Show that $S=\mathbb{Q} \cap [0,1]$. Can you give a hint to show the following exercise?
Let $S \subset [0,1]$ such that $0, 1\in S$ and for all $n\in \mathbb{N}$, $s_1,...,s_n \in S$ distinct, then $\dfrac{s_1+...+s_n}{n}\in S$.
Show that $S=\mathbb{Q} \cap [0,1]$.
Thank you

 A: I assume as stated in the comment it is a subset relation.
One can easily see we have all $a/2^b\in [0,1]$. Let $A=\{a/2^b\in [0,1]\}$.
Let $r=p/q$ be a rational number. We want to show that we can express $r$ as sum of exactly $q$ distinct elements in $A$. It suffices to show $p=a_1+\dots +a_q$.
I will illustrate that we can always do this by an example. Let $r=7/9$. First set $a_1=a_2=0$, and the rest of them to be $1$. Then their sum is exactly $p$.
Now, take some large $N_1$, and change to become $a_1=a_2=0+5/2^{N_1}$ and $a_i=1-2/2^{N_1}$ for all other $i$.
Now I want to change values of $a_i$'s to make them all distinct without really changing their sum.
Since $|\{a_1,a_2\}|$ is even, pick some large enough $N_2$ to change $a_1$ to $a_1 - 1/2^{N_2}$ and $a_2$ to $a_2+1/2^{N_2}$. Note that $N_2$ is chosen large enough to avoid touching other points or the boundary.
Since $|\{a_3,a_4,\dots, a_9\}|$, so we keep $a_3$ the same, and then pair other points up. We do similar things to make them all different, but keeping the sum the same.
