cap [0,1)$. Then $A$ is nothing but $B \cup C$. Note that $B \subseteq (1, + \infty )$, so that $B,C$ are disjoint.
Let's show that $A-A \subseteq \Bbb{Q} \setminus \{ -1, 1\}$. It is enough to show that $1 \notin A-A$ (in fact, if $-1 \in A-A$, then $-1=a_1-a_2$, and so $1 = a_2-a_1 \in A-A$). Let $a_1-a_2$ be an element of $A-A$ , with $a_1, a_2 \in A$ and $a_2<a_1$: we need to show that $a_1-a_2 \neq 1$. Divide three cases:


*

*if both $a_1, a_2 \in C$, then $a_1-a_2 < 1$.

*if both $a_1, a_2 \in B$, then $$a_1-a_2 = (1+r_1+r_1+ \dots + r_m + \dots +r_n)- (1+ r_1 +r_1 + \dots + r_m) =\\= r_{m+1} + \dots + r_n > 1$$

*if $a_1 \in B$ and $a_2 \in C$, then
$$a_1-a_2 = (1+ r_1+r_1 + \dots + r_n)-a_2 > (1+r_1+r_1 \dots + r_n) - 1 =\\= r_1 +r_1+ \dots + r_n > 1$$
In any case, $a_1-a_2 \neq 1$, so we proved the first inclusion.
For the other inclusion $\Bbb{Q} \setminus \{ -1, 1\} \subseteq A-A$, we divide four cases:


*

*for all $x \in \Bbb{Q} \cap (- \infty , -1)$ we have that $-x = r_k$ for some $k$. Then *
$$x = -r_k = (1+r_1 +r_1 + \dots + r_{k-1})-(1+r_1 +r_1 + \dots + r_{k}) \in A-A$$

*for all $x \in \Bbb{Q} \cap (-1,0)$ we have $-x \in C$, hence
$$x= 0-(-x) \in A-A$$

*for all $x \in \Bbb{Q} \cap [0,1)$ we have $x \in C$, hence
$$x= x-0 \in A-A$$

*for all $x \in \Bbb{Q} \cap (1, + \infty )$ we have that $x = r_k$ for some $k$. Then *
$$x = r_k = (1+r_1 +r_1 + \dots + r_k)-(1+r_1 +r_1 + \dots + r_{k-1}) \in A-A$$
*Note that here we have $r_1+ \dots + r_{k-1}$, which has to be intended as the empty sum ($=0$) for $k=1$.
In any case $x \in A-A$, and so we proved the other inclusion.
