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

Show that $S=\mathbb{Q} \cap [0,1]$.

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    Do $s_1\ldots s_n$ have to be distinct? If not, then the problem is nearly trivial... – 2017-01-20
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    Yes, they are distinct, thank you – 2017-01-20
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    If you choose $S=[0,1]$, it's true that $0,1\in S$ and that the average of $s_1,\cdots,s_n$ belongs to $S$ whenever $s_1,\cdots,s_n\in S$ because $[0,1]$ is convex ! So the conclusion $S=\mathbb{Q}\cap [0,1]$ is not always true. – 2017-01-20
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    @Adren I don't think that $S$ can be chosen. It is a given set. – 2017-01-20
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    @AnuragA The point is that the conclusion $S=\mathbb{Q}\cap [0, 1]$ cannot be derived from the facts stated, since different sets *also* have those same properties. The problem is presumably asking to show that $S\supseteq\mathbb{Q}\cap [0, 1]$. – 2017-01-20
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    @NoahSchweber you are right. I misread what Aden stated. Thanks. – 2017-01-20
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    Dear friends, thank you so much for your answers, but the problem says to show the equality. – 2017-01-20
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    @Joe Well, in that case, the problem is incorrect. Maybe there's an additional hypothesis, or you copied it incorrectly? – 2017-01-20
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    Let me see it again – 2017-01-20
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    That is the question, is in portuguese – 2017-01-20
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    Why don't you try to successively prove in steps that rationals of certain type are in $S$. For instance, $1, 1/2$ are there. The average with zero implies $1/4$ is there. Average of this with $0$ =1/8 is there... So, all powers 1/2^n are in S. Notice also, that if we fix the left at some 1/2^n and take averages similar to above will imply that all 1/2^n+1/2^m are in S for all m and n. You said hint, so here is one. – 2017-01-20
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    I was doing exactly that, but i did not get farther than that, i will try one more time. Thank you very much – 2017-01-20
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    @Joe I'm not sure if this approach will be fruitful, but maybe playing around with it will give you an idea or get you closer to a strategy that will work. First show that all rationals of the form $r/2^n\in S$, for any $r,n\in\Bbb Z_{\geq 0}$ with $0\leq r\leq 2^n$. Next, say $p_1 = 2$, $p_2 = 3$, and so on, so $p_m$ is the $m$th prime, and suppose that you know that $r/p_i^n\in S$ for all $r,n\in\Bbb Z_{\geq 0}$, $0\leq r\leq p_i^n$ and for all $i < m$. Use this to show [somehow!] that $r/p_m^n\in S$ for any $r,n\in\Bbb Z_{\geq 0}$, $0\leq r\leq p_m^n$. – 2017-01-20
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    You might also try a similar approach where instead of moving from smaller prime denominators to larger ones (and then to all denominators) a variation where you use strong induction on the denominator $n$. (I.e., you know you have all fractions with denominator less than $n$ in the interval $[0,1]$ and bootstrap your way to fractions with denominator $n$ in $[0,1]$.) I'm not sure this will work on its own either, maybe some combination of the two or another variant I haven't considered will do the job. – 2017-01-20

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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.