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For a finite set $X$, we write $\sum X$ to be the sum of all the numbers in $X$. Suppose we have a set $S \subseteq \mathbb{R}$ such that $-100\le \sum X \le 100$ for all finite subsets $X \subseteq S$. How to show that $S$ is countable?

I do not really know where to begin. Hints would be nice. Thank you!

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For each $n\in\mathbb N$ consider the sets $$ S\cap(-\infty,-1/n) \quad\text{and}\quad S\cap(1/n,\infty) $$ There are countably many of these sets, and their union is $S\setminus\{0\}$. How large can each of them be?

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  • $\begingroup$ Where do you get that each set is finite? $\endgroup$ – Elliot G Apr 11 '16 at 16:44
  • $\begingroup$ @ElliotG: Imagine what happens $S\cap (1/n,\infty)$ has at least $100n$ elements ... $\endgroup$ – Henning Makholm Apr 11 '16 at 16:47
  • $\begingroup$ I'm not seeing how we can say anything about $S\cap (1/n,\infty)$. How do we know it isn't uncountable. Even if it is finite, the sum could still be zero. $\endgroup$ – Elliot G Apr 11 '16 at 16:50
  • $\begingroup$ @ElliotG: How can a sum of $100n$ elements of $(1/n,\infty)$ be zero? $\endgroup$ – Henning Makholm Apr 11 '16 at 16:51
  • $\begingroup$ Ok I see your point there. Still not seeing how this proves $S\cap (1/n,\infty)$ is countable thought $\endgroup$ – Elliot G Apr 11 '16 at 16:54
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This is strongly related to The sum of an uncountable number of positive numbers. Using this you know that all but countably many elements of S are 0. Since 0 appears at most once you are done.

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