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Let $B$ be a set of positive real numbers with the property that adding together any finite subset of elements of $B$ always gives a sum of 2 or less. Show $B$ must be finite or countable.

I do not know where to start with this proof. Any help is appreciated.

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  • $\begingroup$ Can we perhaps get a better title? $\endgroup$
    – Asaf Karagila
    Dec 4 '17 at 7:52
  • $\begingroup$ Duplicate. $\endgroup$
    – Akash Gaur
    Dec 5 '17 at 9:48
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For each $n\in\mathbb N$, let$$B_n=\left\{b\in B\,\middle|\,b\geqslant\frac2n\right\}\subset B.$$Of course, $B_n$ can have no more than $n-1$ distinct elements; otherwise, the sum of $n$ distinct elements of $B_n$ would be grater than $2$.

But$$B=\bigcup_{n\in\mathbb N}B_n.$$Since $\mathbb N$ is countable and each $B_n$ is finite, $B$ is countable.

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  • $\begingroup$ Cute, I like it. Did you think of this just for this question or have you answered similar questions (so you had this arrow ready to shoot)? $\endgroup$ Sep 27 '17 at 2:00
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    $\begingroup$ @martycohen Nice metaphor! Yes, I had it ready. :-) $\endgroup$ Sep 27 '17 at 6:15
  • $\begingroup$ You have showed one example of $B$ that is countable and satisfies the problem statement. But you have to show that no such uncountable $B$ exists. $\endgroup$ Dec 9 '19 at 2:50
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    $\begingroup$ @ShreyasPimpalgaonkar No. I provided no example. What I did was to prove that, for whatever set $B$ for which the given property holds, $B$ is finite or countable. $\endgroup$ Dec 9 '19 at 6:57
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To proceed by contradiction, assume that $B$ is not countable. Choose some $b \in B$, since B is not countable, there must exist a sum of elements , call it c such that $c\gt nb$ for some $n\in \mathbb{N}$. (Simply consider the next n elements that are greater than b, their sum is greater than nb.) Choosing $n \geq \frac{2}{b}$. This implies $\exists~c\in B$ with $c\gt2$. This is absurd since summing this with anything else in B is clearly not less than 2, as the supposition states it should be. $\square$

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